1) A 1200 kg car pulls a 480 kg trailer. The car exerts a horizontal force of 35

Question

1) A 1200 kg car pulls a 480 kg trailer. The car exerts a horizontal force of 3540 N against the ground in order to accelerate. What force does the car exert on the trailer? Assume an effective friction coefficient of 0.190 for the trailer. 2) If the coefficient of kinetic friction between a 30.9 kg crate and the floor is 0.313, what horizontal force is required to move the crate at a steady speed across the floor? What horizontal force is required if ?k is zero 3)box of mass 12.7 kg is being pulled along a level floor as shown. The coefficient of kinetic friction between the floor and the box is 0.17. The person pulling the box is exerting a force of 37 Newtons at an angle of 27 degrees above the horizontal. Calculate the magnitude of the force of the floor pushing up on the box. Calculate the magnitude of the force of friction acting on the box. Calculate the magnitude of the acceleration of the box. (use m/s^2 for the units) 4) Drag-race tires in contact with an asphalt surface have a very high coefficient of static friction. Assuming a constant acceleration and no slipping of tires, calculate the coefficient of static friction needed for a drag racer to cover 1.14 km in 11.4 s, starting from rest. 5) A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.230 and the push imparts an initial speed of 4.30 m/s? 6) The carton shown in figure below lies on a smooth plane tilted at an angle ? = 21.7

Explanation / Answer

1. Friction force = mu*mg = 0.19*1200*10 = 2280 N

    Total force = 3540 N = Friction force + force applied

     force applied = 3540 - 2280 = 1260N

2. Force applied should be limiting friction = mu*mg = 0.313*30.9*10 = 96.7 N

      Same as this force is for limiting friction

3. Normal force = mg - Fsin27 = 12.7*10 - 37sin27 = 110.2 N

    acc = (Fcos27 - mu*N) / m = (37cos27 - 0.17*110.2) / 12.7 = 1.12 m/s^2

4. s = ut + 1/2*a*t^2

    NOT CLEAR

5. Force to stop =mu*mg = 0.23*10*m = 2.3m

    ma = 2.3m

    a = 2.3 m/s^2

    s = v^2/2a = 4.3^2/2*2.3 = 4.02 m

6. a) gsin21.7 - mu*gcos21.7 = a

         a = 10sin21.7 - 0.14*10*cos21.7 = 2.4 m/s^2

         v = sqrt(2gh) = sqrt(2*10*9.31sin21.7) = 8.3 m/s

7. Friction force = mgsin37 - ma = 16.8( 10sin37 - 0.27) = 96.57 N

     mu = 96.57/(mgcos37) = 96.57/(16.8*10*cos37) = 0.72

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