Question

A force of F = (2.00ˆi + 3.00ˆj) N is applied to an object that is pivoted about a fixed axle aligned along the z coordinate axis. The force is applied at the point r = (4.00ˆi + 5.00ˆj) m.
Find:

(a) the magnitude of the net torque about the z axis

(b) the direction of the torque vector τ

Answer 1

Explanation:

It is given that,

Force applied to object, $F=(2i+3j)\ N$

Position, $r=(4i+5j)\ m$  

(b) The cross product of force and position vector is used to find the net torque about the z axis. It is given by :

$\tau=F\times r$

$\tau=(2i+3j) \times (4i+5j)$

$\tau=\begin{pmatrix}0&0&-2\end{pmatrix}$

or

$\tau=(-2k)\ N-m$

The torque is acting in -z axis.

(a) The magnitude of torque is given by :

$|\tau|=\sqrt{(-2)^2}$

$|\tau|=2\ N-m$

Hence, this is the required solution.