Question

A block weighing 82.5 N rests on a plane inclined at 28.8° to the horizontal. The coefficient of the static and kinetic frictions are 0.25 and 0.12 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping? What is the minimum magnitude of F that will start the block moving up the plane? What is the magnitude of F is required to move the block up the plane at constant velocity?

Answer 1

Part A.

Gravitation Force on the block along the slope will be:

Fg = W*sin A

W = weight of Block & A = 28.8 deg

Fg = 82.5*sin 28.8 deg = 39.74 N

Now to stop block from slipping there will be friction force in upward direction, which will be

Ff = us*N = us*W*cos A

Ff = 0.25*82.5*cos 28.8 deg = 18.07 N

So Net force on block is

Fnet = Fg - Ff = 39.74 - 18.07 = 21.67 N (downward)

So there should be 21.67 N force parallel to the plane in upward direction to prevent the block from slipping.

Part B.

To start moving the plane in upward direction, friction force will be in downward direction, So net force required will be

Fnet = Fg + Ff

Fnet = 39.74 + 18.07 = 57.81 N

Part C.

Now once block starts moving there will be kinetic friction on block, So to move the block upward with constant velocity (means acceleration zero, which gives net force zero)

Fnet = F - Fg - Fk

F = Force required to move the block upward with constant velocity

Fk = kinetic friction force = uk*N = uk*W*cos A

Fk = 0.12*82.5*cos 28.8 deg = 8.67 N

So,

Fnet = 0 = F - Fg - Fk

F = Fg + Fk

F = 39.74 + 8.67 = 48.41 N

Please Upvote.

Comment below if you've any query.