You are lowering two boxes, one on top of the other, down the ramp by pulling on

Question

You are lowering two boxes, one on top of the other, down the ramp by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 20.0 . The coefficient of kinetic friction between the ramp and the lower box is 0.492, and the coefficient of static friction between the two boxes is 0.848. A) What force do you need to exert to accomplish this? B) What is the magnitude of the friction force on the upper box? C) What is the direction of the friction force on the upper box?

Explanation / Answer

angle =tan-1(2.5/4.75)=27.75DEGREE
suppose 48kgblock =m1, 32kgblock=m2
a) force on m1 due m2

32gcos()Xs in downward direction

frictional force of floor on the m1 is

(48+32) g cos()Xk pointing upward

48 gsin() down ward

0=F-32gcos()Xs-48 gsin()+(48+32) g cos()Xk

F=113.09N

b) frictional force 32gcos()*s=136.53N

c) frictional forece direction upward

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