What is a two force member

A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear. This will result in all two force members being in either tension or compression as shown in the diagram below.

The forces acting on two force members need to be equal, opposite, and collinear for the body to be in equilibrium.

Why the Forces Must Be Equal, Opposite, and Collinear:

Imagine a beam where forces are only exerted at each end of the beam (a two force member). The body has some non-zero force acting at one end of the beam, which we can draw as a force vector. If this body is in equilibrium, then we know two things: 1) the sum of the forces must be equal to zero, and 2) the sum of the moments must be equal to zero.

In order to have the sum of the forces equal to zero, the force vector on the other side of the beam must be equal in magnitude and opposite in direction. This is the only way to ensure that the sum of the forces is equal to zero with only two forces.

In order to have the sum of the moments equal to zero, the forces must be collinear. If the forces were not collinear, then the two equal and opposite forces would form a couple. This couple would exert a moment on the beam when there are no other moments to counteract the couple. Because the moment exerted by the two forces must be equal to zero, the perpendicular distance between the forces (d) must be equal to zero. The only way to achieve this is to have the forces be collinear.

In order to have the sum of the moments be equal to zero, the forces acting on two force members must always be collinear, acting along the line connecting the two points where forces are applied

Why are Two Force Members Important:

By identifying two force members, we greatly reduce the number of unknowns in our problem. In two force members we know that the forces must act along the line between the two connection points on the body. This means that the direction of the force vectors is known on either side of the body. Additionally, we know the forces are equal and opposite, so if we determine the magnitude and direction of the force acting on one side of the body, we automatically know the magnitude and direction of the force acting on the other side of the body.

The forces in two force members will always act along the line connecting the two points where forces are applied.

Two force members are also important in distinguishing between trusses, and frames and machines. When we analyze trusses using either the method of joints or the method of sections, we will assume everything is a two force member. If this assumption is incorrect, this will cause serous problems in the analysis. By making this assumption though, we can use some shortcuts that will make truss analysis easier and faster than the analysis of frames and machines.

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A truss is an engineering structure that is made entirely of two force members. In addition, statically determinate trusses (trusses that can be analyzed completely using the equilibrium equations), must be independently rigid. This means that if the truss was separated from its connection points, no one part would be able to move independently with respect to the rest of the truss.

Trusses are made entirely of two force members. This means that each member will either be in tension or compression as shown above.

Trusses can be broken down further into plane trusses and space trusses. A plane truss is a truss where all members lie in a single plane. This means that plane trusses can essentially be treated as two dimensional systems. Space trusses on the other hand have members that are not limited to a single plane. This means that space trusses need to be analyzed as a three dimensional system.

The members of these trusses all lie in a single plane. These roof trusses are an example of a plane truss. Image by Riisipuuro CC-BY-SA 3.0.

This bridge consists of two plane trusses connected by members called stringers. Adapted from image by ToddC4176 CC-BY-SA 3.0

This roof supporting truss does not lie in a single plane. This is an example of a space truss. Image by IM3847 CC-BY-SA 4.0.

The power line tower also does not lie in a single plane and is therefore a space truss. Image by Anders Lagerås CC-BY-SA 2.5.

Analyzing Trusses

When we talk about analyzing a truss, we are usually looking to identify not only the external forces acting on the truss structure, but also the forces acting on each member internally in the truss. Because each member of the truss is a two force member, we simply need to identify the magnitude of the force on each member, and determine if each member is in tension or compression.

To determine these unknowns, we have two methods available: the method of joints, and the method of sections. Both will give the same results, but each through a different process.

The method of joints focuses on the joints, or the connection points where the members come together. We assume we have a pin at each of these points that we model as a particle, we draw out the free body diagram for each pin, and then write out the equilibrium equations for each pin. This will result in a large number of equilibrium equations that we can use to solve for a large number of unknown forces.

The method of sections involves pretending to split the truss into two or more different sections and then analyzing each section as a separate rigid body in equilibrium. In this method we determine the appropriate sections, draw free body diagrams for each section, and then write out the equilibrium equations for each section.

The method of joints is usually the easiest and fastest method for solving for all the unknown forces in a truss. The method of sections on the other hand is better suited to targeting and solving for the forces in just a few members without having to solve for all the unknowns. In addition, these methods can be combined if needed to best suit the goals of the problem solver.

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The method of joints is a process used to solve for the unknown forces acting on members of a truss. The method centers on the joints or connection points between the members, and it is usually the fastest and easiest way to solve for all the unknown forces in a truss structure.

Using the Method of Joints:

The process used in the method of joints is outlined below:

In the beginning it is usually useful to label the members and the joints in your truss. This will help you keep everything organized and consistent in later analysis. In this book, the members will be labeled with letters and the joints will be labeled with numbers.

Treating the entire truss structure as a rigid body, draw a free body diagram, write out the equilibrium equations, and solve for the external reacting forces acting on the truss structure. This analysis should not differ from the analysis of a single rigid body.

Assume there is a pin or some other small amount of material at each of the connection points between the members. Next you will draw a free body diagram for each connection point. Remember to include:
- Any external reaction or load forces that may be acting at that joint.
- A normal force for each two force member connected to that joint. Remember that for a two force member, the force will be acting along the line between the two connection points on the member. We will also need to guess if it will be a tensile or a compressive force. An incorrect guess now though will simply lead to a negative solution later on. A common strategy then is to assume all forces are tensile, then later in the solution any positive forces will be tensile forces and any negative forces will be compressive forces.
- Label each force in the diagram. Include any known magnitudes and directions and provide variable names for each unknown.

Write out the equilibrium equations for each of the joints. You should treat the joints as particles, so there will be force equations but no moment equations. With either two (for 2D problems) or three (for 3D problems) equations for each joint, this should give you a large number of equations.

In planar trusses, the sum of the forces in the x direction will be zero and the sum of the forces in the y direction will be zero for each of the joints.

\[\sum \vec{F}=0\]
\[\sum F_x=0\]	\[\sum F_y=0\]

In space trusses, the sum of the forces in the x direction will be zero, the sum of the forces in the y direction will be zero, and the sum of forces in the z direction will be zero for each of the joints.

\[\sum \vec{F}=0\]
\[\sum F_x=0\]	\[\sum F_y=0\]	\[\sum F_z=0\]

Finally, solve the equilibrium equations for the unknowns. You can do this algebraically, solving for one variable at a time, or you can use matrix equations to solve for everything at once. If you assumed that all forces were tensile earlier, remember that negative answers indicate compressive forces in the members.

Use the method of joints to find the forces in all members of the truss shown below. Remember to specify tension or compression.

Use the method of joints to find the forces in all members of the truss shown to the right. Remember to specify tension or compression.

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The method of sections is a process used to solve for the unknown forces acting on members of a truss. The method involves breaking the truss down into individual sections and analyzing each section as a separate rigid body. The method of sections is usually the fastest and easiest way to determine the unknown forces acting in a specific member of the truss.

Using the Method of Sections:

The process used in the method of sections is outlined below:

In the beginning it is usually useful to label the members in your truss. This will help you keep everything organized and consistent in later analysis. In this book, the members will be labeled with letters.

Treating the entire truss structure as a rigid body, draw a free body diagram, write out the equilibrium equations, and solve for the external reacting forces acting on the truss structure. This analysis should not differ from the analysis of a single rigid body.

Next you will imagine cutting your truss into two separate sections. The cut should travel through the member that you are trying to solve for the forces in, and should cut through as few members as possible (The cut does not need to be a straight line).

Next you will draw a free body diagram for either one, or both sections that you created. Be sure to include all the forces acting on each section.
- Any external reaction or load forces that may be acting at the section.
- An internal force in each member that was cut when splitting the truss into sections. Remember that for a two force member, the force will be acting along the line between the two connection points on the member. We will also need to guess if it will be a tensile or a compressive force. An incorrect guess now though will simply lead to a negative solution later on. A common strategy then is to assume all forces are tensile, then later in the solution any positive forces will be tensile forces and any negative forces will be compressive forces.
- Label each force in the diagram. Include any known magnitudes and directions and provide variable names for each unknown.

Write out the equilibrium equations for each section you drew a free body diagram of. These will be extended bodies, so you will need to write out the force and the moment equations.

For 2D problems you will have three possible equations for each section, two force equations and one moment equation.

\[\sum \vec{F}=0\]	\[\sum \vec{M}=0\]
\[\sum F_x=0\]	\[\sum F_y=0\]	\[\sum M_z=0\]

For 3D problems you will have six possible equations for each section, three force equations and three moment equations.

\[\sum \vec{F}=0\]
\[\sum F_x=0\]	\[\sum F_y=0\]	\[\sum F_z=0\]

\[\sum \vec{M}=0\]
\[\sum M_x=0\]	\[\sum M_y=0\]	\[\sum M_z=0\]

Finally, solve the equilibrium equations for the unknowns. You can do this algebraically, solving for one variable at a time, or you can use matrix equations to solve for everything at once. If you assumed that all forces were tensile earlier, remember that negative answers indicate compressive forces in the members.

Use the method of sections to find the forces acting on members BD and CE. Be sure to indicate if the forces are tensile or compressive.

Use the method fo sections to find the forces acting on members AC, BC, and BD of the truss. Be sure to indicate if the forces are tensile or compressive.

Use the method of sections to find the forces in members AB and DE. Be sure to indicate if the forces are tensile or compressive.

Use the method of sections to find the forces in members AC, BC, CD, and CE. Be sure to indicate if the forces are tensile or compressive.

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A frame or a machine is an engineering structure that that contains at least one member that is not a two force member.

This beam is connected to other members (where normal forces would exist) at more than two locations. This beam is therefore not a two force member.

This beam has two connection points, but a force is acting on a third point. Therefore the beam has forces acting on it at more than two locations and it is not a two force member.

A frame is a rigid structure, while a machine is not rigid. This means that no part can move relative to the other parts in a frame, while parts can move relative to one another in a machine. Though there is a difference in vocabulary in describing frames and machines, they are grouped together here because we use the same process to analyze both of these structures.

