Question

The horizontal acceleration that should be given to a smooth inclined plane of angle ${\sin ^{ - 1}}\left( {\dfrac{1}{l}} \right)$ to keep an object stationary on the plane, relative to the inclined plane is
A. $\dfrac{g}{{\sqrt {{l^2} - 1} }}$
B. $g\sqrt {{l^2} - 1}$
C. $\dfrac{{\sqrt {{l^2} - 1} }}{g}$
D. $\dfrac{g}{{\sqrt {{l^2} + 1} }}$

Answer 1

Use Newton’s second law of motion. Apply Newton’s second law of motion to the block in the horizontal direction and determine the horizontal acceleration that should be given to the object to keep it stationary. Derive this horizontal acceleration in terms of $l$.

Formula used:
The equation for Newton’s second law of motion is
${F_{net}} = ma$
Here, ${F_{net}}$ is the net force on the object, $m$ is the mass of the object and $a$ is the acceleration of the object.

Complete step by step answer:
An object is stationary on an inclined plane.

Let $m$ be the mass of the object and $\theta$ is the angle of inclination with the horizontal.

Let $a$ is the horizontal acceleration of the object.

Draw a free body diagram of the object stationary on the inclined plane.

In the above free-body diagram, $mg$ is the weight of the object, $mg\sin \theta$ and $mg\cos \theta$ are the horizontal and vertical components of the weight respectively, $\theta$ is the angle of inclination with the horizontal, $ma$ is the horizontal force acting on the object to keep it stationary, $mg\cos \theta$ and $mg\sin \theta$ are the horizontal and vertical components of the horizontal force acting on the object respectively.

Apply Newton’s second law of motion to the object in the horizontal direction.
$mg\sin \theta = ma\cos \theta$
$\Rightarrow g\sin \theta = a\cos \theta$

Rearrange the above equation for the horizontal acceleration $a$ on the object.
$a = \dfrac{{g\sin \theta }}{{\cos \theta }}$
$\Rightarrow a = g\tan \theta$
$\Rightarrow a = g\dfrac{{\sin \theta }}{{\cos \theta }}$
$\Rightarrow a = g\dfrac{{\sin \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}$

Substitute $\dfrac{1}{l}$ for $\sin \theta$ in the above equation.
$a = g\dfrac{{\dfrac{1}{l}}}{{\sqrt {1 - {{\left( {\dfrac{1}{l}} \right)}^2}} }}$
$\Rightarrow a = g\dfrac{{\dfrac{1}{l}}}{{\sqrt {\dfrac{{{l^2} - 1}}{{{l^2}}}} }}$
$\Rightarrow a = \dfrac{g}{{\sqrt {{l^2} - 1} }}$

Therefore, the horizontal acceleration of the object is $\dfrac{g}{{\sqrt {{l^2} - 1} }}$.

So, the correct answer is “Option A”.

Note:
Newton’s second law of motion is applied to the stationary object only in the horizontal direction as the required acceleration is along the horizontal direction.