If force F = -3i+j+5k acts along r = 7i+3j+k then the torque acting at that point is (a) (14i-38j+16k) (b) (-14i+34j-16k) (c) (21i+4j+4k) (d) (4i+4j+6k)

Answer: (a). Torque τ = r with r = (7,3,1), F = (-3,1,5). τx = 3(5)-1(1) = 14, τy = -[7(5)-1(-3)] = -38, τz = 7(1)-3(-3) = 16 So the answer is (a) 14i-38j+16k.

If force F = -3i+j+5k acts along r = 7i+3j+k then the torque acting… | MHT CET 2025 (22 Apr, Shift 1)

Q4MHT CET 2025 · 22 APR · SHIFT 1Easy

Torque, Couple & Equilibrium of Rigid Bodies

If force $\vec{F} = -3\hat{i}+\hat{j}+5\hat{k}$ acts along $\vec{r} = 7\hat{i}+3\hat{j}+\hat{k}$ then the torque acting at that point is

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Answer: (a)

Torque $\vec{\tau} = \vec{r}\times\vec{F}$ with $\vec{r} = (7,3,1)$ , $\vec{F} = (-3,1,5)$ .
$\tau_x = 3(5)-1(1) = 14$ , $\tau_y = -[7(5)-1(-3)] = -38$ , $\tau_z = 7(1)-3(-3) = 16$
So the answer is (a) $14\hat{i}-38\hat{j}+16\hat{k}$ .

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