A block of mass is on an inclined plane of angle . The coefficient of friction between the block and the plane is and . The block is held stationary by applying a force parallel to the plane. [2010]

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 $P_{1} = m g (\sin θ - μ \cos θ)$ to $P_{2} = m g (\sin θ + μ \cos θ)$ , the frictional force $f$ versus $P$ graph will look like [2010]

Smart Achievers · Accepted Answer

(1)

According to question, $θ$ , so block has a tendency to move down the incline. Force $P$ is applied upwards along the incline to keep the block stationary.

Here, at equilibrium, $P + f = m g \sin θ \Rightarrow f = m g \sin θ - P$

Now as $P$ increases, $f$ decreases linearly with respect to $P$ .

When $P = m g \sin θ$ , $f = 0$ .

When force $P$ is increased further, the block has a tendency to move upwards along the incline and hence frictional force acts downwards along the incline.

Here, at equilibrium, $P = f + m g \sin θ$

$\Rightarrow f = P - m g \sin θ$

Now as $P$ increases, $f$ increases linearly w.r.t $P$ .

Hence graph (1) correctly depicts the situation.

Solution

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