This stool contains non-two force members (the legs) and no part can move relative to the other parts (it is rigid). Therefore this is a frame. Image by Besceh31 CC-BY-SA 2.5.

This pair of locking pliers contains non-two force members and has parts that can move relative to one another (it is not rigid). Therefore this is a machine. Image by Duk CC-BY-SA 3.0.

Analyzing Frames and Machines

When we talk about analyzing frames or machines, we are usually looking to identify both the external forces acting on the structure and the internal forces acting between members within the structure.

The method we use to analyze frames and machines (no special name here) centers around the process of breaking the structure down into individual components and analyzing each component as a rigid body. Where the components are connected, Newton's Third Law states that each body will exert an equal and opposite force on the other body. Each component will be analyzed as an independent rigid body leading to equilibrium equations for each component, but because of Newton's Third Law, some unknowns may show up acting on two bodies.

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The process used to analyze frames and machines, involves breaking the structure down into individual components in order to solve for the forces acting on each component. Sometimes the structure as a whole can be analyzed as a rigid body, and each component can always be analyzed as a rigid body.

The Process Used to Analyze Frames and Machines:

The process used to analyze frames and machines is outlined below:

In the beginning it is usually useful to label the members in your structure. This will help you keep everything organized and consistent in later analysis. In this book, we will label everything by assigning letters to each of the joints.

Next you will need to determine if we can analyze the entire structure as a rigid body. In order to do this, the structure needs to be independently rigid. This means that it would be rigid even if we separated it from its supports. If the structure is independently rigid (no machines, and only some frames will be independently rigid), then analyze the structure as a single rigid body to determine the reaction forces acting on the structure. If the structure is not independently rigid then skip this step.

Next you will draw a free body diagram for each of the components in the structure. You will need to include all forces acting on each member, including...
- First add any external reaction or load forces that may be acting at the components.
- Second identify any two force members in the structure. At their connection points they will cause a force with an unknown magnitude but a known direction (the forces will act along the line between the two connection points on the member).
- Next add in the reaction forces (and possibly moments) at the connection points between non-two force members. For forces with an unknown magnitude and direction (such as in pin joints) the forces are often drawn in as having unknown x and y components (x, y and z for 3D truss problems).
- Remember that the forces at each of the connection points will be a Newton's Third Law pair. This means that if one member exerts some force on some other member, then the second member will exert an equal and opposite force back on the first. When we draw out our unknown forces at the connection points, we must make sure that the forces acting on each member are opposite in direction.

Write out the equilibrium equations for each component you drew a free body diagram of. These will be extended bodies, so you will need to write out the force and the moment equations.

For 2D problems you will have three possible equations for each component, two force equations and one moment equation.

\[\sum \vec{F}=0\]	\[\sum \vec{M}=0\]
\[\sum F_x=0\]	\[\sum F_y=0\]	\[\sum M_z=0\]

For 3D problems you will have six possible equations for each component, three force equations and three moment equations.

\[\sum \vec{F}=0\]
\[\sum F_x=0\]	\[\sum F_y=0\]	\[\sum F_z=0\]

\[\sum \vec{M}=0\]
\[\sum M_x=0\]	\[\sum M_y=0\]	\[\sum M_z=0\]

Finally, solve the equilibrium equations for the unknowns. You can do this algebraically, solving for one variable at a time, or you can use matrix equations to solve for everything at once. If any force turns out to be negative, that indicates that the force actually travels in the opposite direction from what is indicated in your initial free body diagram.

Find all the forces acting on each of the members in the structure below.

If two 150 N forces are exerted on the handles of the bolt cutter shown below, determine the reaction forces (FR1 and FR2) exerted on the blades of the bolt cutter (this will be equal to the cutting forces exerted by the bolt cutters).

A 100 lb force is exerted on one side of a TV tray as shown below. Assuming no friction forces at the base, determine all forces acting on each of the three parts of the TV tray.

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Page 8

So far in statics, we have focused on the external forces acting on on bodies. These external forces are the forces applied on one body by other surrounding bodies. Whenever we have external forces acting on a body though, a set of internal forces and internal moments will develop within the body to balance out the external forces. In order to understand how a body will deform and if it will break under loading, it is important to understand these internal forces and moments.

External forces, as shown in the top diagram are exerted on a body by other surrounding bodies. Internal forces and moments, as shown in the bottom diagram exist within a body to balance and counteract the external forces.

The first step in determining the internal forces, is to define the location we are looking at within the body. Usually this location is defined by examining a specific cross section that we are examining. It is important to clearly define this cross section early on, since the internal forces will vary from one cross section the next, even if all the external forces are the same.

No matter how complex the external loading conditions, we will only have two possible types of internal forces at any given cross section and two types of internal moments at any given cross section. These forces and moments are...

Normal Forces (also sometimes called axial forces) act perpendicular to the cross section being examined and will tend to stretch or compress the body.
Shear Forces act parallel to the cross section and will tend to push deform the body along the cross section in the direction of the force.
Torsional Moments are moments where the moment vector is perpendicular to the cross section being examined and will tend to twist the body.
Bending Moments are moments where the moment vector is parallel to the cross section being examined and will tend to bend the body.

Normal forces (N) exist perpendicular to the cross section and will tend to stretch or compress the body. Shearing forces (V1 and V2) exist parallel to the cross section and will tend to deform the body along the cross section in the direction of the force.

Torsional moments (T) have moment vectors perpendicular to the cross section and will tend to twist the body. Bending moments (M1 and M2) have moment vectors parallel to the body and will tend to bend the body.

Determining Internal Forces and Moments

To determine the internal forces and moments in a body we have different approaches:

The first approach is just a static analysis of half of the body, similar to the approach used in the method of sections for trusses. Start by finding all external forces, draw a free body diagram of half of the body which is split at the cross section of interest, and then solve for all the unknown internal forces at the cross section. This method is usually faster for finding the forces and moments at a set location, though if you want to find the forces and moments at a second location you will need to do the analysis all over again.

The second approach is a graphical method that can be used to solve for the normal forces, shear forces, torsional moments, and bending moments across a range of cross sections. This method generally takes longer, and we can only graph one type of force or moment at a time, but it will allow us to graph the internal forces or moments along the entire length of the beam or shaft.

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As a review, internal forces are the forces that exist within a body in order to balance out the external forces on that body. Understanding these internal forces will be essential step towards determining how bodies deform or even break under loading.

To determine the internal forces and moments within a body, we generally have two approaches. We can either split the body into sections and perform an equilibrium analysis to determine the forces and moments at one specific location, or we can use graphical approaches to determine the forces and moments along the length of a body, beam, or shaft. In this section we are going to focus on the first method, which will be the most straightforward method for determining the internal forces and moments at a single point.

Determining Internal Forces and Moments via Equilibrium Analysis

The procedure to determine the internal forces and moments via equilibrium analysis consists of four steps.

Solve for all external forces.
Draw a free body diagram of one half of the body, cutting the original body in half at the point of interest.
Write out the equilibrium equations based on the free body diagram.
Solve the equilibrium equations for the unknown internal forces and moments.

The first step when determining the internal forces is to solve for any unknown external forces. This is just done using an equilibrium analysis for a rigid body as described in previous chapters, and is necessary because the equations we generate to solve for the internal forces will only just be sufficient to solve for the new unknowns. If we carry forward unknown external forces, we will wind up with more unknowns than equations and the equations will be unsolvable. One shortcut we can take however is that later we will only be analyzing one half of the body, and so we can skip solving for unknowns on the half of the body that we do not plan on using.

Next we will want to imagine cutting the body in half, and we will want to draw a free body diagram of just the one half of he body. Which half you choose is up to you, but the side with fewer external loads on it will make the analysis easier. Make sure you include any external forces that are acting on the chosen half of the body, and we will be adding an unknown normal force, and unknown shear force, and an unknown internal bending moment at the cross section surface.

If we wish to determine the internal forces acting at cross section a-a, we can either draw a free body diagram of everything above the cross section or we can draw a free body diagram of everything below the cross section. In either case, make sure to draw in all external forces on that half of the body as well as the exposed internal forces and moments.

In 2D problems we will always has two unknown forces (normal force N and shearing force V) and one unknown bending moment (Mr). If we are analyzing a 3D system, there will be three unknown forces (one normal and two shearing forces) as well as three unknown moments (one torsional moment and two bending moments) at the cut.

In 3D problems, there will be three unknown internal forces at the cross section and three unknown internal moments.

After we get the free body diagram of one half of the body, the next step to write out the equilibrium equations for that one half of the body. Treat the section of the body just like any other rigid body and set the sum of the forces and the sum of the moments equal to zero.

Finally, once you have the equilibrium equations set up, you simply need to solve them for the unknown internal forces and moments. As long as you have solved for the external forces earlier, you should be able to solve for all the internal forces at this point.

A mounting bracket with the dimensions shown below is subjected to a 200 lb force. Determine all internal forces and moments at cross section a-a.

A hook with a 12 mm curve at the bottom is subjected to a 60N force as shown. Determine all internal forces and moments at cross section a-a.

A mounting bracket with the dimensions shown below is subjected to a 700 N load and two 350 N reaction forces. Determine all internal forces and moments at points B and C.

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As an alternative to splitting a body in half and performing an equilibrium analysis to find the internal forces and moments, we can also use graphical approaches to plot out these internal forces and moments over the length of the body. Where equilibrium analysis is the most straightforward approach to finding the internal forces and moments at one cross section, the graphical approaches are the most straightforward approaches to find the internal forces or the internal moments across the entire length of a beam, shaft, or other body. This may be useful in complex loading scenarios where it may not be obvious where the maximum internal forces or internal moments exist. As a trade off however, we will need to plot out each type of internal load separately (one plot for internal axial forces, one for internal shear forces, one for internal torques, and one for internal bending moments).

