Question

A 3.0 meters long solid shaft makes an angle of 35 degrees with the horizontal. A vertical force of 50 N
is applied 0.60 m from the upper end. What is the magnitude of the torque due to the vertical force
about each end?

Answer 1

Draw the diagram:

The torques from the vertical force around the upper end is the lever times vertical component of the force:

$$\tau_u=-F\text{ cos}\theta\cdot l=50\text{ cos}35°\cdot0.6=-24.6\text{ Nm}.$$

Around the lower end:

$$\tau_l=F\text{ cos}\theta\cdot (L-l)=50\text{ cos}35°(3-0.6)=\\=98.3\text{ Nm}.$$