Question

A block stays in equilibrium on an inclined plane it is easiest to push along?

Answer 1

Answer:

Downwards push is easiest .

Explanation:

It is easiest to push along the inclined plane downwards .

Answer 2

Notice that I have defined a rotated set of axes and I labelled them x’ and y’. The x’-axis is parallel to the inclined plane and the y’-axis is perpendicular to the plane. I chose positive x’-axis down the plane. The component of the weight (mg) acting down the plane is found by resolving the weight into components as shown in my example below using θ=30 :

So the component of the weight acting down the plane is (mg)sinθ. The friction force acts opposite the direction of motion (up the plane) as shown on my free body diagram. To determine Ffric , write a static equilibrium equation in the x’ direction. Static equilibrium is a fancy way of saying all the forces balance:

ΣFx′=0

(mg)sinθ−Ffric=0

∴ Ffric=(mg)sinθ

This result only applies if the angle of the inclined plane is less than the maximum possible angle when slippage occurs. So the force of friction will increase with increasing θ until the box begins to slip. If slippage is impending, then we have the maximum possible friction force given by:

Ffric=μsFN

In this case, first find the normal force between the ramp and the object. To determine FN , write a static equilibrium equation in the y’ direction:

ΣFy′=0

FN−(mg)cosθ=0

FN=(mg)cosθ

Now substitute the normal force into:

Ffric=μsFN  

Ffric=μs(mg)cosθ