A force F → = 4 i ^ + 3 j ^ + 4 k ^ is applied on an intersection point of x = 2 plane and x -axis.…

A force F=4i^+3j^+4k^ 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

τ=r×F

r=2i^-2i^+3j^+4k^=-3j^-4k^

and F=4i^+3j^+4k^

τ=r×F=i^j^k^0-3-4434

=i^-12+12-j^0+16+k^0+12

=-16i^+12k^

 |τ|=162+122=20

Asked in: JEE Main 2021 (16 Mar Shift 2)

Practice more Rotational Motion questions on Aicharya