Torque & Equilibrium Lab: A study of Simple Machines

__________________________________________________________________________________________________ Prof. Steven Strauss, MS, PE The Torque & Equilibrium Lab PHY-110-L Week 8

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Name: ___________________________ Grade: ____ of 50

Torque & Equilibrium Lab: A study of Simple Machines

Simulator: Torque & Equilibrium

• This is a simulation that provides an investigation Into Static Equilibrium. It provides a guided inquiry into rotational equilibrium and foundations for discussion of torque.

Learning Goals: You will use a simulation to:

• Develop an interactive experience exploring the concept of Torque and Static Equilibrium.

• Develop an understanding of how a simple machine- the lever- works to distribute forces.

Background A body is truly in equilibrium when it has no tendency to turn or move. This means no translation and no rotation. There are two conditions of equilibrium:

1. Net Force = 0 ∑Fi= 0 • Sum of forces on left = sum of forces on right • Sum of forces upwards = sum of forces downwards

2. Net Torque = 0 ∑τ i =0

• Sum of clockwise torques = sum of counterclockwise torque Pre-Lab Discussion: Find the definition of the following terms using your textbook or computer. Write these definitions in the provided space(s) below. (5 points)

Force: ___________________________________________________________________________

Torque: __________________________________________________________________________

Lever Arm: _______________________________________________________________________

Fulcrum: _________________________________________________________________________

Equilibrium: ______________________________________________________________________

https://phet.colorado.edu/sims/html/balancing-act/latest/balancing-act_en.html


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Part I Investigation: Achieving Static Equilibrium by balancing the board at the pivot point. Procedure – Go to URL here. We will use this Simulation environment to complete today’s activities. Initially select the “Intro” pane. To the right is an initial screen shot of the website.

As shown in the figure to the right:

• Check to enable all of the boxes in the Show window • In the Position window enable the Rulers radial button

Note: many students find it is easier to put on objects while the plank has its supports enabled.

To disable the supports toggle the switch from left to right

Experiment 1A: Observe the following situations below and predict what will be needed to bring all objects into Equilibrium. Test your prediction and record your actual results. Note: no prediction can ever be marked wrong.

Scenario A (2 points): Place a 5kg mass on the left side of the Fulcrum/pivot point at the 1 meter point. Assess the required position of another 5kg object placed on the right side of the Fulcrum to achieve Static Equilibrium by calculating the applied Force and the applied Torque of this object in the table below.

Note: use g = 9.81 m/sec2 ; the acceleration due to gravity Observations and calculations- Scenario A

Left of the Pivot Point Right of the Pivot Point Mass m

(kg) Force

F=mg (N) Distance from

Pivot d (m) Torque τ=Fd

Mass m (kg)

Force F=mg (N)

Distance from Pivot d (m)

Torque τ=Fd

5 _____ 1 _____ 5 _____ _____ _____ a) Hypothesis: ______________ Meters on the right side b) What is required to achieve Static Equilibrium? ______________________________

c) Actual Results: _________ Meters on the right side

https://phet.colorado.edu/sims/html/balancing-act/latest/balancing-act_en.html


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Scenario B (2 points): Same as Scenario A but this time the heavier 10kg Trash can is used on the right hand side. How does this change the placement of the right-hand side object? 5kg mass on the left side of the Fulcrum/pivot point at the 1 meter point. Assess the required position for the 10kg Trash Can as it is placed on the right side of the Fulcrum relative to achieving Static Equilibrium. Calculating the applied Force and the applied Torque of this object in the table below.

Note: use g = 9.81 m/sec2; the acceleration due to gravity Observations and calculations-Scenario B


(kg) Force



Mass m (kg)

Force F=mg (N)


Torque τ=Fd

5 _____ 1 _____ 10 _____ _____ _____ a) Hypothesis: ______________ Meters on the right side

b) Actual Results: _________ Meters on the right side

Scenario C (5 points): Place a 5kg mass on the left side of the Fulcrum/pivot point at the 1 meter point. A second 5kg object is also placed on the left side of the Fulcrum at the 2 meter point. Assess the required position for the 10kg Trash Can as it is placed on the right side of the Fulcrum relative to achieving Static Equilibrium. Calculating the applied Force and the applied Torque of this object in the table below.