In complex loading situations, such as the loads on this horizontal ceiling beam, it may be difficult to know where the internal forces and internal moments will be greatest. By plotting out the internal forces and moments, we will be able to more easily identify these maximum internal loads and we can design the beam accordingly to withstand these loads.

The internal axial force (N) and the internal toque (T) act along the length of the beam or shaft.

In this section, we will be focusing on the methods used to generate the plots for the internal axial forces, and the internal torques. This will be the force and moment acting along the length of the beam or shaft. The axial force plot is used primarily for vertical columns or cables supporting multiple loads along its length. The torque diagram is used primarily for shafts supporting multiple inputs and outputs. Each of these plots will have a different practical application, but we have grouped them together here because the process used to generate each of the two plots is very similar.

Creating the Axial Force Diagram

The axial force diagram will plot out the internal axial (normal) forces within a beam, column, or cable that is supporting multiple forces along the length of the beam itself. This can be thought of as the internal tension or compression forces. The most relevant practical scenarios that match this description will be main support columns in a multi-story building, or hanging cables that are used to support multiple loads.

The main vertical support columns in this building will support multiple load forces that act along the length of the column Image by Jun Seita CC-BY-NC 2.0.

To create the axial force plot for a body, we will use the following process.

Solve for all external forces acting on the body.
Draw out a free body diagram of the body horizontally. In the case of vertical structures, rotate the body so that it sits horizontally and all the forces act horizontally
Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal axial forces, with positive numbers indicating tension and negative numbers indicating compression.
Starting at zero at the left side of the plot, you will move to the right, pay attention to forces in the free body diagram above. As you move right in your plot, keep steady except...
- Jump upwards by the magnitude of the force for any forces in our free body diagram to the left.
- Jump downwards by the magnitude of the force for any forces to the right.
- You can ignore any moments or vertical forces applied to the body.
By the time you get to the right end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.

The free body diagram of a support column (rotated to be horizontal). Is shown above the corresponding axial force diagram for that column

To read the plot, you simply need to find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Again, positive numbers represent an internal tension at that location and negative numbers represent an internal compression at that location.

Creating the Torque Diagram

The torque diagram will plot out the internal torsional moment within a shaft that is supporting multiple inputs and/or outputs along its length. The most relevant practical scenarios that match this description are shafts within complex gear or pulley driven systems.

This line shaft has a single input delivering torques to multiple outputs via pulleys. Image by Wtshymanski CC-BY-SA 4.0

To create the torque diagram for a shaft, we will use the following process.

Solve for all external moments acting on the shaft.
Draw out a free body diagram of the shaft horizontally, rotating the shaft if necessary, so that all torques act around the horizontal axis.
Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal torsional moment, with positive numbers indicating an internal torsional moment vector to the right and negative numbers indicating an internal torsional moment to the left.
Starting at zero at the left side of the plot, you will move to the right, pay attention to moments in the free body diagram above. As you move right in your plot, keep steady except...
- Jump upwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point left.
- Jump downwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point right.
- You can ignore any forces in the free body diagram or moments not about the x axis.
By the time you get to the right end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.

A free body diagram of a shaft is shown above the corresponding toque diagram. Remember to use the right hand rule to determine if the moment vector is pointing to the right or left.

To read the plot, you simply need to take the find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot.

A wooden beam is subjected to the forces shown below (forces applied at base of vector). Draw the axial force diagram for the beam.

A cable is anchored to the ceiling and subjected to the forces shown below (assume are forces applied at base of the vector). Draw the axial force diagram for the cable.

A steel shaft is subjected to the torques shown below. Draw the torque diagram for this shaft.

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The internal shearing forces (V) and the internal bending moments (M) act in the plane of the cross sections we are taking.

In this section, we will be focusing on the methods used to generate the plots for the internal shearing forces, and the internal bending moments. This will be the force and moment acting in the plane of the cross section. In cases where we have a horizontal beam and primarily vertical forces (such as in the diagram above), we will specifically be looking at vertical shearing forces (V1) and bending moments about a horizontal axis (M2), and the shear and moment diagrams will plot these elements respectively. We will group these plots together, because they will often be used together and because we will need to create the shear diagram in order to create the moment diagram.

Creating the Shear Diagram

The shear diagram will plot out the internal shearing forces within a beam, or other body that is supporting multiple forces perpendicular to the length of the beam or body itself. The shear and moment diagrams are both used primarily in the analysis of horizontal beams in structures, such as floor joists, ceiling joists, and other horizontal beams used in construction.

The horizontal beams under this bridge surface will be supporting load forces perpendicular to the length of the beam. To analyze the internal shearing forces and moments in these beams we could use shear and moment diagrams. Image by Alethe CC-BY-SA 3.0

To create the shear force diagram, we will use the following process.

Solve for all external forces acting on the body.
Draw out a free body diagram of the body horizontally. Leave all distributed forces as distributed forces and do not replace them with the equivalent point load.
Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal shear force.
Starting at zero at the left side of the plot, you will move to the right, pay attention to forces in the free body diagram above. As you move right in your plot, keep steady except...
- Jump upwards by the magnitude of the force for any point forces going up.
- Jump downwards by the magnitude of the force for any point forces going down.
- For any uniform distributed forces you will have a linear slope where the magnitude of the distributed force is the slope of the line (positive slopes for upwards distributed forces, negative slopes for downwards distributed forces).
- For non-uniform distributed forces, the shape of the shear diagram plot will be the integral of the force function.
- You can ignore any moments or horizontal forces applied to the body.
By the time you get to the right end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.

The free body diagram of a horizontal beam. Is shown above the corresponding shear force diagram for that beam.

Positive and negative internal shearing forces

To read the plot, you simply need to find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Positive numbers represent an upwards internal shearing force to the right of the cross section and a downwards force on the left, and negative numbers indicate a downwards internal shearing force to the right of the cross section and a upwards force on the left. A visual of these forces can be seen in the diagram to the right.

Creating the Moment Diagram

The moment diagram will plot out the internal bending moment within a horizontal beam that is subjected to multiple forces and moments perpendicular to the length of the beam. For practical purposes, this diagram is often used in the same circumstances as the shear diagram, and generally both diagrams will be created for analysis in these scenarios.

To create the moment diagram for a shaft, we will use the following process.

Solve for all external forces and moments, create a free body diagram, and create the shear diagram.
Lined up below the shear diagram, draw a set of axes. The x-axis will represent the location (lined up with the shear diagram and free body diagram above), and the y-axis will represent the internal bending moment.
Starting at zero at the left side of the plot, you will move to the right, pay attention to shear diagram and the moments in the free body diagram above. As you move right in your plot, the moment diagram will primarily be the integral of the shear diagram, except...
- Jump upwards by the magnitude of the moment for any negative (clockwise) moments.
- Jump downwards by the magnitude of the moment for any positive (counter-clockwise) moments.
- You can ignore any forces in the free body diagram.
By the time you get to the right end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.

A free body diagram of a beam is shown above the shear and moment diagrams for that beam.

Positive and negative internal bending moments

To read the plot, you simply need to take the find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Positive internal moments would cause the beam to bow downwards (think a smile shape) negative internal moments will cause the beam to bow upwards (think a frown shape). You can also see the positive and negative internal moments in the figure to the right.

A lighting gantry is hanging from the ceiling via two cables and supporting several stage lights as shown below. Assume the gantry itself has a negligible weight. Draw the shear and moment diagram for the gantry.

Draw the shear and moment diagrams for the beam shown below.

A horizontal wooden beam in the lobby of a new hotel is going to be supported and loaded as shown below. Draw the shear and moment diagrams for the beam.

Page 12

Page 13

Dry friction is the force that opposes one solid surface sliding across another solid surface. Dry friction always opposes the surfaces sliding relative to one another and can have the effect of either opposing motion or causing motion in bodies.

Dry friction occurs between the bottom of this training sled and the grassy field. The dry friction would oppose the motion of the sled along the field in this case. Image by Avenue CC-BY-SA 3.0

Dry friction occurs between the tires and the road for this motorcycle. The dry friction force for this motorcycle is what allows it to accelerate, deccelerate, and turn. Public Domain image by Takisha Rappold.

The most commonly used model for dry friction is coulomb friction. This type of friction can further be broken down into static friction and kinetic friction. These two types of friction are illustrated in the diagram below. First imagine a box sitting on a surface. A pushing force is applied parallel to the surface and is constantly being increased. A gravitational force, a normal force, and a frictional force are also acting on the box.

As the pushing force increases, the static friction force will be equal in magnitude and opposite in directionuntil the point of impending motion. Beyond this point, the box will begin to slip as the pushing force is greater in magnitude than the kinetic friction force opposing it.

Static friction occurs prior to the box slipping and moving. In this region the friction force will be equal in magnitude and opposite in direction to the pushing force itself. As the magnitude of the pushing force increases so does the magnitude of the friction force.

If the magnitude of the pushing force continues to rise, eventually the box will begin to slip. As the box begins to slip the type of friction opposing the motion of the box changes from static friction to what is called kinetic friction. The point just before the box slips is known as impending motion. This can also be thought of as the maximum static friction force before slipping. The magnitude of the maximum static friction force is equal to the static coefficient of friction times the normal force existing between the box and the surface. This coefficient of friction is a property that depends on both materials and can usually be looked up in tables.

Kinetic friction occurs beyond the point of impending motion when the box is sliding. With kinetic friction, the magnitude of the friction force opposing motion will be equal to the kinetic coefficient of friction times the normal force between the box and the surface. The kinetic coefficient of friction also depends upon the two materials in contact, but will almost always be less than the static coefficient of friction.

A 500 lb box is sitting on concrete floor. If the static coefficient of friction is .7 and the kinetic coefficient of friction is .6:

What is the friction force if the pulling force is 150 lbs?
What pulling force would be required to get the box moving?
What is the minimum force required to keep the box moving once it has started moving?

A 30 lb sled is being pulled up an icy incline of 25 degrees. If the static coefficient of friction between the ice and the sled is .4 and the kinetic coefficient of friction is .3, what is the required pulling force needed to keep the sled moving at a constant rate?

