A force F⃗ = 4î + 3ĵ + 4k̂ 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)

Solution

Physics sir · 4 steps

Measure r⃗ from the point you take torque about, not from the origin, then cross it with F⃗.

1. Point of application. On the x-axis, y = 0 and z = 0. The plane x = 2 cuts it at (2, 0, 0), so the force acts there.
2. Position vector. Torque about (2, 3, 4) needs r⃗ from that point to the point of application: r⃗ = (2-2)î + (0-3)ĵ + (0-4)k̂ = -3ĵ - 4k̂.

   τ⃗ = r⃗ × F⃗ = |î, ĵ, k̂; 0, -3, -4; 4, 3, 4| = 0 î - 16 ĵ + 12 k̂

3. Expand. The î term is (-3)(4) - (-4)(3) = 0. The ĵ term is -[(0)(4) - (-4)(4)] = -16. The k̂ term is (0)(3) - (-3)(4) = 12.
4. So. |τ⃗| = √(0² + 16² + 12²) = √400 = 20.

The common slip: 10. Taking r⃗ = 2î from the origin gives τ⃗ = -8ĵ + 6k̂ and a magnitude of 10, which is torque about the origin, not about (2, 3, 4).

The idea to keep. Torque about a point P always uses r⃗ = r⃗_(application) - r⃗_P. Only when P is the origin does r⃗ equal the position vector of the point of application.

Answer from the NTA final answer key. Solution by PrepWiser.