Question

A particle, in equilibrium, is subjected to four forces

$F_1=-10k,F_2=U[\frac{4}{13}i-\frac{12}{13}j+\frac{3}{13}k],$

$F_3=V[-\frac{4}{13}i-\frac{12}{13}j+\frac{3}{13}k],$

$F_4=W[\cos \theta i+\sin \theta j]$

Solve for U and V as functions of $\theta .$

• A. $U=\frac{65}{3}(1-4\cot \theta ),V=\frac{65}{3}(1+4\cot \theta ).$
• B. $U=\frac{65}{3}(1+4\cot \theta ),V=\frac{65}{3}(1-4\cot \theta ).$
• C. $U=\frac{65}{3}(1-3\cot \theta ),V=\frac{65}{3}(1+3\cot \theta ).$
• D. $U=\frac{65}{3}(1+3cot\theta ),V=\frac{65}{3}(1-3\cot \theta ).$

Answer 1

The correct option is C $U=\frac{65}{3}(1-3\cot \theta ),V=\frac{65}{3}(1+3\cot \theta ).$

The resultant of the forces is given by $F_1+F_2+F_3+F_4$

$=(\frac{4}{13}U-\frac{4}{13}V+W\cos \theta )i$$+(-\frac{12}{13}U-\frac{12}{13}V+W\sin \theta )j+(\frac{3}{13}U+\frac{3}{13}V-10)k.$

Since the forces are in equilibrium

But $i,j,k$ are non-coplanar, we must have each of the coefficient in R.H.S. equal to zero.

$\frac{4}{13}U-\frac{4}{13}V+W\cos \theta =0;-\frac{12}{13}U-\frac{12}{13}V+W\sin \theta =0$ and $\frac{3}{13}U-\frac{3}{13}V-10=0.$

The above equations can be re-written as $U-V=-\frac{13}{4}W\cos \theta ...\left(1\right)$

$U+V=\frac{13}{12}W\sin \theta ...\left(2\right)$

$U+V=\frac{130}{3}...\left(3\right)$

Equating (2) and (3), we get

$W=40\cos \sec \theta$

From (1) and (2) on putting for W and adding and subtracting, we get

$U=\frac{65}{3}(1-3\cot \theta ),V=\frac{65}{3}(1+3\cot \theta ).$