Note: use g = 9.81 m/sec2; the acceleration due to gravity Observations and calculations-Scenario C


(kg) Force



Mass m (kg)

Force F=mg (N)


Torque τ=Fd

5 _____ 1 _____ 10 _____ _____ _____ 5 _____ 2 _____

Total _____ _____ a) Hypothesis: ______________ Meters on the right side



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Scenario D (5 points): Place a 5kg mass on the left side of the Fulcrum/pivot point at the 0.75 meter point. A 10kg object is also placed on the left side of the Fulcrum at the 0.5 meter point. Assess the required position for a 5kg object as it is placed on the right side of the Fulcrum relative to achieving Static Equilibrium. Calculating the applied Force and the applied Torque of this object in the table below.

Note: use g = 9.81 m/sec2; the acceleration due to gravity Observations and calculations-Scenario D


(kg) Force



Mass m (kg)

Force F=mg (N)


Torque τ=Fd

5 _____ 0.75 _____ 5 _____ _____ _____ 10 _____ 0.5 _____

Total _____ _____ a) Hypothesis: ______________ Meters on the right side


Scenario E (5 points): Place the 10kg mass on the left side of the Fulcrum/pivot point at the 1.25 meter point. A 5kg object needs to be also placed on the left side of the Fulcrum at a yet to be determined point. If a single 5kg object is placed on the right side of the Fulcrum at the 2 meter point assess where to place the 5kg object on the left side- or if it is even possible to achieve to do so- to achieve Static Equilibrium. Calculate the applied Forces and the applied Torques for each side in the table below.

Note: use g = 9.81 m/sec2 ; the acceleration due to gravity Use x for the unknown variable in the table below

Observations and calculations-Scenario E


(kg) Force



Mass m (kg)

Force F=mg (N)


Torque τ=Fd

10 _____ 1.25 _____ 5 _____ 2 _____ 5 _____ tbd _____

Total _____ _____


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a) Can static Equilibrium be achieved here? [Y/N] ______ b) Why or why not? _______________________________________________

___________________________________________________________________ Experiment 1B (5 points) Select the “Balancing Lab” pane. To the right is an initial screen shot of the website.


• Check to enable all of the boxes in the Show window • In the Position windown enable the Rulers radial button

Use the table below to place weights on the right-side of the Fulcrum. Calculate the Torque on the right side given the weights and positioning of the weights. Assess where you will need to place a 20kg object on the left-hand side of the Fulcrum to bring the system into Equilibrium. Calculate your answer first and then test your calculation with the simulator. Observations and calculations-Experiment 1B(a)


(kg) Force



Mass m (kg)

Force F=mg (N)


Torque τ=Fd

20 _____ _____ _____ 5 _____ 0.5 _____ 10 _____ 0.75 _____ 10 _____ 1.0 _____ 10 _____ 1.5 _____

Total _____ _____

a) What is the Calculated Torque of the right side? ________________________ b) What is the required Torque of the left side to achieve Equilibrium? _________ c) Where does the 20kg object on the left side need to be placed? ___________

Meters


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Experiment 1C (5 points) Stay in the “Balancing Lab” pane. Hit the reset simulation button As shown in the figure to the right:

• Check to enable all of the boxes in the Show window • In the Position window enable the Rulers radial button

In the Bricks box hit the right arrow until you get to the last pane that has “Mystery Objects”: E, F, G, H.

