A certain physicist has traded his box car in for a new wedge car. The front of

Question

A certain physicist has traded his box car in for a new wedge car. The front of the car's wedge forms an angle of 65 degrees, so that when the car travels down a horizontal road, this is the angle the top surface makes with respect to the horizontal. Unfortunately, a box of mas 3 kg has been left on the car (as shown) which will have a tendency to slide up or down the ramp-shaped car as the car accelerates. For this problem, take the positive direction of acceleration to be the direction the car is facing (to the right, in the figure). Assuming the car's surface is frictionless and the box is not sliding up or down the car (thus the box is accelerating along with the car), determine the normal force of the car's surface on the box: and the the acceleration of the car: For the remaining parts, the wedge car ha not been washed in quite some time, so that the coefficient of static friction between the car and the box is 0.2; If the coefficient of static friction is 0.2 and the box is on the verge of sliding down the car, determine the normal force of the car's surface on the box: and the the acceleration of the car: (A) If the coefficient of static friction is 0.2 and the box is on the verge of sliding up the car. determine the normal force of the car's surface on the box: N and the the acceleration of the car:

Explanation / Answer

m = 3 kg , theta =65 degrees

A) N = mgcos(65) = 3*9.8 * cos(65)

Normal force = 12.42 N

a = gsin(65) =9.8sin(65)

a = 8.88 m/s^2

B) us =0.2

N = mgcos(65) = 3*9.8 * cos(65)

Normal force = 12.42 N

Net force along parallel to plane is ma

mgsin(65) - Fs = ma

Fs =usN

mgsin(65) -usmgcos(65) = ma

(9.8*sin(65)) - (0.2*9.8*cos(65)) = a

a =8.05 m/s^2

C) us =0.2

N = mgcos(65) = 3*9.8 * cos(65)

Normal force = 12.42 N

Net force along parallel to plane is ma

mgsin(65) + Fs = ma

Fs =usN

mgsin(65) + usmgcos(65) = ma

(9.8*sin(65)) + (0.2*9.8*cos(65)) = a

a = 9.7 m/s^2

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