Answer
a) $\tau=26.95N.m$
b) $\theta=34^o$
Work Step by Step
The lever arm between the weight $W$ and the rotation axis is $(0.55m)\sin\theta$. Therefore, the torque created by $W$ is $$\tau=W(0.55m)\sin\theta=(49N)(0.55m)\sin\theta=26.95\sin\theta$$
a) When $\theta=90^o$, $\tau=26.95\sin90=26.95N.m$
b) $\tau=15N.m$ when $$\sin\theta=\frac{15}{26.95}=0.557$$ $$\theta=34^o$$
