Question

Find the horizontal force and a force inclined at an angle of ${60}^{\circ }$ to the vertical whose resultant is a vertical force p kg wt.

• A. $F_2=3P$ $F_1=2\sqrt{3}P.$
• B. $F_2=2\sqrt{3}P$ $F_1=3P.$
• C. $F_2=\sqrt{3}P$ $F_1=2P.$
• D. $F_2=2P$ $F_1=\sqrt{3}P.$

Answer 1

Answer: (D)

$F_2=2P$ $F_1=\sqrt{3}P.$

$Pj=F_{1}i+(-F_2\cos {30}^{\circ }i+F_2\sin {30}^{\circ }j)$

Equating the coefficients of i and j, we get $F_1-F_2\cdot \frac{\sqrt{3}}{2}=0$ and $F_2\cdot \frac{1}{2}=P.$

$\therefore F_2=2P$ and $F_1=\sqrt{3}P.$