Minimum force needed?

Question

You want to move a 200-kg piano up a 30 degree incline. The static coefficient of friction is .80 and the kinetic coefficient is .30.
a) What is the minimum force you need to push with to get the piano moving?
b) How much force is needed to keep the piano moving up the incline at constant speed?

Answer 1

I will not simply answer your homework question for you but I will give you a few pointers, some things to think about to hopefully make this problem easier.

First of all, in both cases, break forces up into components paralell and perpendicular to the plane. The weight of the piano would have both a paralell and perpendicular component, Normal force would be all perpendicular, frictional force all paralell, and the force applied is also paralell.

For part a.: What force needs to be overcome to begin motion? How big is that force?

Part b.: What is the net force on an object needed to maintain a constant speed (no acceleration)? What forces (both magnitude and direction) are acting on the piano to keep it moving at a constant speed?

Answer 2

a) The minimum force will be just slightly more than the sum of the force of static friction and the component of the weight of the piano along the incline. So, solve first for the force of static friction using the formula:

f=uR, where f is the force of static friction, u is the coefficient of static friction, and R is the force normal to the incline. In this problem, R is equal to 200gcos30.

f=0.80*200gcos30
=1358N downward along the incline

Then solve for the component of the weight of the piano along the incline. This is equal to:

200gsin30=980N downward along the incline

Add the two forces: 1358+980=2338N
downward along the incline

A force slightly more than 2338N upward along the incline will move the piano.

b)As soon as the piano moves, the force of friction becomes smaller because now we use the coefficient of kinetic friction which is smaller. The component of the weight of the piano along the incline will remain the same, i.e.980N. Now solve for the force of kinetic friction using the same formula:

f=uR
=0.30*200gcos30
=509N downward along the incline

Again get the sum of the two forces:

Total=509+980
=1489N downward along the incline.

To keep the piano moving along the incline at constant speed, you need to apply a force equal to 1489N upward along the incline.

Answer 3

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