Lab 06 - Forces and Friction_updated (1)

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PHYS 1: Spring 2023 Spring 2023 Lab 6 The Ramp: Forces and Motion Simulation

PHYS 1: Spring 2023 Spring 2023 Instructions : Familiarize yourself with “ The Ramp: Forces and Motion Simulation ” using different scenarios as shown in the figures on the above page. Click a tab at the top to view a scene. For example, click the Introduction Tab , to view the first figure above. In that figure, you can select the object from the drop-down menu at the center bottom. You can also drag the box or insert the value in the space provided for the applied force to move the box. If you want to see the effect of friction and mass of an object on the force applied click the next tab, friction . You can also “ change gravity ” using the sliding bar just below the object’s mass. If you want to observe graphs that represent what you are doing on the box, check the force graphs tab. In doing all of these, you need to understand the controls to the right of the simulations. These controls help you where you can hide or show the free body diagram, to use the ice or wood surface , display force vectors and/or sum of forces, and more controls to specify the object’s position, and define the angle of the ramp . Once you understand how to operate the simulation, you can predict the results based on the theoretical background (for example, the basics explained in the introduction below) and verify these using the simulation. Introduction: When an object is dragged across a horizontal (or inclined) surface, the force of friction that must be overcome by any parallel (to the surface) force ( F l l ) depends on the normal force ( Fn ) and the coefficient of friction (μ). Depending on the situation the normal force can be expressed in different ways. For example, if the force applied is parallel to the surface as shown below, the normal force is equal to the weight in the case where the object is on the horizontal surface, and the normal force is equal to the weight times the cosine of the angle of the inclined in the case of the inclined surface. Note that the friction changes as the normal force expression changes. If the box moves with constant velocity, the net force is zero. Note the general equation and the subsequent two equations in relation to the state of motion described and the figure above. F f = μF n (1) Horizontal surface: F f = μmg (2) Inclined surface: F f = μmgcosθ (3) mgsin F n F n F f F app F app mg mg mgcos mgsin F f

PHYS 1: Spring 2023 Spring 2023 From the above discussion, it is clear that a force equal to or greater than the static frictional force is required to move a stationary object. When an object is placed in an inclined plane, the object will start sliding once the plane is raised to a particular angle of inclination. This angle is called the critical angle. Let us consider a block resting on a plane that is inclined at an angle θ as shown in Figure 2. The weight of the block which acts vertically downwards can be resolved into normal ( ࠵?࠵? cosθ ) and tangential ( ࠵?࠵? sinθ ) components. In the absence of friction, the block would slide down the inclined plane even for a small angle of inclination due to the tangential force. However, if the block does not move, there must exist a frictional force directed opposite to the tangential force to hold it in place. If we were to raise the angle of the inclined plane, there would exist an angle at which the block will just begin to slide downward. At this point, the tangential force is just slightly greater than the frictional force (in fact, we assume them to be equal), and it is at this point where we will be able to determine the value of maximum frictional force acting between the two objects as follows: ࠵? ! ,#$% = ࠵?࠵? sinθ ………. (4) The Normal force and the maximum frictional force are related as follows: ࠵? !,#$% = ࠵? ! ࠵? & = ࠵? ! ࠵?࠵? cosθ ……. (5) Equating equation (3) and (4) gives, ࠵? ! = tanθ ……. (6) This is one of the methods we can use to determine the coefficient of static friction. Procedure: Open Ramp: Force and Motion https://phet.colorado.edu/sims/cheerpj/motion-series/latest/motion- series.html?simulation=ramp-forces-and-motion 1. Click the Introduction tab. If you have been playing with simulation before, click on “Reset All”. Figure 2: Object in an inclined plane

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PHYS 1: Spring 2023 Spring 2023 a. Select the small crate from the drop-down option. b. Change the values of the ramp angle to zero so that the surface is horizontal c. Adjust the object position to be at 6.0m and click on the ‘ Play ’ button. d. Now, slowly increase the ramp angle until you reach an angle when the small crate is about to move. Angle = 28.1 degree [2 Points] e. Keeping the same angle, what minimum applied force do you need to move the crate along the horizontal surface? (Input a value of Applied Force and find a number where the cart just starts to move). Applied force while on Horizontal Surface = 500 N [2 Points] If you did it correctly, you will notice the cart will stop after going a little higher on the ramp! Now keep on increasing the value of Applied Force such that the Cart is able to move up the inclined plane. Applied force while on Inclined Plane = 600 N [2 Points] The difference between these two forces = 100 N [2 Points] Why do you think we need this additional force to move this cart up the ramp? [2 Points] As the incline increases, gravity does not increase but the component parallel to the angle will – therefore more force would be needed. How does the direction of frictional force differ when the cart was moving down from the inclined plane versus when it is being pushed up and why? [2 Points] The frictional force acts in the opposite direction of the applied force, therefore as an object is pushed upward, the friction force will essentially act downward.

