Question

A particle is in equilibrium when the forces $\overrightarrow{F_1}=-\overrightarrow{10k}$, $\overrightarrow{F_2}=\frac{u}{13}(\overrightarrow{4i}-\overrightarrow{12j}+\overrightarrow{3k})$, $\overrightarrow{F_3}=\frac{\upsilon }{13}(-\overrightarrow{4i}-\overrightarrow{12j}+\overrightarrow{3k})$ and $\overrightarrow{F}=w(\cos \theta \overrightarrow{i}+\sin \theta \overrightarrow{j})$ act on it, then

• A. $u=65(1-3\cot \theta )$
• B. $\upsilon =\frac{65}{3}+65\cot \theta$
• C. $w=40\text{cosec}\theta$
• D. None of these

Answer 1

Answer: (B), (C)

The correct options are
B $\upsilon =\frac{65}{3}+65\cot \theta$
C $w=40\text{cosec}\theta$

A particle can only be in equilibrium when the net force acting upon it is $0$.
So, we check for each of the $\hat{i},\hat{j},\hat{k}$ components and equate them with $0.$
For the $\overline{i}$ component, $\frac{4u}{13}-\frac{4v}{13}+w\cos \theta =0$
For the $\hat{j}$ component, $\frac{-12u}{13}-\frac{12v}{13}+w\sin \theta =0$
For the $\hat{k}$ component, $\frac{3u}{13}+\frac{3v}{13}=10$
We get $u+v=\frac{130}{3}$ ........ $\left(i\right)$
$\Rightarrow w=40\text{cosec}\theta$ ..... $\left(ii\right)$
Thus we have $u-v=-130\cot \theta$ ..... $\left(iii\right)$
Solving simultaneously $\left(i\right)$ and $\left(ii\right),$ we get

$u=\frac{65}{3}-65\cot \theta$ and $v=\frac{65}{3}+65\cot \theta$