#### Answer

a) 40 Nm
b) 1.3 kN

#### Work Step by Step

a) The torques are due to the arm and due to the weight. Thus, we find:
$\tau = rFsin\theta \\ \tau = (.21)(4.2)(9.81)sin(105)+(.56)(6)(9.81)sin105 \approx 40 \ Nm$
b) $\tau = rF sin \theta \\ F = \frac{\tau}{rsin\theta}=\frac{40.19}{(.18)sin(170)}\approx 1.3 \ kN$