A force of $40\ \text{N}$ acts on a point $B$ at the end of an L-shaped object as shown in the figure. The angle $\theta$ that will produce the maximum moment of the force about point $A$ is given by:
Answer: (C) $\tan\theta = \dfrac12$
Take $A$ as origin with $x$ to the right and $y$ upward. Then $B$ is at $\vec r = (2, -4)\ \text{m}$.
The force at $B$ points downward to the left, making angle $\theta$ with the horizontal:
$$\vec F = 40(-\cos\theta,\ -\sin\theta)$$
Moment about $A$:
$$\tau = r_xF_y - r_yF_x = 2(-40\sin\theta) - (-4)(-40\cos\theta) = -(80\sin\theta + 160\cos\theta)$$
The magnitude $80\sin\theta + 160\cos\theta$ is maximum when
$$\frac{d}{d\theta}(80\sin\theta + 160\cos\theta) = 80\cos\theta - 160\sin\theta = 0 \;\Rightarrow\; \tan\theta = \frac12$$
(Then $\vec F$ is perpendicular to $AB$, which has the largest lever arm $AB = \sqrt{20}\ \text{m}$.)
Solution by Sreeraj P, M.Sc Physics