Question

Calculate the net torque (magnitude and direction) on the beam inthe figure below about the following axes.

(a) an axis through O, perpendicularto the page
1 N·m 2---Direction---clockwisecounterclockwise

(b) an axis through C, perpendicular to the page
3 N·m 4---Direction---counterclockwiseclockwise

Answer 1

Concepts and reason

The concepts required to solve the problem is torque.

Initially, find the magnitude of the net torque on the beam about an axis passing through the point O using the relation between torque, force and the distance from the axis where the torque acts. From the sign of the torque, find the direction of the torque.

Then find the magnitude of the net torque on the beam about an axis passing through the point C using the relation between torque, force and the distance from the axis where the torque acts. Again, find the direction of the torque based on the sign of the torque.

Fundamentals

Torque is defined as the rotational effect of force. When a force acts on a body at a distance away from the axis, then the force causes rotation. The torque is given by the equation,

$\vec \tau = \vec r \times \vec F$

Here, $\vec r$ is the position vector from the position where the torque is calculated to the position where the force acts and $\vec F$ is the force vector.

Consider the direction of the positive torques is counter clockwise and the negative torque is clockwise.

In Scalar form the torque is given by,

$\tau = rF\sin \theta$

Here, $F$is the magnitude of the force, $r$is the perpendicular distance from the axis to the point where the force acts, and$\theta$ is the angle between the force vector and the lever arm.

(a.1)

There are three forces acting on the beam. The net torque on the beam about an axis passing through O is the sum of the torque due to the force$30\,{\rm{N}}$, $25\,{\rm{N}}$and$10\,{\rm{N}}$.

$\begin{array}{c}\\\tau = - \left( {30\,{\rm{N}}} \right)\left( {0\,{\rm{m}}} \right)\sin 45^\circ + \left( {25\,{\rm{N}}} \right)\left( {2.0\,{\rm{m}}} \right)\sin \left( {90^\circ - 30^\circ } \right) - \left( {10\,{\rm{N}}} \right)\left( {4.0\,{\rm{m}}} \right)\sin 20^\circ \\\\ = 29.6\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

(a.2)

The direction of the torque is determined by the sign of the torque. Here, the torque of the beam is,

$\tau = 29.6\,{\rm{N}} \cdot {\rm{m}}$

As, the sign of the torque is positive, the rotation is counter clockwise.

(b.1)

The net torque on the beam about an axis passing through C is the sum of the torque due to the force$30\,{\rm{N}}$, $25\,{\rm{N}}$and$10\,{\rm{N}}$.

$\begin{array}{c}\\\tau = \left[ {\left( {30\,{\rm{N}}} \right)\left( {2.0\,{\rm{m}}} \right)\sin 45^\circ } \right] + \left[ {\left( {25\,{\rm{N}}} \right)\left( {0\,{\rm{m}}} \right)\cos 30^\circ } \right] - \left[ {\left( {10\,{\rm{N}}} \right)\left( {2.0\,{\rm{m}}} \right)\sin 20^\circ } \right]\\\\ = 35.6\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

(b.2)

The direction of the torque is determined by the sign of the torque. Here, the torque of the beam is,

$\tau = 35.6\,{\rm{N}} \cdot {\rm{m}}$

As, the sign of the torque is positive, the rotation is counter clockwise.

Ans: Part a.1

The net torque on the beam about an axis passing through O is$29.6\,{\rm{N}} \cdot {\rm{m}}$.

Part a.2

The direction of the net torque on the beam about an axis passing through O is counter clockwise.

Part b.1

The net torque on the beam about an axis passing through C is$35.6\,{\rm{N}} \cdot {\rm{m}}$.

Part a.2

The direction of the net torque on the beam about an axis passing through C is counter clockwise.