Question

A uniform bar has a mass of 2.2 kg and the angle shown is 30 degrees. Calculate the net torque about the point P for the bar. Do not ignore the weight of the bar.

Answer 1

The free body diagram of the bas is shown below:

The torque is given by:

`\tau=r_(\perp)F`

where F is the force and:

`r_(\perp)`

is the perpendicular distance between the rotation axis and the line of action of the force, which we called the moment arm. We know that the torque follows the principle of superposition then to find the total torque we need to add the torque each force exert. Before we do this we need to find the correct sign of the torque so we need to remember that if a force makes an object rotate counterclockwise then the torque is positive; otherwise it is negative. In this case Force two will exert a positive torque while the other two forces acting on the bar exert a negative torque.

For force one and the weight the moments arms are 5 m and 2.5 m, respectively. This comes from the fact that the forces are perpendicular to the rod.

For force two the moment arm is given as:

`r_(\perp)=2\sin 30=1`

Hence the total torque is given by:

`\begin{gathered} \tau=F_2r_(\perp2)+Wr_(\perp w)+F_1r_(\perp1) \\ \tau=(150)(1)-(9.8)(2.2)(2.5)-(20)(5) \\ \tau=-3.9 \end{gathered}`

Therefore, the total torque on the bar is -3.9 Nm (this means that the bar will rotate clockwise).