Question

Calculate the torque (magnitude and direction) about point O due to the force$\overrightarrow{\mathbf{F}}$in each cases sketched in Fig. In each case, both the force $\overrightarrow{\mathbf{F}}$and the rod lie in the plane of the page and the rod has length 4.00 m, and the force has magnitude $\mathit{F}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{N}\mathbf{.}$

(a)

(b)

(c)

(d)

(e)

(f)

Short answer

(a) Torque is 40 Nm directed out of the plane of the figure.

(b) Torque is 34.6 Nm directed out of the plane of the figure.

(c) Torque is 20 Nm directed out of the plane of the figure.

(d) Torque is 17.3 Nm directed into the plane of the figure.

(e) Torque is zero.

(f) Torque is zero.

Step-by-step solution

Given that

Length of the rod$=4m$

Force applied $F=10N$

Formulas used

The magnitude of torque $\tau =Fl$where

F is force and l is the distance

$l=r\sin \theta$where

r is distance and$\theta$ is the angle between r and F

Right-hand thumb rule: Take your right hand and point it in direction of the position vector and turn the fingers in the direction of the force and the thumb will point toward the direction of the torque.

Find Torque

(a)

Force$F=10N$

Distance$r=4m$

Angle $\theta =90°$

So

$$\begin{gathered} l=r\sin \theta \\ =\left(4m\right)\sin 90{}^{\circ } \\ =4m \end{gathered}$$

Hence Torque $\tau =\left(10N\right)\left(4m\right)=40Nm$

This force is producing counterclockwise rotation, so by right-hand thumb rule, the torque is directed out of the plane of the figure.

Find Torque

(b)

Force$F=10N$

Distance$r=4m$

Angle $\theta =120°$

So

$$\begin{gathered} l=r\sin \theta \\ =\left(4m\right)\sin 120{}^{\circ } \\ =\left(4\right)\left(\frac{\sqrt{3}}{2}\right) \\ =3.464m \end{gathered}$$

Hence Torque $\tau =\left(10N\right)\left(3.464m\right)=34.6Nm$

This force is producing counterclockwise rotation, so by right-hand thumb rule, the torque is directed out of the plane of the figure.

Find Torque

(c)

Force$F=10N$

Distance$r=4m$

Angle $\theta =30°$

So

$$\begin{gathered} l=r\sin \theta \\ =\left(4m\right)\sin 30{}^{\circ } \\ =\left(4\right)\left(\frac{1}{2}\right) \\ =2m \end{gathered}$$

Hence Torque$\tau =\left(10N\right)\left(2m\right)=20Nm$

This force is producing counterclockwise rotation, so by right-hand thumb rule, the torque is directed out of the plane of the figure.

Find Torque

(d)

Force $F=10N$

Distance$r=2m$

Angle $\theta =60°$

So

$$\begin{gathered} l=r\sin \theta \\ =\left(2m\right)\sin 60{}^{\circ } \\ =\left(4\right)\left(\frac{\sqrt{3}}{2}\right) \\ =1.732m \end{gathered}$$

Hence Torque $\tau =\left(10N\right)\left(1.732m\right)=17.3Nm$

This force is producing clockwise rotation, so by right-hand thumb rule, the torque is directed into the plane of the figure.

Find Torque

(e)

Force $F=10N$

Distance $r=0m$

Angle $\theta =60°$

So

$$\begin{gathered} l=r\sin \theta \\ =0 \end{gathered}$$

Hence Torque $\tau =\left(10N\right)\left(0m\right)=0$

Find Torque

(f)

Force $F=10N$

Distance$r=0m$

Angle$\theta =180°$

So

$$\begin{gathered} l=r\sin \theta \\ =0 \end{gathered}$$

Hence Torque $\tau =\left(10N\right)\left(0m\right)=0$