A plastic box is sitting on a steel beam. One end of the steel beam is slowly raised, increasing the angle of the surface until the box begins to slip. If the box begins to slip when the beam is at an angle of 41 degrees, what is the static coefficient of friction between the steel beam and the plastic box?

A 60 kg wheelbarrow with the dimensions shown below is subjected to a pulling force. If there is assumed to be no friction at the wheel at B, and the static coefficient of friction at A is assumed to be .4, what is the expected pulling force needed to get the wheelbarrow moving?

A cabinet with a weight of 30 lbs is sitting on a 15 degree incline as shown below. The cabinet has a pair of rubber feet at the front (A), and a pair of frictionless wheels at the back (B). What is the minimum value for the static coefficient of friction for the rubber feet at the front to ensure the cabinet does not roll down the incline?

Page 14

Imagine a box sitting on a rough surface as shown in the figure below. Now imagine that we start pushing on the side of the box. Initially the friction force will resist the pushing force and box will sit still. As we increase the force pushing the box however, one of two things will occur.

The pushing force will exceed the maximum static friction force and the box will begin to slide across the surface (slipping).
Or, the pushing force and the friction force will create a strong enough couple that the box will rotate and fall on it's side (tipping).

As the pushing force increases on the box, it will either begin to slide along surface (slipping) or it will begin to rotate (tipping)

When we look at cases where either slipping or tipping may occur, we are usually interested in finding which of the two options will occur first. To determine this, we usually determine both the pushing force necessary to make the body slide and the pushing force necessary to make the body tip over. Whichever option requires less force is the option that will occur first.

Determining the Force Required to Make an Object "Slip":

A body will slide across a surface if the pushing force exceeds the maximum static friction force that can exist between the two surfaces in contact. As in all dry friction problems, this limit to the friction force is equal to the static coefficient of friction times the normal force between the body. If the pushing force exceeds this value then the body will slip.

If the pushing force exceeds the maximum static friction force (us * FN) then the body will begin to slide.

Determining the Force Required to Make an Object "Tip"

The normal forces supporting bodies are distributed forces. These forces will not only prevent the body from accelerating into the ground due to gravitational forces, but they can also redistribute themselves to prevent a body from rotating when forces cause a moment to act on the body. This redistribution will result in the equivalent point load for the normal force shifting to one side or the other. A body will tip over when the normal force can no longer redistribute itself any further to resist the moment exerted by other forces (such as the pushing force and the friction force).

At rest (A) the normal force is a uniformly distributed force on the bottom of the body. As a pushing force is applied (B) the distributed normal force is redistributed, moving the equivalent point load to the right. This creates a couple between the gravity force and the normal force that will counter the couple exerted by the pushing force and the friction force. If the pushing force becomes large enough (C), the couple exerted by the gravitational force and the normal force will be unable to counter the couple exerted by the pushing force and the friction force.

The easiest way to think about the shifting normal force and tipping is to imagine the equivalent point load of the distributed normal force. As we push or pull on the body, the normal force will shift to the left or right. This normal force and the gravitational force create a couple that exerts a moment. This moment will be countering the moment exerted by the couple formed by the pushing force and the friction force.

Because the normal force is the direct result of physical contact, we cannot shift the normal force beyond the surfaces in contact (aka. the edge of the box). If countering the moment exerted by the pushing force and the friction force requires shifting the normal force beyond the edge of the box, then the normal force and the gravity force will not be able to counter the moment and as a result the box will begin to rotate (aka. tip over).

The body will tip when the moment exerted by the pushing and friction forces exceeds the moment exerted by the gravity and normal forces. For impending motion the normal force will be acting at the very edge of the body.

A box with the dimensions shown below is placed on a flat surface. We increase the magnitude of the pushing force until the box either tips over or begins to slide. Which will occur first? Assume the center of mass of the box is at the center point of the box.

A box with the dimensions shown below is placed on a flat surface. We increase the magnitude of the pushing force until the box either tips over or begins to slide. What is the largest value of d that will allow the box to slide along the surface before it tips over. Assume the center of mass of the box is at the center point of the box.

A pushing force is applied to the box as shown below. The magnitude of the force is increased until the box either tips over or begins to slide. Which will occur first? Assume the center of mass of the box is at the center point of the box.

A 30 lb box with the dimensions shown below is placed on a surface and the angle of that surface is slowly increased until the box either tips over or begins sliding. What will occur first and what angle will this occur at? Assume the center of mass of the box is at the center point of the box.

Page 15

A wedge is a thin wedge-shaped object that is used to force two objects apart or to force one object away from a close surface. Wedges have the effect of allowing users to create very large normal forces to move objects with relatively small input forces. The friction forces in wedge systems also tend to be very large though, and can reduce the effectiveness of wedges.

A hammer is used to push this wedge into the crack in this log. The normal forces are pushing the two halves of the log apart while the friction forces are opposing the pushing force. Adapted from image by Luigi Chiesa CC-BY-SA-3.0.

To analyze a wedge system we will need to draw free body diagrams of each of the bodies in the system (the wedge itself and any bodies the wedge will be moving). We need to be sure that we include the pushing force on the wedge, normal forces along any surfaces in contact, and friction forces along any surfaces in contact.

The top diagram shows a wedge being used to push a safe away from a wall. The first step in analyzing the system is to draw free body diagrams of the wedge and the safe. Remember that all normal forces will be perpendicular to the surfaces in contact and that all friction forces will be parallel to the surfaces in contact.

After we draw the free body diagram we can work to simplify the problem. It is usually assumed that the wedge and the bodies will be sliding against one another, so each friction force will be equal to the kinetic coefficient of friction between the two surfaces times the associated normal force between the two forces. This reduces the number of unknowns and will usually allow us to solve for any unknown problems.

By replacing each of the friction forces with the kinetic coefficient of friction times the normal force, we can reduce the number of unknowns in our analysis.

With our simplified diagram we will assume that the bodies are all in equilibrium and we will write our equilibrium equations for the two bodies. By solving the equilibrium equations, we can solve for any unknowns we have.

A heavy safe is being pushed away from a wall with a wedge as shown below. Assume the wedge has an angle of 5 degrees, the coefficient of friction (static and kinetic) between the wedge and the safe is .16, and the coefficients of friction (static and kinetic) between the wedge and the wall and the safe and the floor are both .35. What is the pushing force required to move the safe out from the wall?

A wedge as shown below is being used to lift the corner of the foundation of a house. How large must the pushing force be to exert a one ton (2000 lb) lifting force?

Page 16

A power screw (also sometimes called a lead screw) is another simple machine that can be used to create very large forces. The screw can be thought of as a wedge or a ramp that has been wound around a shaft. By holding a nut stationary and rotating the shaft, we can have the nut sliding either up or down the wedge in the shaft. In this way a relatively small moment on the shaft can cause very large forces on the nut.

The screw in this cider press is rotated with the handle on the top. The stationary nut on the frame of the press forces the shaft downward as it is turned. Public Domain image by Daderot.

By rotating the screw in this scissor jack, the user can move the nut (on the left end) closer or further from the anchor on the right end, which will raise or lower the car. Public Domain image by nrjfalcon1.

Static Analysis of Power Screws:

The easiest way to analyze a power screw system is to turn the problem into a 2D problem by "unwrapping" the ramp from around the shaft. To do this we will need two numbers. First we will need the diameter of the shaft, and second we will need either the threads-per-inch/centimeter or the pitch of the screw. The threads-per-inch tells you how many threads you have per inch/centimeter of screw. With a single thread design (most screws) this will also be the number of times the thread wraps around the screw in one inch/centimeter. The pitch on the other hand gives you the distance between two adjacent threads. Either of these numbers can be used to find the other.

Once we have these numbers, we can imagine unwrapping the ramp from around the screw and ending up with a ramp in one of the two situations below. In either case, we can use the inverse tangent function to find the lead angle, which can be thought of as the angle of the thread that the nut is climbing up. Finding the lead angle is the first step in analyzing a power screw system.

The lead angle of a screw is the angle of the thread that the nut will be climbing up as the shaft rotates.

Once we find the lead angle, we can draw a free body diagram of the "nut" in our unwrapped system. Here we include the pushing force which is pushing our nut up the incline, the load force which is the force the nut exerts on some external body, the normal force between the nut and screw, and the friction force between the nut and the screw.

The free body diagram of the "nut" in our power screw system.

If our screw is pushing a load at some constant rate, then we can assume two things: First the nut is in equilibrium, so we can write out the equilibrium equations for the nut. Second the nut is sliding, indicating that the friction force will be equal to the normal force times the kinetic coefficient of friction.

\[F_f= \mu_k*F_N\]

\[\sum F_x=F_{push}-F_N*sin\left ( \theta \right )-\mu _k*F_N*cos\left ( \theta \right )=0\]

\[\sum F_y=-F_{load}+F_N*cos\left ( \theta \right )-\mu _k*F_N*sin\left ( \theta \right )=0\]

We can then simplify the equations above into a single equation relating the load force and the pushing force.

\[F_{push}=\frac{sin\left ( \theta \right )+\mu_k*cos\left ( \theta \right )}{cos\left ( \theta \right )-\mu_k*sin\left ( \theta \right )}*F_{load}\]

In reality the pushing force is not a single force at all. It is the forces preventing the nut from rotating with the screw. The cumulative pushing force will really cause an equal and opposite moment to the input moment that is spinning the shaft.

The pushing force is really just a representation of the forces keeping the nut from rotating. If the nut is not rotating then they must cause an equal and opposite moment to the torque that is driving the screw.