The object of this Experiment is to assess what the Unknown Mass of the Mystery Objects: F and G are. To do this you will need to initially place Object F on the Plank (let’s say on the left side…) at any distance you want (but I recommend placing it at position =1 meter) and then cycle back to the first pane (Bricks) and put the Bricks of various weights and positions on the right side allowing you to establish the equilibrium point and assess the Mass of the Mystery Object under study. Record your results below. Repeat for Object G. Mystery Object F:

To establish Equilibrium: ∑F right = ______________ N ∑τ right = ______________ Mass of Mystery Item F = ________ kg

Mystery Object G:

To establish Equilibrium: ∑F right = ______________ N ∑τ right = ______________ Mass of Mystery Item G = ________ kg


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Pulling it all together (2 points)

1. Now imagine a different set up…

a) What would the scale read if the plank is in static equilibrium? The 15 kg mass is

located at 3 m away from the pivot and the scale is 4m away. Scale reading is: ____________ kg.

b) What would be your answer to the scenario in the previous problem if the mass on the plank is 5 kg? ____________ kg

Part II (4 points) Investigation: Pulleys & their effect on required Force Reduction . Background The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system.

𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝒂𝒂 𝑨𝑨𝑨𝑨𝑨𝑨𝑴𝑴𝑴𝑴𝑨𝑨𝑴𝑴𝑨𝑨𝑴𝑴 = 𝑳𝑳𝑳𝑳𝑴𝑴𝑨𝑨 𝑭𝑭𝑳𝑳𝑭𝑭𝑴𝑴𝑴𝑴𝑬𝑬𝑬𝑬𝑬𝑬𝑳𝑳𝑭𝑭𝑨𝑨 𝑭𝑭𝑳𝑳𝑭𝑭𝑴𝑴𝑴𝑴

The work (W) done by a constant force of magnitude F on a point which undergoes a displacement (s) in a straight line in the direction of the force is:

𝑾𝑾𝑳𝑳𝑭𝑭𝑾𝑾 = 𝑭𝑭𝑳𝑳𝑭𝑭𝑴𝑴𝑴𝑴 ∗ 𝑫𝑫𝑴𝑴𝑫𝑫𝑫𝑫𝒂𝒂𝑴𝑴𝑴𝑴𝑴𝑴𝑫𝑫𝑴𝑴𝑴𝑴𝑨𝑨 𝑳𝑳𝑬𝑬 𝑨𝑨𝑴𝑴𝑴𝑴 𝑶𝑶𝑶𝑶𝑶𝑶𝑴𝑴𝑴𝑴𝑨𝑨

A one-wheel pulley allows the direction of force to be changed. Pulling down 1m of rope raises the load by 1m.

The force required - and work done - remains the same. Therefore:


= 𝟏𝟏𝟏𝟏 = 1


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Why is this useful? Is it easier or harder to pull down or up (typically)? It is typically easier to pull down because your body weight aids in the process!

But using multiple pulleys in a system makes things even easier. In a 2-pulley system pulling down 1m of rope raises the load by 0.5m.

Therefore:


= 𝟏𝟏𝟎𝟎.𝟓𝟓

= 2

This means that the Effort Force (the one you exert on the system!) is now ½ of the Load- easier to lift that object!

Procedure – Go to URL here. To the right is an initial screen shot of the website. In general: If a pulley system is perfectly efficient the mechanical advantage is equal to the number of pulleys. This simulation allows you to better understand the effect of multiple pulleys being used in a system. For this section assume that the weight attached to the system is: 20N and that the weight of the loose Pulleys can be fully ignored (i.e. = 0). Assume a perfectly efficient pulley system. Fill in the following table:

Pulleys Required Force (N) MA 1 _____ _____ 2 _____ _____ 4 _____ _____ 6 _____ _____

Note: the single-pulley system is not an available option in this simulation. I’m not sure why, honestly…..

https://www.walter-fendt.de/html5/phen/pulleysystem_en.htm


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Part III (5 Points) Investigation: Incline Planes: To lift, Push, or Pull? That is the question…… Background We sometimes find ourselves in situations where the available forces we have at our disposal are not enough to get the job done. Simple machines are mechanical devices that are used to make work easier. In physics, work done is measured in joules (a unit of energy), and force is measured in Newtons. Energy is expended if the object is moved against a force.

