Question

Consider a bar, of length 3.5 m, shown in the figure, being acted on by three forces, and constrained to rotate about it's left end. The magnitudes of the first two forces are 12 N and 21 N, and the first force is acting on the end of the bar at an angle of 45° as shown. what is torque one due to F1.

Answer 1

Given,

The length of the bas, L=3.5 m

The first force, F₁=12 N

The second force, F₂=21 N

The angle at which the first force is acting on the bar, θ=45°

The torque is the product of the force, the distance at which the force is applied, and the sine of the angle between the force and the bar.

Thus the torque due to the 1st force is given by,

`\tau=F_1L\sin \theta`

On substituting the known values,

`\begin{gathered} \tau=12*3.5*\sin 45^(\circ) \\ =29.7\text{ Nm} \end{gathered}`

Thus the torque due to the first force is 29.7 Nm