A force of 40 N acts on a point B at the end of an L -shaped object, as shown in the figure. The angle θ that will produce maximum moment of the force about point A is given by: (a) tan θ = 1/2 (b) tan θ = 2 (c) tan θ = 4 (d) tan θ = 1/4

Answer: (a). Take A as origin: B is at (2, -4) m. F = -40(cos θ i + sin θ j) (pointing into B from upper right). τz = xFy - yFx = 2(-40sin θ) - (-4)(-40cos θ) = -(80sin θ + 160cos θ) Maximum of 80sin θ + 160cos θ: (d)/(dθ) = 80cos θ - 160sin θ = 0 tan θ = 80/160 = 1/2, so the answer is (a).

A force of 40 N acts on a point B at the end of an L -shaped object… | JEE Main 2018 (15 Apr, Shift 1)

Q37JEE Main 2018 · 15 APR · SHIFT 1Medium

Torque, Couple & Equilibrium of Rigid Bodies

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 maximum moment of the force about point $A$ is given by:

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Answer: (a)

Take A as origin: B is at $(2, -4)$ m. $\vec{F} = -40(\cos\theta\,\hat{i} + \sin\theta\,\hat{j})$ (pointing into B from upper right).
$\tau_z = xF_y - yF_x = 2(-40\sin\theta) - (-4)(-40\cos\theta) = -(80\sin\theta + 160\cos\theta)$
Maximum of $80\sin\theta + 160\cos\theta$ : $\frac{d}{d\theta} = 80\cos\theta - 160\sin\theta = 0$
$\tan\theta = \frac{80}{160} = \frac{1}{2}$ , so the answer is (a).

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