When you pick up a box, the force of gravity is pulling down on the box, causing you to expend energy. When you push the box across a surface, the only force acting against its movement is friction (and negligible air resistance). If the surface is smooth, the friction will be minimal, and it won't take much energy to move the box. But if the surface is rough, the force of friction is greater, and you'll expend more energy trying to move it.

Picture a 1 kg block with sandpaper wrapped around it. Picking up the block would be very easy, because gravity is only exerting about 10 Newtons of force on it. But try to push the block sideways on a wood surface, and you’ll find it more difficult. That’s because sandpaper is a very rough surface, and therefore creates a lot of friction.

On the other hand, a 100 kg block of ice would be really hard to pick up, because gravity is exerting a force of almost 1000 Newtons on it. But let's say it's sitting on a frozen lake. Then the force of friction would be minimal because ice is slippery, and you could probably be able to push it easily!

Procedure Experiment III-A Russ is getting a refrigerator from his uncle. He grabs a friend of his to help him move it out of his uncle’s house. The frig, which is fully loaded, weighs 200 kg. Because Russ is lazy and thinks he and his buddy are strong enough to lift this object onto his pickup truck, they decide not to take anything out and therefore try to lift the entire load. Assume that Russ is capable of exerting 1000N of force and that his friend- who is smaller in stature- is only capable of exerting a force of 850N. If the height of the pickup is 1 meter, and using the constant g=9.81m/s2: a) how much force is required to lift the object onto the track? b) are these guys up to the challenge? c) How much work is actually done?

a) how much force is required to lift the object onto the track? _____________ N b) are these guys up to the challenge? [Y/N] _______ Why? ______________________ c) In a Physics sense how much work is actually done? __________________ J


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So, because Russ had PHY-110, he knows that his life can perhaps be made easier by leveraging a Simple Machine. So, he grabs a ramp that is 3 meters long to allow him to push the 200 kg object onto the truck without having to go into traction the next day! Russ sits down and looks to calculate the required force to push the object up the ramp using the inclined plane. The first thing he does is calculate the angle of the incline. He knows that the height of the ramp from the ground is 1 meter and that the ramp itself is 3 meters. After some quick Trig he comes to the conclusion that the angle is 19.47° – so he calls it 20°! He draws out the scenario because he realizes that it is not as simple as simply lifting an object any longer. Ummm, he says… I know that n is the “Normal Force” and this is the force perpendicular to the surface an object is sliding on and in general the force that opposes the Force due to gravity – in the Vertical direction! The normal force is the force pushing the two surfaces together, and the stronger the normal force, the stronger the force due to friction. He further assumes that the coefficient of Kinetic Friction is 0.2 and the value of the Static Friction is 0.5 (both are reasonable assumed values). So, what force or forces does he need to overcome? Well, initially the guys need to overcome the Static Friction Force as well as the Horizontal direction force due to gravity are the forces he needs to overcome. Why Static Friction? Because Static friction is what keeps the box from moving without being pushed, and it must be overcome with a sufficient opposing force before the box will move. Kinetic friction, on the other hand, - also referred to as dynamic friction- is the force that resists the relative movement of the surfaces once they're in motion. He starts by breaking down the m2g vector into its horizontal and vertical components.

m2g

Fy

Fx

20°


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𝑭𝑭𝑭𝑭 = 𝑫𝑫𝟐𝟐𝑨𝑨 𝐬𝐬𝐬𝐬𝐬𝐬𝜽𝜽 𝑭𝑭𝑭𝑭 = (200 ∗ 9.81 ∗ sin 20° ) = 671 𝑁𝑁

