Question

What is the torque about the origin of the force (5.0ˆi−2.0ˆj+1.0ˆk)N if it is applied at the point whose position is: →r=(−2.0ˆi+4.0ˆj)m?

a) 10.0 N·m
b) 13.0 N·m
c) 15.0 N·m
d) 20.0 N·m

Answer 1

The torque about the origin of the given force can be calculated by finding the cross product of the force vector and the position vector. The torque is ( -4.0î + 8.0ĵ + 20.0k)N.The correct option is d) 20.0 N·m.

Step-by-step explanation:

To calculate the torque about the origin, we need to find the cross product of the force vector and the position vector. The torque equation is given by:

τ = r × F

Given that the force vector is (5.0î - 2.0ĵ + 1.0k)N and the position vector is (-2.0î + 4.0ĵ)m, we can calculate the torque as follows:

τ = (-2.0î + 4.0ĵ) × (5.0î - 2.0ĵ + 1.0k)

= (-2.0 × 1.0)î × î + (4.0 × 5.0)ĵ × î + (-2.0 × 1.0)î × ĵ + (4.0 × 1.0)ĵ × ĵ + (-2.0 × 2.0)î × k + (4.0 × 2.0)ĵ × k

= 0 + 20.0k + 0 + 4.0k + -4.0î + 8.0ĵ

Therefore, the torque about the origin is (0 - 4.0î + 8.0ĵ + 0 + 20.0k)N, which can be written as ( -4.0î + 8.0ĵ + 20.0k)N. The correct option is d) 20.0 N·m.