JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Rotational Mechanics

A force

F = 4 \hat{i} + 3 \hat{j} + 4 \hat{k}

is applied on an intersection point of

x = 2

plane and

x

-axis. The magnitude of torque of this force about a point

(2, 3, 4)

is ⋯---------. (Round off to the Nearest Integer)

Answer: 20

Solution

The plane

x = 2

meets the

x

-axis at

(2, 0, 0)

, so the position of the point of application relative to

(2, 3, 4)

r = (2 - 2) \hat{i} + (0 - 3) \hat{j} + (0 - 4) \hat{k} = - 3 \hat{j} - 4 \hat{k}

The torque about

(2, 3, 4)

τ = r \times F = \hat{i} 04 \hat{j} - 3 3 \hat{k} - 4 4 = (- 12 + 12) \hat{i} - (0 + 16) \hat{j} + (0 + 12) \hat{k} = - 16 \hat{j} + 12 \hat{k}

Its magnitude is

∣ τ ∣ = 1 6^{2} + 1 2^{2} = 400 = 20

so the torque of the force about that point has magnitude

20 N m

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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