Question

Torque, like angular velocity and angular acceleration, is a vector quantity. Most precisely, it is the cross product of the displacement vector, $r$, from the axis of rotation to the point where the force is applied, and the vector for the applied force, $F$.
$T=r\times F$
To determine the direction of the torque vector, use the right-hand rule, curling your fingers around from the $r$ vector over to the $F$ vector. In the example of lifting the lever, the torque would be represented by a vector at $0$ pointing out of the page.

A student exerts a force of $50N$ on a lever at a distance $0.4m$ from its axis of rotation. The student pulls at an angle that is ${60}^o$ above the lever arm. What is the torque experienced by the lever arm?

Answer 1

Torque about a point $\tau =rF$ $sin\theta$ where $\theta$ is the angle between $r$ and $F$

Given : $F=50$ N $r=0.4$ m $\theta ={60}^o$

$\therefore$ $\tau =0.4\times 50\times 0.866=17.32$ N m (out of the page)