Question 275: Laws of Motion

Question

A block of mass m is on an inclined plane of angle θ. The coefficient of
friction between the block and the plane is μ and tan θ > μ. The block is held
stationary by applying a force P parallel to the plane. The direction of force
pointing up the plane is taken to be positive. As P is varied from P
1
=
mg(sinθ − μ cosθ) to P
2
= mg(sinθ + μ cosθ), the frictional force f versus P
graph will look like

Accepted Answer

The forces acting on the block are its weight mg, normal reaction N, applied force P and frictional force f.
Resolve mg along and normal to the inclined plane and apply Newton's second law to get
$$0 = P + f - mg\sin 	heta $$,
which gives
$$f =  - P + mg\sin 	heta $$. ...... (1)
This is a straight line with slope $$-$$1. Substitute the values of P
1
and P
2
in equation (1) to get the frictional force at those points i.e.,
$${f_1} = \mu mg\cos 	heta $$ and $${f_2} =  - \mu mg\cos 	heta $$.

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Laws of Motion

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