Finally, if we replace the pushing force with the moment that is driving the screw in our system (in this case torque T), we can relate the input torque that is driving our screw, to the force that the nut on the screw is pressing forward with. Screw systems are usually designed to allow fairly small input moments to push very large load forces.

\[T=\frac{sin\left ( \theta \right )+\mu_k*cos\left ( \theta \right )}{cos\left ( \theta \right )-\mu_k*sin\left ( \theta \right )}*F_{load}*r_{shaft}\]

Self Locking Screws

Imagine we apply a torque to a power screw to lift a body, then when we get the load to the desired height we stop applying that toque to let the body sit where it is. If we were to redraw our free body diagram from earlier for the new situation, we would find two things.

The pushing force is missing (since a torque is no longer applied to the shaft).
The friction is now fighting against the nut sliding back down the ramp.

Without the pushing force from an applied torque, the friction force acts to prevent the nut from sliding down the ramp.

With this new free body diagram, there are two possible scenarios that could occur:

The friction force is large enough to keep the nut from sliding down the ramp, meaning everything will remain in static equilibrium if released.
The friction force will not be sufficient to keep the nut from sliding down the ramp, meaning that the load would begin to fall as soon as the torque is removed from the shaft.

With power screw applications such as a car jack, the second option could be very dangerous. It is therefore important to know if a power screw system is self locking (option 1 from above) or not self locking (option 2 from above).

To define the boundary between self locking systems and non-self locking systems, we use something called the self locking angle. As intuition would tell us, slipping does not occur on very gentle slopes (small lead angles) while it does occur on very steep slopes (large lead angles). The angle at which the nut would begin to slip is known as the self locking angle.

To find the self locking angle, we will assume impending motion (relating the friction force to the normal force) and leave the lead angle as an unknown. This lets us create the free body diagram as shown below and gives us the equilibrium equations below.

To find the self locking angle we will assume impending motion. We then draw our free body diagram (above) and write out our equilibrium equations (below) accordingly.

\[\sum F_x=-F_N*sin\left ( \theta \right )+\mu_s*F_N*cos\left ( \theta \right )=0\]

\[\sum F_y=-F_{load}+F_N*cos\left ( \theta \right )+\mu_s*F_N*sin\left ( \theta \right )=0\]

Using the x equilibrium equation as a starting point, we can solve for theta (eliminating the normal force all together in the process). This new equation shown below gives us the self locking angle.

\[\theta_{locking}=tan^{-1}\left ( \mu_s \right )\]

Systems with lead angles smaller than this will be self locking, while systems with lead angles larger than this will not be self locking.

The power screw below is being used to lift a platform with a weight of 12 pounds. Based on the information below...

What is the required torque on the shaft to lift the load?
Would the load fall if the toque was removed from the shaft?

You are designing a screw lift system that will be used in auto repair shops as shown below. Each screw lift will be required to support loads up to 20kN.

Assuming a coefficient of friction of .1, what is the lead angle at which the system will be self-locking?
Assuming a 5cm diameter screw, and an available motor torque of 60Nm, what is the maximum pitch that could be used to lift the 20kN load?

Page 17

A bearing is the machine element used to support a rotating shaft Bearing friction is the friction that exists between the rotating shaft and bearing that is supporting that shaft. Though many types of bearings exist (plain, ball, roller, hydrodynamic) in this course we will only be looking at plain bearings (also sometimes called journal bearings).

Some friction will exist in the bearings on this train car. Public Domain image by ds_30.

In a plain bearing as the one shown here, a friction force will oppose the rotation of the spinning shaft in the stationary bearing. This force will cause a small moment opposing the rotation of the shaft.

A plain bearing consists of a circular shaft fitted into a slightly larger circular hole as shown to the right. The shaft will usually be rotating and will be exerting some sort of load (F load) onto the bearing. The bearing will then be supporting the shaft with some normal force and a friction force will exist between the bearing surface and the surface of the shaft. Sometimes the rotation of the bearing will cause the shaft to climb up the side of the bearing, causing the angle of the normal and friction forces to change, but this climbing is usually small enough that it is neglected.

If we assume that the climbing in the bearing is negligible, the normal of the bearing on the shaft will be equal and opposite to loading force of the shaft on the bearing. Furthermore, if the shaft is rotating relative to the bearing, then the friction force will be equal to kinetic coefficient of friction times the normal force of the bearing on the shaft.

If the climb angle is assumed to be small, then...

\[F_{load}=F_{N}\]

\[F_{f}=\mu _{k}* F_{N}\]

The friction force will then be exerting a moment about the center of the shaft opposing the rotation of the shaft. Unless some other moment is keeping the shaft spinning, this moment will eventually slow down and stop the rotation in the shaft. The magnitude of this moment will be equal to the magnitude of the friction force times the radius of the shaft.

\[M_{f}=r_{shaft}* \left( \mu _k* F_{N} \right)\]

Another important factor to keep in mind is that most bearings are lubricated. This can significantly lower the coefficients of friction (this the the main reason for using lubrication actually). When performing calculations, it is import to know if the bearing is lubricated and if so using an appropriate friction coefficients.

A cart and its load weigh a total of 200 lbs. There are two wheels on the cart each with a diameter of 2 feet and each supporting half of the 200 pound load. The bearing attaching each wheel to the cart is a lubricated steel on steel (kinetic coefficient of friction = .05) journal bearing with a shaft one inch in diameter. What is the moment due to friction from each bearing? Assuming there are no other sources of friction, what magnitude must the pulling force have in order to keep the cart moving at a constant speed?

Page 18

Disc friction is the friction that exists between the end of a rotating shaft or other rotating body and a stationary surface. Disc friction will tend to exert a moment on the bodies involved, resisting the relative rotation of the bodies. Disc friction is applicable to a wide variety of designs including end bearings, collar bearings, disc brakes, and clutches.

This orbital sander rotates a circular sanding disc against a stationary surface. The disc friction between the sanding disc and the surface will exert a moment on both the surface and the sander. Image by Hedwig Storch CC-BY-SA 3.0

Hollow Circular Contact Area

To start our analysis of disc friction we will use the example of a collar bearing. In this type of bearing, we will have a rotating shaft traveling through a hole in a surface. The shaft is supporting some load force as shown and a collar is used to support the shaft itself. In this case we will have a hollow circular contact area between the rotating collar and the stationary surface.

In a collar bearing we will have a hollow circular contact area between the rotating collar and the stationary surface.

The friction force at any point in the contact area will be equal to the normal force at that point times the kinetic coefficient of friction at that point. If we assume a uniform pressure between the collar and the surface and a uniform coefficient of friction then we will have the same friction force exerted at all points. This does not translate however into an equal moment exerted by each point. Points further from the center of rotation will exert a larger moment than points closer to the center of rotation because they will have a larger moment arm.

\[F=\mu _k\; F_{N}\]

\[F_i=F_0\]

\[M_i\neq M_0\]

Though the friction forces at any point will be the same, points along the outside surface of the contact area will exert a larger moment than points along the inside surface of the contact area.

To determine the net moment exerted by the friction forces, we will need to use calculus to sum up the individual moments over the entire contact area. The moment at each individual point will be equal to the kinetic coefficient of friction, times the normal force pressure at that point (p), times the distance from that point to the center of rotation (r).

\[M=\int_{A}dM=\int_{A}\mu _k*p*r*dA\]

To simplify things we can move the constant coefficient of friction, and the constant normal force pressure term outside of the integral. We can also replace the pressure term with the load force on the bearing over the contact area. Finally, so that we can integrate over the range of R values, we can recognize that the rate of change in the area (dA) for the hollow circular areas is simply the rate of change of the r (dr) term time the circumference of the circle at r. These changes lead to the equation below.

\[M=\mu _k \left ( \frac{F_{load}}{\pi \left ( R_{0}^{2}-R_{i}^{2} \right )} \right )\int_{R_i}^{R_0}r*\left ( 2\pi r \right )dr\]

Finally we can evaluate the integral from the inner radius to the outer radius. If we evaluate the integral and simplify we will end up with the final equation below.

\[M=\frac{2}{3}\mu _k\; F_{load} \left ( \frac{R_{0}^{3}-R_{i}^{3}}{R_{0}^{2}-R_{i}^{2}} \right )\]

Solid Circular Contact Area

In cases where we have a solid circular contact area such as with a solid circular shaft in an end bearing or with the orbital sander shown at the top of the page we simply set the inner radius to zero and we can simplify the formula. If we do so, the original formula is reduced to the equation below.

An end bearing has a solid circular contact area between the rotating shaft and the stationary bearing.

\[M=\frac{2}{3}\mu _{k}F_{load}R_{0}\]

Circular Arcs (Disc Brakes)

In cases, such as in disc brakes, we may have a contact area that looks like a section of the hollow circular contact area we had earlier. For this scenario however, we have less area to exert the friction force and cause a moment, but the smaller area also causes higher pressures in that contact area for the same load force. The end result is that these terms cancel each other out and we end up with the same formula we had for the hollow circular contact area when examining a single brake pad. Most disc brakes however have a pair of pads however (one on each side of the rotating disc), so we will need to double the moment in our equation for the usual pair of pads.

The contact area in disc brakes is often approximated as a circular arc with a given contact angle (theta)

Single Brake Pad:	\[M=\frac{2}{3}\mu _k\; F_{load} \left ( \frac{R_{0}^{3}-R_{i}^{3}}{R_{0}^{2}-R_{i}^{2}} \right )\]
Brake Pad on Each Side:	\[M=\frac{4}{3}\mu _k\; F_{load} \left ( \frac{R_{0}^{3}-R_{i}^{3}}{R_{0}^{2}-R_{i}^{2}} \right )\]

The calculations above show that the contact angle (theta) is irrelevant to the stopping power of the brakes in theory. In practice however, larger brake pads can slightly increase the stopping power of the brakes and have other advantages such as better heat disapation.

A disc sander is pressed against a wooden surface with a force of 50N. Assuming the kinetic coefficient between the sanding pad and the wood is .6 and the diameter of the sanding disc is .2 meters, what is the torque the motor must exert to keep the disc spinning at a constant rate?

Image by Hedwig Storch CC-BY-SA 3.0.

