Question

If the system is in equilibrium $(\cos 53°=\frac{3}{5})$, then the value of $P$ is

• A. $16N$
• B. $4N$
• C. $\sqrt{208}N$
• D. $\sqrt{232}N$

Answer 1

The correct option is C

$\sqrt{208}N$

Explanation for correct option:

Given: $\cos 53°=\frac{3}{5}$

We have to find the value of $P$.

As we know that, A line's vector equation is an equation that is satisfied by a vector with its head at one of the line's points.

The sum of all vector forces will be $0$ when the system is at equilibrium.

$\Rightarrow$ $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{P}=0$

We can rewrite it,

$\Rightarrow$ $\overrightarrow{P}=-\left(\overrightarrow{A}+\overrightarrow{B}\right)$

Here, $\overrightarrow{P}$ will have the same magnitude as the sum of $\overrightarrow{A}$ and $\overrightarrow{B}$, but in the opposite direction.

Therefore,

$\Rightarrow$ $\left|\overrightarrow{P}\right|=\sqrt{A^2+B^2+2\overrightarrow{A}·\overrightarrow{B}}$

$\Rightarrow$ $\left|\overrightarrow{P}\right|=\sqrt{A^2+B^2+2AB(\cos 53°)}$

$\Rightarrow$ $\left|\overrightarrow{P}\right|=\sqrt{{10}^2+6^2+2\times 10\times 6\times \frac{3}{5}}$

$\Rightarrow$ $\left|\overrightarrow{P}\right|=\sqrt{100+36+72}$

$\Rightarrow$ $\left|\overrightarrow{P}\right|=\sqrt{208}N$

The value of $P$ is $\sqrt{208}N$.

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