Lab 5

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Mechanical Engineering

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Dec 6, 2023

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Forces and Friction Name Simranpreet Cheema Lab Procedure – Answer the questions in red. Download and run the Java application “ramp-forces-and-motion_en.jar”. The Introduction screen will look like the screen capture below. 1. On the “Introduction” screen, you will be starting with a crate that has a mass of 100 kg and a coefficient of sliding friction of 0.3 and a coefficient of static friction of 0.5 Draw the Free Body Diagram (a picture showing the forces on the crate) before you apply any force. You do not need to turn the diagram in, just describe below. The crate should be stationary at -6.0m on the horizontal axis.

a) Describe the vector forces in your free body diagram. The vector forces in my diagram include vector force F(N) (natural force) pushing up on the crate, and vector force F(G) (gravity) pushing down on the crate. Make sure that the application is initially “paused”. Also, check the “record” button. If it is not paused, press the pause button, the clear button. Add 100N of applied force, and push the button and record what happens. b) Draw a free body diagram showing all the forces . Describe all of the vector forces on your diagram. The natural force and gravity forces are the same, but there is also frictional force pushing the crate to the left and applied force of 100 N pushing the crate to the right. 2. C alculate how much force would be required to get the crate moving from rest. a) Explain how you calculated this, and give your result. To get the crate to move from rest, the applied force must overcome frictional force. 100 kg = 980.665 N Static friction coef. * Weight in Newtons = 980.665 N * 0.5 = 490.332 N Now try it out by entering the force you calculated (or a tiny bit more, say 5%), and pressing the play button. Make sure you are paused with a clear screen first. What happened?

The crate moved slowly and came to a stop slightly up the ramp. b) What happened as the crate began to move up the ramp? Describe a free body diagram while the crate is on the ramp. What force is working against your applied force? As the crate moved up the ramp, it slowed down and eventually stopped. F(N) is pushing on the crate toward the left, f(a) is pushing on the crate toward the right, and gravity is pushing down on the crate. The force working against the applied force is gravity. c) If the crate were initially barely moving, calculate the MINIMUM force you will need to apply to get the crate up to the top of the ramp. Use a coefficient of friction of 0.3. Explain how you did the calculation and give your prediction. Minimum force needed = applied force exceeding both the force of gravity and kinetic friction. F(g) + F(f) = F(a) F(a)= m*g*Sin(theta) + (friction coef.)*Cos(theta) = 100*9.8*sin(30) + 0.3*cos(30) =744.6 N The crate should reach the top of the ramp if the applied force is greater than 744.6 N. 3. Select the “Friction” tab on the app. Reset and clear all. Make sure you are paused. The applied force should be set at zero. You can use the box on the right labeled “More Controls” to set the initial position of the crate at +8m, up near the top of the ramp. If the system is paused (play/pause button is showing forward arrow), the crate will remain in place. Use the slider bar to set the coefficient of kinetic friction to 1.0.

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a) Draw a free body diagram for the crate. Explain the vectors in your diagram. The forces included are F(N), F(g), and F(f). F(N) pushing on the crate towards the left, F(g) pushing down on the crate, and F(f) pushing up the crate towards the right. b) Press the play button. What happened? Why? The crate slid down the ramp and stayed near the man because gravity was greater than friction. c) Calculate the highest coefficient of static friction the will allow the crate to begin sliding down the ramp when it is released. Explain your calculation and the result you obtained. F= m*g*Sin(theta) – (friction coef.) *Cos(theta) 0=100*9.8*sin (30) – x*100*9.8*cos (30) X*848.7 = 490 X= 0.577 0.577 is the highest coefficient of static friction that will allow the crate to begin sliding down the ramp when released. Any value greater than 0.577 will keep the crate in place. d) Try out your prediction from (c) by using the application. What is your result?

A static friction of 0.5 allowed the crate to move, so my prediction was correct. 3. With everything paused and cleared, place the block near the top of the 30 o ramp (Position = 8.0 m). Set the coefficient of static friction back to 0.5. What is the net force on the block down the ramp? (you can use your free body diagram from 3 (a) above). a) What is the acceleration of the block down the ramp? Describe how you did your calculation. First calculate Fnet- Fnet= F(g) – F(f) =980*Sin(30) – 0.5*980*Cos(30) =65.6 N Acceleration – F=m*a 65.6= 100*a a= 0.656 m/s^2 b) What would be the final velocity of the block at the bottom of the ramp? Final velocity- V(final)^2 = V(initial)^2 + 2*acceleration*position V(final)^2= 0^2 + 2*0.656*8 =3.24 m/s c) What force is acting on the crate once it hits the flat part at the bottom of the ramp? What is the magnitude and direction of that force?

The force acting on the crate once it hits the flat part at the bottom of the ramp is friction. Magnitude- F(f)= u*m*g =0.5*9.8*100 =490N Direction- negative, towards the left d) What will be the acceleration of the crate on the flat part at the bottom of the ramp? Acceleration= F=m*a -490=100*a =-4.9 m/s^2 e) Calculate how far the crate should slide at the bottom. Explain how you did your calculation, including equations used. V(final)^2 = V(initial)^2 + 2*acceleration*position 0=3.24^2 + 2*-4.9*x -10.5= -9.8x X= 1.07 m f) Try it out. How did your calculation compare to the applet result? My calculation was not correct compared to the result. I’m not sure why.

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