Chapter 10 Problem 25

Problem

Calculate the net torque about the axle of the wheel shown. Assume that a friction torque of 0.40 m.N opposes the motion.

Solution

Each force is oriented so that it is perpendicular to its lever arm. Call counterclockwise torques positive. The torque due to the three applied forces is given by the following.

$\tau _{\text{applied}}=\left(28{\text{N}}\right)\left(0.24{\text{m}}\right)-\left(18{\text{N}}\right)\left(0.24{\text{m}}\right)-\left(35{\text{N}}\right)\left(0.12{\text{m}}\right)=-1.8{\text{m.N}}$

Since this is negative, the net torque is clockwise. Thus the frictional torque is counterclockwise.

$\tau _{\text{net}}=\left(28{\text{N}}\right)\left(0.24{\text{m}}\right)-\left(18{\text{N}}\right)\left(0.24{\text{m}}\right)-\left(35{\text{N}}\right)\left(0.12{\text{m}}\right)+0.40{\text{m.N}}=-1.4{\text{m.N}}$

clockwise $1.4{\text{m.N}}$