Question

a body of mass 10 Kg is placed on an inclined surface of angle 30 degree if the coefficient of limiting friction is root 3 find the force required to just push the body up the inclined surface the force is being parallel to the inclined surface

Answer 1

Answer:

Force required to just push the object = 200 N

Answer 2

Given,

m = 10 kg

θ = 30°

μ = √3

In the fig., F is the force required to just pull the block up the inclined plane. And f is the frictional force.

Weight of the body = mg = 10 × 10 = 100 N acting vertically downwards.

Here, normal reaction acting on the block is given by,

R = mg cosθ = 100 × √3 / 2 = 50√3 N

Then, the frictional force is,

f = μR = √3 × 50√3 = 150 N

The acceleration acting on the block is g sinθ = 10 × 1 / 2 = 5 ms^(-2). But this is when the force F is not acting.

The minimum possible force F can make the block rest on the inclined plane and not to slide down. Thus, the acceleration acting on this block, in this case, is zero, not 5 ms^(-2).

Then, from the fig.,

f + 50 - F = 0

since acceleration is 0 ms^(-2). So,

F = f + 50

F = (150 + 50) N

F = 200 N

For the block be just pushed upward on the inclined plane, F > 200 N.