Question

A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from the pivot causing a ccw torque, and a force of 2.60 N is applied at the end of the rod 3.00 m from the pivot. The 2.60-N force is at an angle of 30.0o to the rod and causes a cw torque. What is the net torque about the pivot?

Answer 1

From the definition we have that the Torque corresponds to the Multiplication between the Force (or its respective component) and the radius of distance of the Force to the inertial turning point.

Mathematically this can be expressed,

$\tau = F \times d$

Where,

F = Perpendicular component of force

d = distance from pivot point

The total sum of the torques would be equivalent to

$\tau_{net} = \tau_1 +\tau_2$

According to the values given, torque 1 and 2 would be given by

$\tau_1 = 6*1.2 = 7.2N\cdot m (+)$

$\tau_2 = -5.2sin(30) = -7.8N\cdot m (-)$

Therefore the net Torque is

$\tau_{net} = \tau_1+\tau_2$

$\tau_{net} = 7.2-7.8$

$\tau_{net} = -0.6N\cdot m$

Therefore the net torque about the pivot is -0.6Nm