Question

The uniform bar of length l 40 N and is subjected to four forces shown find the magnitude location and direction of a fith force F needed to keep the bar in equilibrium

Answer 1

$F_x=80\cdot\cos30°=69.282\ (N)$

$F_y=60+70+40-50-80\cdot\sin30°=80\ (N)$

$F=\sqrt{69.282^2+80^2}=105.83\ (N)$

$\sum\tau_a=0\to70\cdot0.2L+40\cdot0.5L+60\cdot L-50\cdot0.8L-$

$-80\cdot xL=0\to x=0.675$

$\alpha=\tan^{-1}(80/69.282)=49.1°$

Answer:

$F=105.83\ (N)$

$\alpha=49.1°$

$x=0.675L$