I need help with part C. I will provide the answers of Part A and a reference video of the model I made for part B. Just please tell me what I should say in the presentation with the relevant data. Thank you! Project -A car on a ramp.

Part A

Solving an inclined plane problem

Draw a free-body diagram

A car of mass, M₁=1000-kg car is on a rough ramp inclined at =20 above the horizontal. A light cord connects it over an ideal pulley at the top edge of the ramp to a hangingcrate9 mass M₂₎. The cable pulls on the car parallel to the surface of the ramp.

The coefficient of static friction between the Asphalt and the tires is 0.8, and the coefficient of kinetic friction is 0.7.

1. What should be the crate's mass to start pulling the car up the ramp? Draw a free-body diagram: Show all the forces (such as the force of gravity, friction, normal force, and force of tension in the cord) acting on the car.

• Show all the forces acting on the crate
• Resolve the force of gravity to its components.
• The free-body diagram should fill the whole page. Label all the force vectors.
• Write the formulas that apply to the problem on the second sheet of paper.
• These formulas should match the force vectors on your sketch.
• Derive a formula for the mass.
• Substitute the given data in the problem to calculate the mass of the crate.

Graphs and analyzing the graph:

2) Repeat the above calculation for the angles between the inclined ramp and the horizontal. Fill the table with your data

in degrees	20	30	40	50	60
Mass of the crate in kg

Use the excel program to draw the graph of mass vs. the angle. Label the two axes and the title of the chart.

Is the relationship linear or curvature?

At what angle mass of the crate is the most?

Describe the relationship between the angle and the mass of the crate.

Part B:

Building a 3-D model of the ramp in the problem.

Watch the video I showed a model of an inclined plane.

Build a model for the problem you solved in part A

Be creative and use recyclable items to build your model.

The Arrows for the force vectors should be movable but not loose.

The size of the arrows should match the rest of the hands.

Your model should be presentable with the correct data.

Part C:

Project Report: Show the solution to the problem as youexplainevery formula and calculation. Add the excel graph to your paper. You need to type the entire report, including the solution.

Record a 1 to 2 minutes video to present your presentable artifact. Explain the role of each vector as you used them in solving the problem.

Your presentation should be in casual or formal dress, not in pajamas.

You can use any social media so you and your partner can be present during the presentation.

M1 = 1000 kq Ms = 0.8 MK = 0.7 - constant velocity DV ( a = 0 ) FN - normal force X We - force of gravity (car ) FT D AFT Wer - force of gravity (crate) FT - tension force Ma A fx - kinetic friction V . When the car is at constant velocity, We V kinetic friction act on it. Wer . The surface of the incline should be the FREE - BODY DIAGRAM x-axis, perpendicular to it should be the y- axis .. The surface of the incline should be the FREE - BODY DIAGRAM X-axis, perpendicular to it should be the y- axis . CAR : Wex = Wc sine A FN We Way = We cos e Wex FT N X V Wey We = Mig We Wey = We cos e V Wex = Wcsine CRATE : . The force of gravity and tension force on the crate A should be along the y- axis . . The tension force acting on the car is equal to the X tension force acting on the crate. Wer Wer = Mag\fM1 = 1000 kq Ms = 0.8 MK = 0.7 - constant velocity DV ( a = 0 ) Fy - normal force X FH We - force of gravity (car ) FT D AFT Wer - force of gravity (crate ) O FT - tension force Ma 4 fx - kinetic friction V . When the car is at constant velocity, We V kinetic friction act on it. Wer FREE - BODY DIAGRAM . The surface of the incline should be the X-axis, perpendicular to it should be the y- axis .the y-axis. CAR : Wex = Wc sine A FN A Way = We cos e Wex FT X Way We = Mig WA- - Wey= We cos O V Wex = Wcsine CRATE : . The force of gravity and tension force on the crate A should be along the y- axis. . The tension force acting on the car is equal to the X tension force acting on the crate. Wer Wer = MagFORMULAS "UP Add all the forces along the x- and y-axis. CAR : Fx = FT + (- Wex) + (-fx) Fy = N + ( - Wey) total force along total force along X -axis y -axis Fx = FT - Wex - fx (1) Fy = N - Wey (2) Fx = Max , then (1) will become M, ax = FT - Wex - FK Since The car is not accelerating along the suface, then ax = 0, 0 = FT - WEx - fx F- - We sine - MX N = 0 FT - Migsine - UKN = 0 (3 ) Also, Fy = May , then (2) will become May = N - Weyay= O, the car has no motion along the y-axis, then O = N - Wcy N - WE COS 0 = 0 N - Migcose = 0 N = Migcose (4) Substitute (4) to (3), we have FT - Mig sing - MKN = 0 FT - Migsin G - MK Migcos = 0 (5 ) CRATE : Fy = Fr + (- Wer) ( 6 ) Fy = Mady , then (6) will become Maay = FT - Wer Also, the crate is not accelerating along the y-axis, then ay = 0, O = FT - Wer Fy - Mag = 0FT - Mag = 0 FT = Mag (7 ) substituting (7) to (5 ) , we have FT - Mig sing - MK Migcose = 0 Mag - Mig sin 9 - MK Migcose = 0 Mag = MK Migcose + Mig sing M2 7 MK M, cose + M, Sing Ma = Mi (Mkcose + sine ) (8 ) 20 : Ma = (1000 kg) [(0.7) cos 20 + sin 20' Ma = 999. 80 kq 30 . Ma = (1000 kg) (0.7) cos 30 + sin 30. ]30 : Ma = (1000 kg) (0.7 ) cos 80 + sin 30- ] Ma = 1106. 22 kg 40 : Ma = (1000 *g) [(0.7) cos 400 + sin 40' ] M2 = 1179. 02 kq 50 : Ma = (1000 *g) [(0.7) cos 500 + sin50' ] M2 = 1216.00 kg 60 : Ma = (1000 kg) [(0.7) cos 60' + sin60' Ma = 1216.03 kg