PHYS 1: Spring 2023 Spring 2023 f. Draw the free body diagram while the cart was on the horizontal floor as well as on the inclined plane in the space below. Do not forget to label your Forces . (You can click on Insert ° Shapes in Microsoft Word and choose an appropriate shape to draw the diagram). [3 + 3 = 6 Points] g. Switch the small crate with the File cabinet and repeat steps ‘b’ through ‘ e ’. compare the angle and the applied forces you found for the small crate. Which quantity didn’t change its value? If so why? [4 Points] Angle: 28.1 degree Applied force (horizontal surface): 250 N Applied force (inclined surface): 500 N The angle did not change as the components associated with the force of gravity will not change. h. Now, change the object to a mysterious object and try to find the angle of the ramp at which the mystery object is about to move. Angle = 13.5 degree [2 Points]

PHYS 1: Spring 2023 Spring 2023 i. What is the mass of the mysterious object? Show your calculation. (Hint: Find the applied force when the cart just begins to move like you did in step ‘e’ and use the concept of the Free body diagram). [4 Points] 14.29 kg j. What will be the minimum applied force needed for the mysterious object to move along the horizontal plane? [2 Points] You would not be able to calculate the minimum force needed as there was no friction constant given. Using the simulation I found that it is 450 N

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PHYS 1: Spring 2023 Spring 2023 k. Check your answers for steps ‘d‘ theoretically by using the force balance concept while the object was in the ramp. [4 Points] Theoretically, all force values make sense. 2. Click the friction tab a. Adjust the static and kinetic coefficients of frictions to 0.4 and 0.3 respectively for a 100 kg object located on Earth. b. Let the ramp angle be 30 degrees. c. If the object starts sliding down the inclined plane from the 6.0 m position, how far can it travel along the horizontal plane? Ä x = 4.8m[2 Points] d. If you want the object to cover 1.0 m on a horizontal plane after it slides down the inclined plane from the 8.0 m position, you need to adjust the coefficient of friction or the angle of the ramp. i. Keeping the angle at 30 degrees, find the static and kinetic coefficients of friction for which Ä x = 1m. [2 Points] Static = 0.6 Kinetic = 0.5 ii. Keeping the static and kinetic coefficients of friction to 0.4 and 0.3, find the ramp angle for which Ä x=1m. [2 Points] Angle: approx. 22 degrees e. Repeat step a-c with position = 8 m. How far does it travel along the horizontal plane? Why is this number different than when released from 6 m? [4 Points] The crate traveled 6.5 meters on the horizontal plane, this is a higher value of the y component of the force of gravity, meaning it will have a higher acceleration.

PHYS 1: Spring 2023 Spring 2023 3. Click the Force graphs tab a. Check the F Friction , F Gravity, and F Sum boxes b. Using steps, a, b, and c of procedure 1, and the ramp angle found in step d run the simulation. (If the object does not slide, add 0.5 degrees to the angle to make it move.) c. What are the values of forces you read from the graphs [5 Points] i. On inclined plane: F Friction = 259.345 N, F Gravity = -461.592 N, F Sum = -202.246N ii. On a horizontal surface : F Friction = 294 N, F Gravity = 0 N, F Sum = 294 N d. Take a screenshot of the graph obtained and paste it below. Based on that graph explain how F Friction , F Gravity, and F Sum varied with time, and why? [5 Points] The values varied with time as the object was accelerating down the incline and eventually reached the horizontal plane. e. Using distances along the inclined plane and horizontal planes and the respective times from the graph, calculate the average speeds of the crate. Show your calculation to get full credit. [6 Points]

PHYS 1: Spring 2023 Spring 2023 1. on the inclined plane : v avg = -2.22 m/s 2. on the horizontal surface : v avg = -1.95 m/s f. Adjust the ramp angle to 30 degrees, select a sleepy dog, and run the simulation with wood and then with ice using the friction control. Using the time information from the graph, and acceleration along the inclined plane, find the speed at the bottom of the inclined plane. Show your calculation. [5 Points] Starting with the dog from 6m i. Wood: v=-1.40 m/s, Ice: v= -4.0 m/s Follow-up Questions: 1. As the angle of the ramp is increased, the normal force increases / decreases / remains the same and the friction force increases /decreases / remains the same. [1 Point] 2. As the angle of the ramp is increased, the force parallel increases /decreases / remains the same. [1 Point] 3. The angle at which the force down the plane was equal to the force of friction (for the cabinet) was tan^-1 (u) = 11.3 degrees [1 Point]

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PHYS 1: Spring 2023 Spring 2023 4. Consider a very low (~ zero) friction, 5.0 kg skateboard on a ramp at an angle of 15 o to the horizontal. What would be the net force that would cause acceleration when the skateboard is allowed to move? Show your work. [4 Points] 12.68 N F = mg x sin(angle) so F = 5(9.8) x (sin(15)) = 12.68N 5. What would the skateboard's acceleration be down the plane? Show your calculation. [2 Points] Force divided by mass = a so 12.68/5 = 2.54 m/s^2 6. Now consider the same no-friction 5.0 kg skateboard on the same 15 o ramp. If a 45kg teenager jumps on, what would be his/her acceleration down the ramp? Show your calculation. [2 Points] Same calculations as before however now you add the masses to equal 50 kg 50(9.8) x sin(15) = 126.82 N 126.82/50 = a = 2.54 m/s 7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10 o ramp. If the coefficient of friction is 0.02 , what is the friction force you must overcome to push the cart up the ramp? Show your calculation. [2 Points] Mass = 15 + 25 = 40 kg Mg = (40)(9.8) = 392N times cos(10) = 386.044 then x by friction 386.044(.02) = 7.72 N 8. Realizing that there is also a force parallel (as a component of weight) you must ALSO overcome, what is the TOTAL force you must apply to push the cart up the ramp at a constant speed? Show your work. [2 Points] Mg x sin(10) = 392sin(10) = 68.07 + force calculated above (7.72) = 75.79N

PHYS 1: Spring 2023 Spring 2023

Lab 06 - Forces and Friction_updated (1)

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