Question

Torque of a force F = 10(î– k) acting on point (1,0,2) about a point (1,1,0) is:
(a) (20i – 10j) units
(b) (10î – 20j)units
(c) (-10î + 20j)units
(a) (10î – 20k )units​

Answer 1

Answer:

10i+20j+10k

Explanation:

Use the simple 3x3 matrix and solve.

+ options are misleading here.

(I guess you are an allen student)

Answer 2

torque of a force F acting on point (1, 0, 2) about a point (1, 1, 0) is 10i + 20j + 10k

It is given that force, F = 10(i - k) acting on point (1, 0, 2) about a point (1, 1, 0).

we have to find the torque of force F acting on point (1, 0, 2) about a point (1, 1, 0).

first find position vector of point (1, 0, 2) with respect to (1, 1, 0).

i.e., r = (1 i + 0 j + 2k) - (1 i + 1 j + 0 k) = -1 j + 2 k

now torque = r × F

= (-1 j + 2 k) × 10(i - k)

= 10 [(-j + 2 k) × (i - k) ]

= 10(i + 2j + k)

= 10j + 20j + 10k

thereforce torque of a force F acting on point (1, 0, 2) about a point (1, 1, 0) is 10i + 20j + 10k