Question

For the truss shown in figure, the force in the member PQ is

• A. 4600 N
• B. 4807.4 N
• C. 5605.55 N
• D. 3000 N

Answer 1

The correct option is B 4807.4 N


$\sum F_x=0$

$H_p=\mathrm{0...............}\left(1\right)$

$\sum Fy=0$

$V_P-5000-2000+R_A=\mathrm{0........}\left(2\right)$

$\sum M_P=0$

$5000\times 2+2000\times 4-R_A\times 6=0$

$\therefore R_A=\frac{18000}{6}=3000N........\left(3\right)$

Substituting (3) in (2)

$V_P=4000N$

Take the joint, P

$\sum F_y=0$

$V_P=F_{PQ}\sin \theta$

(Assume $F_{PQ}$ in downward direction)

$4000=F_{PQ}\times \frac{3}{\sqrt{13}}$

$\therefore F_{PQ}=4807.4N$