Question

$2$ forces $P$ and $Q$are acting parallel to the length and base of an inclined plane respectively would each of them singly support a weight $W$, on the plane, then $\frac{1}{P^2}-\frac{1}{Q^2}=$

• A. $\frac{1}{W^2}$
• B. $\frac{2}{W^2}$
• C. $\frac{3}{W^2}$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{W^2}$

Step 1. Given data

Two forces $P$ and $Q$ are acting parallel to the length and base of the inclined plane.

Step 2. Finding the force, $P$

By using the Lami 's theorem

According to this theorem, if three concurrent forces act on a body keeping it in Equilibrium, the each force is proportional to the sine of the angle between the other two forces.

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

Now, Case $I$- When force $P$ is acting

Using Lami's theorem, we get

$\frac{P}{\sin \left(180°-\alpha \right)}=\frac{W}{\sin 90°}=\frac{R}{\sin \left(90°+\alpha \right)}$ [Where, $W$ is weight, $R$ is the reaction]

$P=W\sin \alpha$ [$\sin \left(180°-\alpha \right)=\sin \alpha ,\sin 90°=1$] [$eq\left(1\right)$]

Step 3. Finding the force, $Q$

Now, Case $II$- When force $Q$ is acting

Again, using Lami's theorem

$\frac{Q}{\sin \left(180°-\alpha \right)}=\frac{W}{\sin \left(90°+\alpha \right)}=\frac{R}{\sin 90°}$

$\frac{Q}{\sin \alpha }=\frac{W}{\cos \alpha }$ [$\sin \left(90°+\alpha \right)=\cos \alpha$]

$Q=W\tan \alpha$ [$\frac{\sin \alpha }{\cos \alpha }=\tan \alpha$] [$eq\left(2\right)$]

Now, value of $\frac{1}{P^2}-\frac{1}{Q^2}$

$\frac{1}{P^2}-\frac{1}{Q^2}=\frac{1}{{\left(W\sin \alpha \right)}^2}-\frac{1}{{\left(W\tan \alpha \right)}^2}$ [from $eq\left(1\right),eq\left(2\right)$]

$\frac{1}{P^2}-\frac{1}{Q^2}=\frac{1}{W^2}\left(\cos e{c}^2\alpha -co{t}^2\alpha \right)$ [$\frac{1}{\sin \alpha }=\cos ec\alpha ,\frac{1}{\tan \alpha }=cot\alpha$]

$\frac{1}{P^2}-\frac{1}{Q^2}=\frac{1}{W^2}$ [$\cos e{c}^2\alpha -co{t}^2\alpha =1$]

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