Problem 516 | Friction

$N_1 = 120 \cos 30^\circ = 103.92 \, \text{ lb}$

$f_1 = 0.60(103.92) = 62.35 \, \text{ lb}$
 

$N_2 = N_1 + 200 \cos 30^\circ$

$N_2 = 103.92 + 173.20$

$N_2 = 277.12 \, \text{ lb}$
 

$f_2 = 0.60(277.12) = 166.27 \, \text{ lb}$
 

$P + 200 \sin 30^\circ = f_1 + f_2$

$P + 100 = 62.35 + 166.27$

$P = 128.62 \, \text{ lb}$           answer
 

$T = f_1 + 120 \sin 30^\circ$

$T = 62.35 + 60$

$T = 122.35 \, \text{ lb}$           answer