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Question

A force F=4^i+3^j+4^K is applied on an intersection point of x=2 plane and xaxis. The magnitude of torque of this force about a point (2,3,4) is (Round off to the Nearest Integer)

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Solution


Given: F=4^i+3^j+4^k

Position of intersection point where force is applied, rB=2^i

Point about which torque to be calculated, rA=2^i+3^j+4^k

Now, from figure rA+r=rB

r=rBrA

r=(3^j4^k)

We know that,

τ=r×F

τ=∣ ∣ ∣^i^j^k034434∣ ∣ ∣

=16^j12^k=(16)2+(12)2

=400=20

Correct Answer: 20

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