Problem 510 | Friction

$N_1 = 60 \, \text{ kN}$

$f_1 = \mu N_1 = 0.10(60) = 6 \, \text{ kN}$

$T_1 = f_1 = 6 \, \text{ kN}$
 

 

$N_2 = 40 \cos 30^\circ = 34.64 \, \text{ kN}$

$f_2 = \mu N_2 = 0.10(34.64) = 3.46 \, \text{ kN}$
 

$W = 40 \sin 30^\circ + T_1 + f_2$

$W = 20 + 6 + 3.46$

$W = 29.46 \, \text{ kN}$           answer