You are lowering two boxes, one on top of the other, down the ramp shown in the

Question

You are lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 14.0 cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.409, and the coefficient of static friction between the two boxes is 0.828.

What force do you need to exert to accomplish this?

What is the magnitude of the friction force on the upper box?

What is the direction of the friction force on the upper box?

Explanation / Answer

here, u1=0.409, u2=0.828, F1=Friction on larger block due to ramp, F2= Friction on smaller, N= normal reaction due to ramp on system of boxes

For smaller box, friction F2 (between the boxes)will act backward as it has to resist downward acceleration

For larger fox friction F1 (due to ramp)will act backward as it has to resist downward acceleration and F2 will act forward as it is action reaction pair

Fexerted = Mg sin theta - F1 + F2

F2 = mg sin theta

F1= u1 N = 0.409* N

N =( M+m)g cos theta

F1= 0.409*(48+32)*9.8* 4.75 /sqrt(4.752 + 2.502)

=0.409*80*9.8*4.75/5.37 = 283.6N

F2= 32*9.8*2.50/5.37 = 146N

Fexerted = Mg sin theta - F1 + F2

=48*9.8*2.5/5.37 - 283.6 +146

=219 + 146 - 283.6= 81.4N

Hence Force exerted is 81.4 N

Friction force on upper block is 146N

Direction of the friction force on upper block is up the ramp

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