A block of mass $m$ is on an inclined plane of angle$\ \ heta$. The coefficient of friction between the block and the plane is $\mu$ and $ an{ heta \mu}.$ The block is held stationary by applying a force $E$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_{1} = mg(\sin{ heta - \mu\cos{ heta)}}$ to $P_{2} = mg(\sin{ heta + \mu\cos{ heta),}}$ the frictional force $f\ versus\ P$ graph will look like [image:5_image26.jpeg]

Question

A block of mass $m$ is on an inclined plane of angle$\ \ 	heta$. The coefficient of friction between the block and the plane is $\mu$ and $	an{	heta > \mu}.$ The block is held stationary by applying a force $E$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_{1} = mg(\sin{	heta - \mu\cos{	heta)}}$ to $P_{2} = mg(\sin{	heta + \mu\cos{	heta),}}$ the frictional force $f\ versus\ P$ graph will look like [image:5_image26.jpeg]

Dr. Abubacker · Accepted Answer

(a) When
$P = mg\ (\sin{	heta - \mu\cos{	heta)}}$
$f = \mu mg\cos	heta$ (upwards)
$when\ \ P = mg\sin	heta$
$f = 0$
$and\ when\ \ P = mg(\sin{	heta + \mu\cos{	heta)}}$
$f = \mu\ mg\cos{	heta\ }$ (downwards)
Hence, friction is first positive, then zero and then negative.

Explanation

Reviewed by Dr. Abubacker

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