𝑭𝑭𝑭𝑭 = 𝑫𝑫𝟐𝟐𝑨𝑨 𝐜𝐜𝐜𝐜𝐬𝐬𝜽𝜽 𝑭𝑭𝑭𝑭 = (200 ∗ 9.81 ∗ cos 20° ) = 1844 𝑁𝑁

So, the Normal Force, n, is simply Fy (opposite sign). Therefore:

𝑭𝑭𝑴𝑴 = 𝑭𝑭𝑭𝑭 = 1844 𝑁𝑁 Russ, being the smart guy he is, know the forces he has to overcome are the Forces of Friction and the 𝑭𝑭𝑭𝑭 force. His pushing Force is the Green Vector arrow in the diagram above or the vector labeled “T” in the initial diagram! The Force to Friction is simply the Normal Force multiplied by the coefficient of Friction. Therefore, the Static Friction component is:

𝑭𝑭𝑬𝑬𝑫𝑫 = 𝑭𝑭𝑴𝑴 ∗ µ𝑫𝑫 = 1844 𝑁𝑁 ∗ 0.5 = 922𝑁𝑁

While the Dynamic/Kinetic Friction component is: 𝑭𝑭𝑬𝑬𝑾𝑾 = 𝑭𝑭𝑴𝑴 ∗ µ𝑾𝑾 = 1844 𝑁𝑁 ∗ 0.2 = 369𝑁𝑁

So, the needed force by Russ and his buddy to get the Frig initially moving up the incline is: (𝑭𝑭𝑬𝑬𝑫𝑫 + 𝑭𝑭𝑭𝑭) = (922 + 671) = 1593 N

a) are these guys now up to the challenge of getting that frig onto the truck? [Y/N] ____ Why or why not? ______________________

b) once they get the object moving, how much force is required to keep the frig moving?

_________ N In conclusion what can you assess about lifting a heavy object relative to pushing that same object with the aid of an incline? ___________________________________________________________

A note of clarification around the Friction Forces: In many cases the Static and Kinetic Friction Force coefficients are assumed to be the same. Kinetic friction (or dynamic friction) is the force that resists the relative movement of the surfaces once they're in motion. The static friction between two surfaces is always higher than the kinetic friction (at least, in practical, real-world applications).


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To see this scenario in action go to: https://phet.colorado.edu/sims/cheerpj/motion-series/latest/motion-series.html?simulation=ramp-forces-and-motion Initially select the “Force Graphs” pane. To the right is an initial screen shot of the website.


• Check to enable the Show box in the Free Body Diagram window

• Click the Sum of Forces Box in the Vector window

• Set the Object Position to 1.0 meters (simply enter the value)

• Set the Ramp Angle to 20 degrees • Choose the Refrigerator in the Choose an Object window.

o Note that the Frig has a mass of 200 kg, µk = 0.2, and µs = 0.5 • On the left side of the Simulation click to enable each box except for the Fwall box

This set-up very closely approximates the analysis we just did.

Apply an initial force of 1500N into the Forces window. After several seconds Pause the Simulation. Did the object start to move with an applied force of 1500N? [Y/N] _______. Why or why not? ________________________________________________________________.

Hint: review the calculations in the previous section! In the above analysis it was calculated what the needed force by Russ and his buddy was to get the Frig initially moving up the incline. If that value (what is it? ____________ N) is placed in the Forces window and the Simulator re-started does the object actually start moving? [Y/N] ___________ After several seconds again Pause the Simulation.

Take a look at the Graph and especially at the FFriction graph. Describe what happened to the Friction Force once the object was started in Motion. In your description make a statement around the required force to continue pushing the object up the incline once started.

