Question

A horizontal force $F$ is applied to a small object $P$ of mass $m$ on a smooth plane inclined to the horizon at an angle $\theta$. If $F$ is just enough to keep $P$ in equilibrium, then $F=$?

• A. $mg{\cos }^2\theta$
• B. $mg{\sin }^2\theta$
• C. $mg\cos \theta$
• D. $mg\tan \theta$

Answer 1

The correct option is D

$mg\tan \theta$

Step 1. Given data

A horizontal force $F$ is applied to a small object $P$ of mass $m$ on a smooth plane inclined to the horizon at an angle $\theta$

Step 2. Explanation for the correct option(s):

Lami’s Theorem states, “When three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces”. Referring to the above diagram, consider three forces $A,B,C$acting on a particle or rigid body making angles $\alpha ,\beta$ and $\gamma$ with each other.

In the mathematical or equation form, it is expressed as,

$\frac{A}{\sin \alpha }=\frac{B}{\sin \beta }=\frac{C}{\sin \gamma }$

By applying Lami’s theorem at $P$, we get

$\begin{array}{rcl}\frac{R}{(\sin 90°)}& =& \frac{F}{\sin (180°–\theta °)}=\frac{mg}{\sin (90°+\theta °)}\\ \frac{R}{1}& =& \frac{F}{\sin \theta }=\frac{mg}{\cos \theta }\end{array}$

[$R$ is reaction, $g$is acceleration due to gravity]

Thus, $F=mg\tan \theta$

Hence, option $\left(D\right)$ is correct.