Problem 309 | Equilibrium of Concurrent Force System

$P \cos 25^\circ = N \sin 55^\circ$

$P = 0.9038N$
 

$\Sigma F_V = 0$

$P \sin 25^\circ + N \cos 55^\circ = 400$

$(0.9038N) \sin 25^\circ + N \cos 55^\circ = 400$

$0.9556N = 400$

$N = 418.60 \, \text{ lb}$           answer
 

$P = 0.9038(418.60)$

$P = 378.34 \, \text{ lb}$           answer

$P \cos 30^\circ = 400 \sin 55^\circ$

$P = 378.35 \, \text{ lb}$           (ok!)
 

$\Sigma F_y = 0$

$N = P \sin 30^\circ + 400 \cos 55^\circ$

$N = 378.35 \sin 30^\circ + 400 \cos 55^\circ$

$N = 418.60 \, \text{ lb}$           (ok!)

$\dfrac{P}{\sin 55^\circ} = \dfrac{N}{\sin 65^\circ} = \dfrac{400}{\sin 60^\circ}$

$P = 378.35 \, \text{ lb}$           (ok!)

$N = 418.60 \, \text{ lb}$           (ok!)