_______________________________________________________________________________

_______________________________________________________________________________

https://phet.colorado.edu/sims/cheerpj/motion-series/latest/motion-series.html?simulation=ramp-forces-and-motion

https://phet.colorado.edu/sims/cheerpj/motion-series/latest/motion-series.html?simulation=ramp-forces-and-motion


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Experiment III-B Explore: Is pulling something easier than pushing? Let’s say we have a heavy box sitting there. We don’t want to lift it- we just want to take it to the other side of the room. We can either push it or pull it- but is one easier relative to the other? It sorta depends…. I know- not a great Physics answer…. Let’s assume you are trying to push this heavy object. Typically, you try to get leverage but nearly always push it near the top of the object. Fk is the Friction Force. As we know it is the Friction coefficient multiplied with the Normal Force (N). Here we don’t have an incline like we did previously, so the Normal force is Force due to gravity- or is it? Well, turns out it isn’t…. As you likely can imagine the Force you are pushing with is at an angle- so you need to re-do that Force in terms of its x- and y-components. Doing that we see that the Free Body Diagram for Pushing an object looks like the following figure:

OK- interesting…. What this shows is that when you are pushing this object there is one component of “Pushing Force” (Fy) that also adds to the weight of the object (mg). So, the Normal Force is:

𝑭𝑭𝑴𝑴 = 𝑫𝑫𝑨𝑨+ 𝑭𝑭𝑫𝑫𝑴𝑴𝑴𝑴𝜽𝜽 We know that Fk, the Friction Force, is the Friction coefficient multiplied with the Normal Force (N). And now let’s look at what happens when we pull something.

When we pull something the Force arrow is in the opposite direction as the Pushing case- makes sense- right? So that diagram looks like this now: So, because this again is assumed to be pulled from near the top of the object – and at an angle- we need to break the (pushing) force down into its x- and y-components. Doing that yields a Free Body Diagram for Pulling that looks like the following:

F

F


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Notice now that F*sinθ acts upwards now and in the opposite direction of the weight m*g vector- therefore the Pulling vector would work to decrease the Normal Force N.

𝑭𝑭𝑴𝑴 = 𝑫𝑫𝑨𝑨− 𝑭𝑭𝑫𝑫𝑴𝑴𝑴𝑴𝜽𝜽 Again, we know that Fk, the Friction Force, is the Friction coefficient multiplied with the Normal Force (N). Since the Normal Force (N) is smaller in the Pulling case than the Pushing case it stands to reason that the frictional force is reduced in the Pulling case relative to the Pushing case! Conclusion: it is easier to Pull than to Push! But…..

The answer really should be: Pulling is easier if you are taller than the object! Otherwise, pushing might be easier….. why? When you pull an object smaller than you, a part of the force we apply acts upward and reduces the weight of the object. But if you push, the vertical component of your force increases the weight and makes it tough(er). We just analyzed this scenario and showed that to be valid. However, when you push a taller (than you) object, you reduce its weight and find moving it easier. Again- it depends on where you apply the force! Remember that it’s hard to move a heavy object because of friction. The force of friction that trys to oppose your push or pull is proportional to the weight of the body! If we had a completely friction-less surface, then pushing and pulling won't make any difference!


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Pulling it all together (5 points) 1. The two diagrams below depict the free-body diagram for a 1000-kg roller coaster on the

first drop of two different roller coaster rides. Use the above principles of vector resolution to determine the net force of the roller coaster cars. Assume a negligible effect of friction and air resistance.

Mass = 1000 kg g = 9.81 m/s2 No Friction Analysis Diagram Fnorm (N) Fnet (N)

A _______ _______ B _______ _______

If the Force due to Friction is now no longer negligible, what would be the effects of that on Diagram A? Assume Coefficient of Friction is 0.2.

Diagram A (with Friction): Fnorm = _________ N Fnet = __________ N

2. Three lab partners - David, Robert, and Thomas - are discussing an incline problem (see

diagram below). They are debating the value of the normal force. David claims that the normal force is 250 N; Robert claims that the normal force is 433 N; while Thomas claims that the normal force is 500 N. While all three answers seem reasonable, only one is correct. Indicate which answer is correct? __________

a) David: 250N b) Robert: 433N c) Thomas: 500N d) None of them

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Torque & Equilibrium Lab: A study of Simple Machines