In the disc brake setup shown below, a pair of brake pads is pressed into the rotor with a force of 300 lbs. If the kinetic coefficient of friction between the brake pads and the rotor is .4, find the stopping torque exerted by the brake pads.

Page 19

In any system where a belt or a cable is wrapped around a pulley or some other cylindrical surface, we have the potential for friction between the belt or cable and the surface it is in contact with. In some cases, such as a rope over a tree branch being used to lift an object,the friction forces represent a loss. In other cases such as a belt driven system, these friction forces are put to use transferring power from one pulley to another pulley.

In many belt driven systems, the belt friction keeps the pulley from slipping relative to the belt. This allows us to use belts to transfer forces from one pulley to another pulley. Image by Kiilahihnakone CC-BY-SA 3.0

If we were to pass a rope over a tree branch to help lift an object such as this bear bag, the rope would experience belt friction resisting the sliding of the rope relative to the surface of the tree branch. Image by Virginia State Park Staff CC-BY 2.0.

For analysis, we will start a flat, massless belt passing over a cylindrical surface. If we have an equal tension in each belt, the belt will experience a non-uniform normal force from the cylinder that is supporting it.

In a frictionless scenario, if we were to increase the tension on one side of the rope it would begin to slide across the cylinder. If friction exists between the rope and the surface though, the friction force will oppose with sliding motion, and prevent it up to a point.

With equal tensions on each side of the belt, only a non-uniform normal force exists between the belt and the surface.

With unequal tensions, a friction force will also be present opposing the relative sliding of the belt to the surface.

Friction in Flat Belts

For a flat belt, the belt or cable will interact with the bottom surface. For a V belt, the belt or cable will interact with the sides of a groove.

A flat belt is any system where the pulley or surface only interacts with the bottom surface of the belt or cable. If the belt or cable instead fits into a groove, then it is considered a V belt.

When analyzing systems with belts, we are usually interested in the range of values for the tension forces where the belt will not slip relative to the surface. Starting with the smaller tension force on one side (T1) we, can increase the second tension force (T2) to some maximum value before slipping. For a flat belt, the maximum value for T2 will depend on the value of T1, the static coefficient of friction between the belt and the surface, and the contact angle between the belt and the surface (Beta) given in radians, as described in the equation below.

The maximum value of T2 before the belt starts slipping can be determined with the equation above.

Friction in V Belts

A V belt is any belt that fits into a groove on a pulley or surface. For the V belt to be effective, the belt or cable will need to be in contact with the sides of the groove, but not the base of the groove as shown in the diagram below. With the normal forces on each side, the vertical components must add up the the same as what the flat belt would have, but the added horizontal components of the normal forces, which cancel each other out, increase the potential for friction forces.

The equation for the maximum difference in tensions in V belt systems is similar to the equation in flat belt systems, except we use an "enhanced" coefficient of friction that takes into account the increased normal and friction forces possible because of the groove.

In a V belt, the "enhanced" coefficient of friction takes into account the coefficient of friction between the two materials as well as the groove angle.

As we can see from the equation above, steeper sides to the groove (which would result in a smaller angle alpha) result in an increased potential difference in the tension forces. The trade-off with steeper sides however is that the belt becomes wedged in the groove and will require force to unwedge itself from the groove as it leaves the pulley. This will cause losses that decrease the efficiency of the belt driven system. If very high tension differences are required, chain driven systems offer an alternative that is usually more efficient.

Torque and Power Transmission in Belt Driven Systems

In belt driven systems there is usually an input pulley and one or more output pulleys. To determine the maximum torque or power that can be transmitted by the belt, we will need to consider each of the pulleys independently, understanding that slipping occurring at either the input or the output will result in a failure of the power transmission.

A belt driven system with a single input and a single output.

The first step in determining the maximum torque or power that can be transmitted in the belt drive is to determine the maximum possible value for T2 before slipping occurs at either the input or output pulley(again slipping at either location cannot occur). To start we will often be given the "resting tension". This is the tension in the belt when everything is stationary and before power is transferred. Sometimes machines will have adjustments to increase or decrease the resting tension by slightly increasing or decreasing the distance between the pulleys. If we turn on the machine and increase the load torque at the output, the tension on the one side of the pulleys will remain constant as the resting tension while the tension on the other side will increase. Since the resting tension is constant and always the lower of the two tensions, it will be the T1 tension the equations discussed in earlier sections.

Though it is often wise to check, assuming the pulleys are the same material (and therefore the same coefficients of friction), it is often assumed that the belt will first slip at the smaller of the two pulleys in a single input, single output belt system. This is because the smaller pulley will have a the smaller contact angle (Beta), while all other values remain the same.

Once we have the maximum value for T2, we can use that to find the torque at the input pulley and the torque at the output pulley. Note that these two values will not be the same unless the pulleys are the same size. To find the torque, we will simply need to find the net moment exerted by the two tension forces, where the radius of the pulley is the moment arm.

Maximum input torque before slipping

\[M_{max}=\left ( T_{2max}-T_1 \right )\left ( r_{input} \right )\]

Maximum output torque before slipping

\[M_{max}=\left ( T_{2max}-T_1 \right )\left ( r_{output} \right )\]

To find the maximum power we can transfer with the belt drive system, we will use the rotational definition of power, where the power is equal to the torque times the angular velocity in radians per second. Unlike the torque, the power at the input and the output will be the same assuming no inefficiencies.

\[P_{max}=\left ( M_{input max} \right )\left ( \omega _{input} \right )=\left ( M_{output max} \right )\left ( \omega _{output} \right )\]

A steel cable supports a 60 kg mass and is then run a quarter of the way around a steel cylinder and supported by a pulling force as shown in the diagram below. The static coefficient of friction between the cable and the steel cylinder is .3.

What is the minimum pulling force required to lift the mass?
What is the minimum pulling force required to keep the mass from falling?

A V belt pulley as shown below is used to transmit a torque. If the diameter of the pulley below is 5 inches, the resting tension in the belt is 20 lbs, and the coefficient of friction between the belt material and the pulley is .4, what is the maximum torque the pulley can exert before slipping?

A flat belt is being used to transfer power from a motor to an alternator as shown in the diagram below. The coefficient of friction between the belt material and the pulley is .5. If we require a power of 100 Watts (Nm/s) while the input is rotating at a rate of 1000 rpm and the output is rotating at a rate of 1428.6 rpm, what is the required resting tension in the belt?(Assume contact angles of approximately 180 degrees)

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Imagine we have a particle that is moving along a single axis. At any given point in time, this particle will have a position, which we can quantify with a single number which we will call x. This value will measure the distance from some set origin point to the position of the particle. If the particle is moving over time, we will need a function to describe the position over time x(t). This is an equation where if we plug a value for t, it will give us the position at that time.

The position of a particle in one dimension can be described with a single number. If the position is changing over time, we will use the function x(t) to describe the position at any given point in time.

The velocity of the particle is then the rate of change of the position over time. If the particle is not moving, then position is not changing over time and the velocity is zero. If the particle is moving, we will first need to find the equation for position x(t), and then take the derivative of the position equation to find the velocity equation v(t). Velocity differs from speed in that the velocity has a direction (either positive or negative for now) while the speed is simply the magnitude of the velocity (always a positive number).

Next up is the acceleration, which is the rate of change of the velocity over time. If the velocity is not changing, the acceleration will be zero. If the velocity does change over time, then we will need to take the derivative of the velocity equation v(t) to find the acceleration equation a(t). The acceleration is then also the double derivative of the position equation over time. Like the velocity the acceleration has both a magnitude and a direction.

To simplify the notation, we often use a dot to indicate a time derivative. This makes the the velocity (the derivative of x) x dot, and the acceleration (the derivative of the derivative of x) x double dot. These relationships and their shorthand notations are all shown below.

Position:	\[x(t)\]
Velocity:	\[v(t)=\frac{dx}{dt}=\dot{x}\]
Acceleration:	\[a(t)=\frac{dv}{dt}=\frac{d^2x}{dt^2}=\ddot{x}\]

If we instead start with the equation for acceleration, we can take the integral of that equation a(t) to find the equation for velocity v(t). Unlike the derivatives though we will have an extra step in this process because whenever we integrate we wind up with a constant of integration (which we will usually call C). When we integrate the acceleration equation to find the velocity equation, this constant will be the initial velocity (the velocity at time = 0)

Next we can take the integral of the velocity equation v(t) to find the position equation x(t). With this integration we will again wind up with a constant of integration, which in this case will be the initial position (the position at time = 0). These relationships are shown below.

Acceleration:	\[a(t)\]
Velocity:	\[v(t)=\int a(t)\]
Position:	\[x(t)=\int v(t)=\int \int a(t)\]

Constant Acceleration Systems:

In cases where we have a constant acceleration (often due to a constant force), we can start with a constant value for a(t) = a, and work the integrals from there. Along the way we will add the initial velocity and the initial position as the constants of integration to wind up with the formulas below.

Acceleration:	\[a(t)=a\]
Velocity:	\[v(t)=at+v_{0}\]
Position:	\[x(t)=\frac{1}{2}at^{2}+v_{0}t+x_{0}\]

If we take the equations for the position and the velocity from above then solve both of them for t and set those equations equal to one another we can actually wind up with another equation that directly relates position, velocity, and acceleration without needing to know the time.

\[v^{2}-v{_{0}}^{2}=2a(x-x_{0})\]

It is important to remember that these equations are only valid when the acceleration is constant. When that is not the case, you will need to use calculus to find the derivatives or integrals based on the equations for position, velocity, and acceleration that you do know.

You are in a van that steadily accelerates from 20 m/s to 35 m/s over the course of 10 seconds. What is your rate of acceleration?

You are in a van that steadily accelerates from 20 m/s to 35 m/s over the course of 10 seconds. How many meters did you travel in those ten seconds?

In a rocket sled deceleration experiment, a manned sled is decelerated from a speed of 200 mph (89.4 m/s) to a stop at a constant rate of 18Gs (176.6 m/s2). How long does it take for the sled to stop? How far does the sled travel while decelerating?

