Question

A force of (2i-4j+2k) N act at a point
(3i+2j; -4k) m from the origin. The
magnitude of torque is
(a) zero (b) 0.244 Nm
(c) 24.4 Nm (d) 2.444 Nm​

Answer 1

Answer:

c) 24.41 Nm

Explanation:

Torque=force*radius

cross multiply the vectors

|i. j k|

|2 -4 2|

|3 2 -4|

=12i +14j +16k

manitude=√(144+196+256)=24.4Nm

Answer 2

The magnitude of torque is 24.4 N-m.

Explanation:

It is given that,

Force acting on the particle, $F=(2i-4j+2k)\ N$

Position of the particle, $r=(3i+2j-4k)\ m$

Let $\tau$ is the torque acting on the particle. It is given by :

$\tau=F\times r$

$\tau=(2i-4j+2k)\ N\times (3i+2j-4k)\ m)$

$\tau=\begin{pmatrix}12&14&16\end{pmatrix}$

$\tau=12i+14j+16k$

$|\tau|=\sqrt{12^2+14^2+16^2}$

$\tau=24.4\ N-m$

So, the magnitude of torque is 24.4 N-m. Hence, this is the required solution.

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Torque

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