Question

Q 4. ​A bar 2 m long makes an angle of 30^o with the horizontal. A vertical force of 40 N is applied at a point 0.4 m from the upper end. Calculate the torque due to this force about each end.

Answer 1

Dear Student,
Expression for torque:

$\tau =Fr\cos \theta$

Torque due to the force about lower end:

$$\begin{gathered} {\tau }_1=\left(40\right)\left(2-0.4\right)\cos {30}^0 \\ =40\times 1.6\times \cos {30}^0 \\ =55.4\mathrm{N}.\mathrm{m} \end{gathered}$$

Torque due to the force about upper end:

$$\begin{gathered} {\tau }_2=\left(40\right)\left(0.4\right)\cos {30}^0 \\ =40\times 0.4\times \cos {30}^0 \\ =13.9\mathrm{N}.\mathrm{m} \end{gathered}$$
Regards