Problem 313 | Equilibrium of Concurrent Force System

$\Sigma F_x = 0$

$F \cos 60^\circ + 300 = P \cos 15^\circ + 400 \cos 30^\circ$

$F = 1.9318P + 92.82$
 

 

$\Sigma F_y = 0$

$F \sin 60^\circ + P \sin 15^\circ = 200 + 400 \sin 30^\circ$

$(1.9318P + 92.82) \sin 60^\circ + P \sin 15^\circ = 200 + 400 \sin 30^\circ$

$1.6730P + 80.38 + 0.2588P = 200 + 200$

$1.9318P = 319.62$

$P = 165.45 \, \text{ lb}$           answer
 

$F = 1.9318(165.45) + 92.82$

$F = 412.44 \, \text{ lb}$           answer