An object is released from rest at the top of a tall building of unknown height. Using a precision stopwatch, you note that it takes 2.5 seconds for the object to hit the ground. Assuming the standard rate of acceleration of 32.2 ft/s2 and negligible air resistance, what is the estimated height of the building in feet?

A metallic particle is accelerated in a magnetic field such that its velocity over time is defined by the function v(t) = 4t2 - 12, where time is in seconds and velocity is in meters per second. If we assume that the particle has an initial position of zero (x0 = 0), what are the equations that describe the acceleration and position over time?

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In continuous motion, we used a single mathematical function each to describe the position, velocity, or acceleration over time. If we cannot describe the motion with a single mathematical function over the entire time period, that motion is considered non-continuous motion. In cases such as this we will use different equations for different sections of the overall time period.

For an example of non-continuous motion, imagine a car that accelerates for a few seconds, then holds a steady speed for a few seconds, then puts on the brakes and comes to a stop over the final few seconds. There is no one mathematical function we can use to describe the motion for the full time period, but if we break the motion into three pieces, then we can come up with an equation for each section of the motion.

As this car accelerates, its velocity steadily goes up for a few seconds, then it holds constant for a few seconds, then it steadily goes down for a few seconds. Since we would need a separate equation for each section of this motion, this is considered non-continuous motion.

Analyzing the first time period will be exactly the same as analyzing a continuous function. We will initially need to identify the mathematical function to describe position, or velocity, or acceleration for that first time period. Next we take derivatives to move from position to velocity to acceleration or take integrals to move from acceleration to velocity to position. Whenever we take an integral, we need to remember to include the constant of integration which will represent the initial velocity or the initial position (in the velocity and position equations respectively).

For the second, third (and all following) sections, we will do much the same process. We will start by identifying an equation for the position, or velocity, or acceleration for that time period. From there we again take derivatives or integrals as appropriate, but now the constants of integration will be a little more complicated. Those constants still represent initial velocities and and positions in a sense, but they will be the velocity and position when t=0, not the velocity and position at the start of that section.

To find the constants of integration, we are instead going to have to use the transition point, which is the point in time where we are moving from one set of equations to the next. Even though the equations are changing, we cannot have an instantaneous jump in either the position or velocity. An instantaneous jump in either position or velocity would require infinite acceleration, which is physically impossible.

This figure shows the acceleration, velocity, and position plots from the car earlier on. Though there are noticeable jumps in the acceleration moving from one section to the next, there are no jumps in the velocity or position as we move from one section to the next. This is because jumps in those plots would require infinite accelerations.

To find the velocity equations for the second time frame (or third or fourth etc.), we start by integrating the acceleration equation for that same time period. This will lead to an equation with an unknown constant of intergeneration. To solve for that constant we look back to the velocity equation for the previous time frame and solve for the velocity at the very end of this prior time period. Since it can't jump instantaneously, this is also the velocity at the start of the current time period. Using this velocity, along with the time t at the transition point, we can solve for the last unknown in the current velocity equation (the constant of integration).

To find the position equation for the second time frame (or third or fourth etc.), we start by integrating the velocity equation for the same time period (you will need to solve for the unknowns in the velocity equation first, as discussed above). After integration, we should have one new constant of integration in the position equation. Just as we did with the velocity equations, we will use the position equation from the prior time frame to solve for the position at the transition point, then use that value along with the known time t to solve for the unknown constant in the current position equation.

A car accelerates from rest at a rate of 10 m/s2 for 10 seconds. The car then immediately begins decelerating at a rate of 4 m/s2 for another 25 seconds before coming to a stop. Find the equations for the acceleration, velocity, and position functions over the full 35 second time period and plot these functions.

A plane with an initial speed of 95 m/s touches down on a runway. For the first second the plane rolls without decelerating. For the next 5 seconds reverse thrust is applied, decelerating the plane at a rate of 4 m/s2. Finally, the brakes are applied with reverse thrust increasing the rate of deceleration to 8 m/s2. How long does it take for the plane to come to a complete stop? How far does the plane travel before coming to a complete stop?

A satellite records the velocity function shown below over a sixty second time period. During that same time period determine the acceleration and position functions and draw these functions on a plot.

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Two Dimensional Motion (also called Planar Motion) is any motion in which the objects being analyzed stay in a single plane. When analyzing such motion, we must first decide the type of coordinate system we wish to use. The most common options in engineering are rectangular coordinate systems, normal-tangential coordinate systems, and polar coordinate systems. Any planar motion can potentially be described with any of the three systems, though each choice has potential advantages and disadvantages.

The rectangular coordinate system (also sometimes called the Cartesian coordinate system) is the most intuitive approach to describing motion. In rectangular coordinate systems we have an x and a y axis. These axes remain fixed to some origin point in the environment and they do not change over time. Instead, the bodies we are analyzing usually move relative to these fixed axes. An example of a body with a rectangular coordinate systems is shown in the figure below.

In the rectangular coordinate system we have a fixed origin point at o, the particle at point p, and the x and y directions, which must be perpendicular to one another. The vector r is the vector going from o to p. The component of this vector in the x direction is x and the component of this vector in the y direction is y. We usually describe position in terms of x and y at any given point in time. The vectors i and j represent unit vectors (vectors with a length of one) in the x and y directions respectively.

Rectangular coordinates work best for systems where all forces maintain a constant direction. The most common example of this is projectile motion, where gravity (the only force in these systems) maintains a constant downward direction. An example of a system where the forces change direction over time would be something like a car going around a curve in the road. In this case the friction force at the tires is going to be rotating with the car. The car problem will therefore be better suited to the use of normal-tangential or polar coordinate systems.

When describing the position of a point in rectangular coordinate systems, we are going to start by describing both x and y coordinates in a vector form. For this, the values x and y represent distances and the unit vectors i and j are used to indicate which distance corresponds with which direction. This may seem redundant, but remember when solving actual problems, x and y will just be numbers.

Position:	\[r_{p/o}(t)=x(t)\hat{i}+y(t)\hat{j}\]

Just as with one dimensional problems, if we take the derivative of the position equation, we will find the velocity equation. If we take the derivative of the velocity equation we will wind up with the acceleration equation. Also like one dimensional problems, we can use integration to move in the other direction, moving from an acceleration equations to a velocity equation to a position equation.

The unit vectors add a new element in two dimensions, but since the unit vectors don't change over time (aka. they are constants), we treat them like we would any other constant for derivatives and integrals. The resulting velocity and acceleration equations are as follows.

Velocity:	\[v(t)=\dot{x}(t)\hat{i}+\dot{y}(t)\hat{j}\]
Acceleration:	\[a(t)=\ddot{x}(t)\hat{i}+\ddot{y}(t)\hat{j}\]

The above equations are vector equations with velocities and accelerations broken down into x and y components. Since the x an y directions are perpendicular, they are also independent (movement in the x direction doesn't impact movement in the y direction and vice versa). This essentially means we can split our vector equation into a set of two scalar equations. To do this we just put everything in front of the i unit vectors in the x equations and everything in front of the j unit vectors in the y equations.

Position:	\[x(t)=...\]	\[y(t)=...\]
Velocity:	\[\dot{x}(t)=...\]	\[\dot{y}(t)=...\]
Acceleration:	\[\ddot{x}(t)=...\]	\[\ddot{y}(t)=...\]

Once we have the everything split into x and y directions we can just use the same processes we used for one dimensional motion to move from x, to x dot, to x double dot and from y, to y dot, to y double dot. The variable linking the two equations is time t, or time in both the x and y equations.

A motorcycle drives off a one meter tall ramp at an angle of 30 degrees as shown below. Determine the equations for the acceleration, velocity, and position over time. How far does the motorcycle travel in the x direction before hitting the ground?

A basketball is thrown towards a hoop that is three feet higher in the y direction and 25 feet away in the x direction. If the ball is thrown at an initial angle of 50 degrees, what must the initial velocity be for the ball to make it into the hoop?

A tennis ball is launched at an angled surface as shown below such that it will bounce up after impact. The tennis ball has an initial velocity of 30 m/s at a 20-degree angle and is launched 20 meters from the base of the surface, which itself is angled at 45 degrees. Determine the coordinates where the tennis ball is expected to impact the surface.

You are designing an artillery system with the aim of hitting a target 20 km away with two shells at the same time.

The first shot will be fired at a 50-degree angle. What is the required initial velocity needed to hit the target? Assume air resistance is negligible.
The second shot will be fired five seconds after the first. What is the required initial velocity and angle to hit the target at the same time as the first shot?

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The normal-tangential coordinate system centers on the body in motion. The origin point will be the body itself, meaning that the position of the particle in the n-t coordinate system is always "zero". The tangential direction (t-direction) is defined as the direction of travel at that moment in time (the direction of the current velocity vector), with the normal direction (n-direction) being 90 degrees counterclockwise from the t-direction. The diagram below shows a particle following a curved path with the current normal and tangential directions.

In the normal-tangential coordinate system the particle itself serves as the origin point. The t-direction is the current direction of travel and the n-direction is always 90 degrees counter-clockwise from the t-direction. The u-t and u-n vectors represent unit vectors in the t and n directions respectively.

Normal-tangential coordinate systems work best when we are observing motion from the perspective of the body in motion, such as being a passenger in a car or plane. In such cases, we would define ourself as the origin point and "forward" would be the tangential direction. An important distinction between the rectangular coordinate system and the normal-tangential coordinate system is that the axes are not fixed in the normal-tangential coordinate system. If we go back to the car example, the" forward" or tangential direction will turn with the car, but the "east" direction or the x-direction will remain constant no matter which way the car is pointed.

The way the coordinate system is defined, the position of the particle is always set to be at the origin point. The velocity is also always set to be in the tangential direction, and thus there is no velocity in the n-direction. The variable "v"is the body's current speed.

