A force F = (i + 2j + 3k) N acts at a point (4i + 3j - k) m. Then the magnitude of torque about the point (i + 2j + k) m will be √(x) N m. The value of x is _______. Numerical answer type. Work it out, then open the answer.

Answer: 195. r = (4,3,-1)-(1,2,1) = (3,1,-2) m, F = (1,2,3) N. τx = 1(3)-(-2)(2) = 7, τy = -[3(3)-(-2)(1)] = -11, τz = 3(2)-1(1) = 5 |τ|² = 49+121+25 = 195, so |τ| = √(195) N m. x = 195

A force F = (i + 2j + 3k) N acts at a point (4i + 3j - k) m. Then the… | JEE Main 2020 (5 Sep, Shift 1)

Q28JEE Main 2020 · 5 SEP · SHIFT 1Easy

Torque, Couple & Equilibrium of Rigid Bodies

A force $\vec{F} = \left(\hat{i} + 2\hat{j} + 3\hat{k}\right)$ N acts at a point

$\left(4\hat{i} + 3\hat{j} - \hat{k}\right)$ m. Then the magnitude of torque

about the point $\left(\hat{i} + 2\hat{j} + \hat{k}\right)$ m will be $\sqrt{x}$ N m.

The value of x is _______.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 195

$\vec{r} = (4,3,-1)-(1,2,1) = (3,1,-2)$ m, $\vec{F} = (1,2,3)$ N.
$\tau_x = 1(3)-(-2)(2) = 7$ , $\tau_y = -[3(3)-(-2)(1)] = -11$ , $\tau_z = 3(2)-1(1) = 5$
$|\vec{\tau}|^2 = 49+121+25 = 195$ , so $|\vec{\tau}| = \sqrt{195}$ N m.
x = 195

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