Question

The resultant of two forces $P$ and $Q$ is $R$. If the direction of $P$ is reversed keeping the direction of $Q$ the same, the result remains unaltered. The angle between $P$ and $Q$ is:

• A. ${90}^{\mathrm{o}}$
• B. $60°$
• C. $45°$
• D. $30°$

Answer 1

The correct option is A

${90}^{\mathrm{o}}$

Step 1. Given data:

$\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}.......\left(1\right)$

Also, $-\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}.......\left(2\right)$

Step 2. Formula used

From the triangle law of vector addition, we know

Resultant of two vectors is given by, $\overrightarrow{R}=\sqrt{P^2+Q^2+2PQ.\cos \theta }..........\left(3\right)$

Step 3. Calculating angle

Squaring both sides of the equation $\left(3\right)$, we get

${\mathrm{R}}^2={\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQcos\theta }.......\left(4\right)$

When the direction of vector P is reversed, i.e. equation $\left(2\right)$, the resultant will be the same as given in the question.

${\mathrm{R}}^2={\left(-P\right)}^2+{\mathrm{Q}}^2-2\mathrm{PQcos\theta }$

Since ${\left(-P\right)}^2=P^2$

Therefore, ${\mathrm{R}}^2=P^2+{\mathrm{Q}}^2-2\mathrm{PQcos\theta }.........\left(5\right)$

Equating $\left(4\right)$ and $\left(5\right)$, we get

$$\begin{gathered} {\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQcos\theta }={\mathrm{P}}^2+{\mathrm{Q}}^2-2\mathrm{PQcos\theta } \\ 2\mathrm{PQcos\theta }=-2\mathrm{PQcos\theta } \\ 4\mathrm{PQcos\theta }=0 \\ \mathrm{cos\theta }=0 \\ \mathrm{\theta }={\cos }^{-1}\left(0\right)={90}^{\mathrm{o}} \end{gathered}$$

Thus, the angle between the two forces is ${90}^{\mathrm{o}}$.

Therefore, the two forces are perpendicular to each other.

Hence, option A is the correct answer.