Position:	\[r_{p/o}(t)=0\hat{u}_{t}+0\hat{u}_{n}\]
Velocity:	\[v(t)=v\hat{u}_{t}\]

To find the acceleration, we need to take the derivative of the velocity function. This may seem simple, but there is a new thing to consider in that the ut unit vector is not constant. The means a change in speed can cause an acceleration or a change in direction (which would change the ut direction) can cause an acceleration. Going back to our car example this makes some intuitive sense. We can feel accelerations, and we would be able to feel acceleration if we suddenly stepped on the gas and increased our speed, but we would also be able to feel the acceleration if we took a tight turn at a constant speed.

Going back to the derivative, we will use the product rule, taking the derivative of one piece at a time.

\[a(t)=\dot{v}\hat{u}_{t}+v\dot{\hat{u}_{t}}\]

V dot is the rate of change of speed of the body, which is called the tangential acceleration. Going back to our car analogy this is the acceleration we would experience from pressing the gas or brake pedals.

The other piece of our derivative is the speed times the derivative of a unit vector, which we will need to analyze further. When thinking about the derivative of a rotating unit vector, we think about rotating the coordinate system by a small amount d theta.

The derivative of a rotating unit vector can be thought of as the change in position of the head of that vector as it rotates about a small angle d theta.

The derivative of the u-t vector is then the change in position of the head of the vector. Using some geometry, we can see that the distance the head of the vector moves is the length of the vector (which is always 1 for a unit vector) times the angle of rotation in radians. The direction the head of the u-t vector travels is roughly the u-n direction. In fact as d theta approaches zero, it becomes exactly the u-n direction. Putting this back into our derivative we wind up with the following equation for acceleration.

\[a(t)=\dot{v}\hat{u}_{t}+v\dot{\hat{u}_{t}}=\dot{v}\hat{u}_{t}+v\dot{\Theta}\hat{u}_{n}\]

Before we arrive at our final set of equations, we have one last potential substitution. If a particle is moving along a curved path, the rate at which it is turning (theta dot) will be equal to the velocity of the particle divided by the radius of the path at that point (v divided by rho). Putting this last substitution in, we have our final set of equations with two equivalent options for calculating the accelerations.

\[a(t)=\dot{v}\hat{u}_{t}+v\dot{\Theta}\hat{u}_{n}=\dot{v}\hat{u}_{t}+\frac{v^{2}}{\rho}\hat{u}_{n}\]

When using these equations it is important to remember that they are acceleration equations. If we want to know the overall acceleration we would need to add the two acceleration components as vectors. Also if we are given an acceleration that is not in the normal or tangential direction, we will first need to break that acceleration vector down into normal and tangential components before using the above equations. Finally, if we want the velocity or acceleration in directions other than the normal and tangential directions, we will need to use a coordinate transformation.

A commercial jetliner is traveling a constant 250 m/s when it executes an emergency 180 degree turn. If the turn takes 20 seconds, what is the acceleration experienced by the passengers? What is the radius of the curve taken by the plane?

Assuming a cloverleaf interchange has a radius of curvature of 80 meters at the tightest part of the turn, what is the fastest a car could travel around this curve without experiencing more than 1/2 a g in acceleration? Assume the car is traveling at a constant speed. If the car was instead increasing speed at a rate of 2 m/s2, what would be the new overall magnitude of the acceleration experienced by the passengers?

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The polar coordinate system uses a distance (r) and an angle (theta) to locate a particle in space. The origin point will be a fixed point in space, but the r-axis of the coordinate system will rotate so that it is always pointed towards the body in the system. The variable "r" is also used to indicate the distance from the origin point to the particle. The theta axis will then be 90 degrees counter clockwise from the r-axis with the variable "theta" being used to show the angle between the r-axis and some fixed axis that does not rotate.. The diagram below shows a particle with a polar coordinate system.

In the polar coordinate system the r direction always points from the origin point to the body. The variable "r" is also used to indicate the distance between the two points. The theta direction will always be 90 degrees counter-clockwise from the r direction. Theta is also used to indicate the angle between the r direction and some fixed axis used for reference. The u-r and u-theta vectors represent unit vectors in the r and theta directions respectively.

Polar coordinate systems work best in systems where a body is being tracked via a distance and an angle such as a radar system tracking a plane. In cases such as this, the raw data from this in the form of an angle and a distance would be direct measures of theta and r respectively. Polar coordinate systems will also serve as the base for extended body motion, where motors and actuators can directly control things like r and theta.

The way the coordinate system is defined, the r-axis will always point from the origin point to the body. The distance from the origin to the point is defined as "r" with no component of the position being in the theta direction.

Position:	\[r_{p/o}(t)=r\hat{u}_{r}+0\hat{u}_{\Theta}\]

To find the velocity, we need to take the derivative of the position function over time. Since the distance r can change over time as well as the direction r changing over time to track the body, we need to worry about the derivative of r as well as the derivative of the unit vector. Like we did with the normal-tangential systems, we will use the product rule and then substitute in a value for the derivative of the unit vector.

Velocity:	\[v(t)=\dot{r}\hat{u}_{r}+r\dot{\hat{u}_{r}}=\dot{r}\hat{u}_{r}+r\dot{\Theta}\hat{u}_{\Theta}\]

To find the acceleration, we need to take the derivative of the velocity function. As all of these terms, including the unit vectors, change over time, we will need to use the product rule extensively. The u-r term will split into two terms, and the u-theta term will split into three terms.

Acceleration:	\[a(t)=\ddot{r}\hat{u}_{r}+\dot{r}\dot{\hat{u}_{r}}+\dot{r}\dot{\Theta}\hat{u}_{\Theta}+r\ddot{\Theta}\hat{u}_{\Theta}+r\dot{\Theta}\dot{\hat{u}_{\Theta}}\]

Again we will need to substitute in values for the derivatives of the unit vectors similar to before, but it is worth mentioning that the derivative of the u-theta vector as it rotates counter clockwise is in the negative u-r direction.

The derivatives of the u-r and u-theta unit vectors. Notice that the derivative of the u-theta vector is in the negative u-r direction.

After substituting in the derivatives of the unit vectors and simplifying the function we arrive at our final equation for the acceleration.

Acceleration:	\[a(t)=\ddot{r}\hat{u}_{r}+\dot{r}\dot{\Theta}\hat{u}_{\Theta}+\dot{r}\dot{\Theta}\hat{u}_{\Theta}+r\ddot{\Theta}\hat{u}_{\Theta}-r\dot{\Theta}^{2}\hat{u}_{r}\]
	\[=(\ddot{r}-r\dot{\Theta}^{2})\hat{u}_{r}+(2\dot{r}\dot{\Theta}+r\ddot{\Theta})\hat{u}_{\Theta}\]

Though this final equation has a number of terms, it is still just two components in vector form. Just as with the normal-tangential coordinate system, we will need to remember that we will need to split the single vector equation into two separate scalar equations. In this case we will have the equation for the terms in the r direction and the equation for the terms in the theta direction.

A radar tracking station gives the following raw data to a user at a given point in time. Based on this data, what is the current velocity and acceleration in the r and theta directions? What is the current velocity and acceleration in the x and y directions?

A spotlight is tracking an actor as he moves across the stage. If the actor is moving with a constant velocity as shown below, what values do we need for the spotlight angular velocity (theta dot) and spotlight angular acceleration (theta double dot) so that the spotlight remains fixed on the actor?

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Dependent motion analysis is used when two or more particles have motions that are in some way connected to one another. The way in which the motion of these particles is connected is known as the constraint. A simple example of a constrained system is shown in the image below. Imagine a pickup truck gets stuck in the sand, and a friend uses a rope to help pull them out. This friend ties one end of the rope to the rear bumper of her car, loops the rope around a bar on the front of the pickup truck, then ties the other end to a stationary tree. In this case the two vehicles will not have the same velocity or acceleration, but their motions are related because they are tied together by the rope. In this case, the rope is acting as the constraint, allowing us to know the velocity or acceleration of one vehicle based on the velocity or acceleration of the other vehicle.

This represents a constrained system. The motion of the car and the pickup truck will be related to one another via rope that is connecting them.

The first thing we will need to do when analyzing these systems is to come up with what is known as the constraint equation. A constraint equation will be some geometric relationship that will remain true over the course of the motion. In the above example, imagine the rope is 50 ft long. Using the tree as a stationary point, we can also say that the length of the rope is the distance from the green car to the tree, plus two times the distance from the pickup truck to the tree (since it must go out to the pickup truck and then back). If we put this into an equation (the constraint equation) we would have the following.

Constraint Equation: Positions	\[L= 50 ft = 2 L_{A}+L_{B}\]

Once we have a constraint equation that works for positions, we can take the derivative of this equation to find another constraint equation that relates velocities. In this case, the length of the rope is constant, and therefore the derivative of the length will be zero. If we take the derivative of the constraint equation again, we wind up with a third constraint equation that relates accelerations.

Constraint Equation: Velocities	\[\dot{L}=0=2 \dot{L}_{A}+\dot{L}_{B}\]
Constraint Equation: Accelerations	\[\ddot{L}=0=2 \ddot{L}_{A}+\ddot{L}_{B}\]

In these equations, it is important to remember that the values represent the changes in length, rather than direct measures of velocities. Though both vehicles would have positive velocities in the example above (velocities to the right), one L dot value will be positive and one will be negative. This is because the truck is getting closer to the tree while the green car is getting further away. A similar situation will occur for the accelerations, where both vehicles would have positive accelerations even if the L double dot values are a mix of positive and negative values.

A truck becomes stuck in the sand at a local beach. To help, a friend takes a 50 ft long rope, ties one end to her car, loops the rope around a bar at the front of the truck, and then ties the other end to a stationary tree as shown below. If the car accelerates at a rate of .2 ft/s2, what will the velocity of the truck be by the time it gets to the tree?

A man has hooked up a pulley, a rope, and a platform as shown below to lift loads up onto a nearby rooftop. If x is currently 15 meters, y is currently 5 meters, and the man is walking away from the building at a rate of .5 meters per second, what is the current velocity of the platform?