Question

A particle in equilibrium is subjected to four forces: $F_1=-10\hat{k},F_2=u(\frac{4}{13}\hat{i}-\frac{12}{13}\hat{j}+\frac{3}{13}\hat{k}),F_3=v(-\frac{4}{13}\hat{i}-\frac{12}{13}\hat{j}+\frac{3}{13}\hat{k})$ and $F_4=w(\cos \theta \hat{i}+\sin \theta \hat{j})$Then:

• A. $u+v=\frac{130}{3},w=40cosec\theta$
• B. $u=\frac{1}{3}-\cot \theta ,v=\frac{1}{3}+\cot \theta ,w=40cosec\theta$
• C. $u=65(\frac{1}{3}-\cot \theta ),v=65(\frac{1}{3}+\cot \theta )w=40cosec\theta$
• D. $u=55(\frac{1}{3}-\cot \theta ),v=55(\frac{1}{3}+\cot \theta ),w=60cosec\theta$

Answer 1

Answer: (A), (C)

$u+v=\frac{130}{3},w=40cosec\theta$
$u=65(\frac{1}{3}-\cot \theta ),v=65(\frac{1}{3}+\cot \theta )w=40cosec\theta$

The particle is in equilibrium means the forces acting on it in all directions are cancelling each other. so we just have to add the components of forces in each direction and equate them to zero.

For $F_1$ $x$ and $y$ components are zero while for $F_4$ the $z$ component is zero, so we put values accordingly.

Equating $x$ component of sum to zero: $\frac{4u}{13}-\frac{4v}{13}+wcos\theta =0$

Equating $y$ component of sum to zero:$\frac{-12u}{13}-\frac{12v}{13}+wsin\theta =0$

Equating $z$ component of sum to zero:$-10+\frac{3u}{13}+\frac{3v}{13}=0$

We get 3 variable and 3 equations so we can easily solve for $u,v$ and $w$

which comes out to be

$u=65(\frac{1}{3}-cot\theta ),v=65(\frac{1}{3}+cot\theta ),w=40cosec\theta$