Answer
(a) $9.56 N \times m$ (b) $8.83N \times m$
Work Step by Step
(a) Torque is equal to $$\tau=rFsin\theta$$ Substituting known values of $r=0.605m$, $F=w=(1.61kg)(9.81m/s^2)=15.8N$ and $\theta=90^{\circ}$ yields a torque of $$\tau=(0.605m)(15.8N)sin(90^{\circ})=9.56N \times m$$ (b) Substituting known values of $r=0.605m$, $F=w=(1.61kg)(9.81m/s^2)=15.8N$ and $\theta=90^{\circ}-22.5^{\circ}=67.5^{\circ}$ yields a torque of $$\tau=(0.605m)(15.8N)sin(67.5^{\circ})=8.83N \times m$$
