Question

Three forces $\overrightarrow{F_1}$, $\overrightarrow{F_2}$ and $\overrightarrow{F_3}$ act on a body. $\overrightarrow{F_1}$ acts vertically downwards with magnitude $\sqrt{3}\text{N}$ , $\overrightarrow{F_2}$ acts horizontally leftwards with magnitude $1\text{N}$ and $\overrightarrow{F_3}$ acts at an angle $\theta$ with horizontal as shown in the figure. Find the magnitude of $\overrightarrow{F_3}$ and $\theta$ for the body to remain in equilibrium.

• A. $2\text{N}$, ${30}^{\circ }$
• B. $2\text{N}$, ${60}^{\circ }$
• C. $1\text{N}$, ${30}^{\circ }$
• D. $1\text{N}$, ${60}^{\circ }$

Answer 1

The correct option is B $2\text{N}$, ${60}^{\circ }$

Given, $F_1=-\sqrt{3}\text{N}$ and $F_2=1\text{N}$
Let the magnitude of $\overrightarrow{F_3}$ be $p\text{N}$.
Let us resolve this in horizontal and vertical direction.
We can write after the resolution
$F_3=\left(p\cos \theta \right)+\left(p\sin \theta \right)$
Now the net force along $x$ must be $0$.
$\Rightarrow p\cos \theta -1=0$ ----------(1)
Similarly, the net force along $y$ must be $0$
$\Rightarrow p\sin \theta -\sqrt{3}=0$ ----------(2)

From eq (1) and (2) we get,
$\tan \theta =\sqrt{3}\Rightarrow \theta ={60}^{\circ }$
Putting this value in equation (1) we get
$p=2\text{N}$.