Question

The resultant $\mathrm{R}$ of two forces $\mathrm{P}$ and $\mathrm{Q}$ act at right angles to $\mathrm{P}$. Then the angle between the forces is:

• A. ${\cos }^{-1}\left(\frac{P}{Q}\right)$
• B. ${\cos }^{-1}\left(-\frac{P}{Q}\right)$
• C. ${\sin }^{-1}\left(\frac{P}{Q}\right)$
• D. ${\sin }^{-1}\left(-\frac{P}{Q}\right)$

Answer 1

The correct option is B

${\cos }^{-1}\left(-\frac{P}{Q}\right)$

Step 1: Given Data

There are two forces $\mathrm{P}$ and $\mathrm{Q}$ and their resultant is $\mathrm{R}$.

The resultant $\mathrm{R}$ of two forces $\mathrm{P}$ and $\mathrm{Q}$ is at a right angle to the $\mathrm{P}$.

$ɑ=\frac{\pi }{2}$ , $ɑ$is the angle between $\mathrm{P}$ and $\mathrm{R}$

Step 2: Formula used,

The direction of the resultant is given by

$\tan ɑ=\frac{\mathrm{Qsin\theta }}{\mathrm{P}+\mathrm{Qcos\theta }}=0$

$ɑ$is the angle between $\mathrm{P}$ and $\mathrm{R}$

$\mathrm{P}$ and $\mathrm{Q}$ are the forces and $\theta$is the angle between the forces $\mathrm{P}$and $\mathrm{Q}$

Step 3: Calculating the angle between the two forces

$$\begin{gathered} \tan \frac{\mathrm{\pi }}{2}=\frac{\mathrm{Qsin\theta }}{\mathrm{P}+\mathrm{Qcos\theta }}=\infty \\ \frac{P+Q\cos \theta }{\mathrm{Qsin\theta }}=\frac{1}{\infty } \\ \frac{P+Q\cos \theta }{\mathrm{Qsin\theta }}=0 \\ P+Q\cos \theta =0 \end{gathered}$$

$$\begin{gathered} \mathrm{Qcos\theta }=-\mathrm{P} \\ \mathrm{cos\theta }=-\frac{\mathrm{P}}{\mathrm{Q}} \\ \mathrm{\theta }={\cos }^{-1}\left(-\frac{\mathrm{P}}{\mathrm{Q}}\right) \end{gathered}$$

Therefore, the angle between the forces is ${\cos }^{-1}\left(-\frac{\mathrm{P}}{\mathrm{Q}}\right)$.

Therefore, the correct answer is (B).