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 18.0cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.489, and the coefficient of static friction between the two boxes is 0.751.

What force do you need to exert to accomplish this?

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

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 18.0cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.489, and the coefficient of static friction between the two boxes is 0.751. What force do you need to exert to accomplish this? What is the magnitude of the friction force on the upper box?

Explanation / Answer

let m= 32 kg

M= 48 kg

angle of ramp A= tan-1 (2.5/4.75)=27.75 deg

frictional force between lower box and floor Fr= Uf (M+m)g cos A

   = .489(32+50)9.8 * cos 27.75

= 339.28

force required, F =(M+m)g SIN (A) - Fr

=(48+32)*9.8 SIN (27.75) - 339.28

=26.166 Newton

coeffecient of friction between box Ub= 0.75

Vertical reaction in upper box from lower box R= mg cos A

=32*9.8* cos 27.75

=277.75 Newton

Static Friction Between Boxes Ff = Ub *R

=.75 *277.75

=208.14 Newton

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