Question

If two forces $P$ and $Q$ act on such an angle that their resultant force $R$ is equal to force $P$, then if $P$ is doubled then the angle between the new resultant force and $Q$ will be

• A. $30°$
• B. $60°$
• C. $45°$
• D. $90°$

Answer 1

The correct option is D

$90°$

Step 1: Given data

Resultant force of $P$ and $Q$ is $R$ and it is equal to $P$.

Step 2: Formulas used

From the law of cosines,

$a^2=b^2+c^2-2bc\cos A$

where $a$, $b$, and $c$ are the three sides of a triangle and $\angle A$ is the angle opposite to side $a$

Step 3: Solution

Let the angle between the forces $P$ and $Q$ is $\alpha$.

From the law of cosines an triangle law of vector addition, we have
$\begin{array}{ccc}& R^2=P^2+Q^2+2PQ\cos \alpha & \\ \Rightarrow & P^2=P^2+Q^2+2PQ\cos \alpha & \left[\because R=P\right]\\ \Rightarrow & \cos \alpha =-\frac{Q}{2P}& \end{array}$

When the $P$ is doubled, let the new resultant make an angle $\theta$ with $Q$.

It can be written as,

$\begin{array}{ccc}& \tan \theta =\frac{2P\sin \alpha }{Q+2P\cos \alpha }& \\ \Rightarrow & \tan \theta =\frac{2P\sin \alpha }{Q+2P\left({\displaystyle \frac{-Q}{2P}}\right)}& \left[\because \cos \alpha =-\frac{Q}{2P}\right]\\ \Rightarrow & \tan \theta =\infty & \\ \Rightarrow & \theta =90°& \end{array}$

Thus, the new resultant is at $90°$ to $Q$.

Hence, option D is correct.