A slab is subjected to two forces F₁ and F₂ of same magnitude F as shown in the figure. Force F₂ is in XY-plane while force F₁ acts along z = axis at the point (2i + 3j).. The moment of these forces about point O will be: (a) (3i - 2j - 3k)F (b) (3i + 2j - 3k)F (c) (3i + 2j + 3k)F (d) (3i - 2j + 3k)F

Answer: (d). F₁ = Fk at (2i+3j): τ₁ = (2i+3j)× Fk = F(3i-2j). F₂ acts at (0,6,0), making 30° with the y-axis and pointing towards -x, -y: F₂ = -Fsin 30 - Fcos 30. τ₂ = 6j×(-F/2i) = 3Fk (the j part gives zero). Total: τ = (3i-2j+3k)F, so the answer is (d).

A slab is subjected to two forces F₁ and F₂ of same magnitude F as… | JEE Main 2019 (11 Jan, Shift 1)

Q34JEE Main 2019 · 11 JAN · SHIFT 1Medium

Torque, Couple & Equilibrium of Rigid Bodies

A slab is subjected to two forces $\vec{F_1}$ and $\vec{F_2}$ of same magnitude F as shown in the figure. Force $\vec{F_2}$ is in XY-plane while force $\vec{F_1}$ acts along z = axis at the point $\left(2\vec{i} + 3\vec{j}\right).$ . The moment of these forces about point O will be:

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Answer: (d)

$\vec{F_1} = F\hat{k}$ at $(2\hat{i}+3\hat{j})$ : $\vec{\tau_1} = (2\hat{i}+3\hat{j})\times F\hat{k} = F(3\hat{i}-2\hat{j})$ .
$\vec{F_2}$ acts at $(0,6,0)$ , making $30^\circ$ with the y-axis and pointing towards $-x$ , $-y$ : $\vec{F_2} = -F\sin 30^\circ\hat{i} - F\cos 30^\circ\hat{j}$ .
$\vec{\tau_2} = 6\hat{j}\times\left(-\frac{F}{2}\hat{i}\right) = 3F\hat{k}$ (the $\hat{j}$ part gives zero).
Total: $\vec{\tau} = (3\hat{i}-2\hat{j}+3\hat{k})F$ , so the answer is (d).

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Torque, Couple & Equilibrium of Rigid Bodies question from JEE Main 2019 (11 Jan, Shift 1)

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