Question

A 3.0-m rod is pivoted about its left end. A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a ccw torque, and a force of 5.2 N is applied at the end of the rod 3.0 m from the piVOT?

Answer 1

Answer:

-0.6Nm

Explanation:

Torque is defined as the perpendicular component of force \times distance from pivot point.

#$T_n_e_t=\sum{T_i}$ where $T_i$ is the torques from different applied forces.

$T_1=6N\times1.2m=7.2Nm$  (+ve, counterclockwise)

$T_2=-5.2N\times sin30\textdegree\times 3m\\=-7.8Nm$(-ve, clockwise)

$T_n_e_t=T_1+T_2\\=(7.2-7.8)Nm\\=-0.6Nm$

Hence, the total torque is -0.6Nm