1. 30. (a) Computing torques about point A, we find
L
Tmax L sin θ = W xmax + Wb .
2
We solve for the maximum distance:
Tmax sin θ − Wb
2 500 sin 30◦ − 200
2
xmax = L= (3.0) = 1.5 m .
W 300
(b) Equilibrium of horizontal forces gives
Fx = Tmax cos θ = 433 N .
(c) And equilibrium of vertical forces gives
Fy = W + Wb − Tmax sin θ = 250 N .