Question

A body of mass 8 kg is in limiting equilibrium over an incline plane of inclination $30^\circ$. If the inclination is made $60^\circ$ the minimum force required to prevent the body from sliding down is $\left( {g = 10m/{s^2}} \right)$
(A) 80 N
(B) $\dfrac{{80}}{{\sqrt 3 }}N$
(C) $\dfrac{{40}}{{\sqrt 3 }}N$
(D) $40\sqrt 3 N$

Answer 1

In the second situation, the forces acting on the object parallel to the surface of the incline plane would be friction, weight component along that axis, and that minimum force required to keep it from sliding down. Use the first state to calculate the coefficient of friction which is acting between the mass and the surface of the incline plane.
Formula used: In this solution we will be using the following formulae;
$f = \mu mg\cos \theta$ where $f$ is the frictional force acting between the surface of an inclined plane and the block, $\mu$ is the coefficient of friction, $m$ is the mass of the block, $g$ is the acceleration due to gravity and $\theta$ is the inclination angle.
$W = mg$ where $m$ is the mass of the block, $g$ is the acceleration due to gravity

Complete Step-by-Step solution:
In the first case, the state is in limiting equilibrium, hence, the frictional force is equal to the weight component parallel to the surface. i.e.
$W\sin \theta = f$
$\Rightarrow mg\sin \theta = \mu mg\cos \theta$ where $\mu$ is the coefficient of friction, $m$ is the mass of the block, $g$ is the acceleration due to gravity and $\theta$ is the inclination angle.
By inserting the angle and simplifying, we have
$\sin 30^\circ = \mu \cos 30^\circ$
$\Rightarrow \mu = \dfrac{{\sin 30}}{{\cos 30}} = \dfrac{1}{{\sqrt 3 }}$
Now, in the second state, the angle was increased to 60 degrees, and a force is required to keep it from falling, then the forces acting up along the surface must be equal to the forces acting downward. Hence
$W\sin \theta = f + F$ where $F$ is the additional force required to make the object to not fall
Hence,
$W\sin \theta = \mu mg\cos \theta + F$
Then,
$F = mg\sin 60^\circ - \left( {\dfrac{1}{{\sqrt 3 }}mg\cos 60} \right)$
$\Rightarrow F = mg\left( {\dfrac{{\sqrt 3 }}{2}} \right) - \left( {\dfrac{1}{{2\sqrt 3 }}mg} \right)$
Simplifying the expression, we have
$F = mg\left( {\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{2\sqrt 3 }}} \right) = mg\left( {\dfrac{{3 - 1}}{{2\sqrt 3 }}} \right)$
Hence, $F = \dfrac{{mg}}{{\sqrt 3 }} = \dfrac{{8 \times 10}}{{\sqrt 3 }}$
$\Rightarrow F = \dfrac{{80}}{{\sqrt 3 }}N$

Hence, the correct option is B

Note: For clarity, the limiting equilibrium in the statement means that the friction required to keep the object from moving is reaching maximum, and any slight increase in the weight component will cause a motion. This is why we could use the same coefficient of friction in state 1 and 2.