When a mechanic tries to remove a hex nut with a wrench, it is easier if the force is applied at the farthest end of the wrench handle. The lever arm is the distance from the pivot point (the hex nut in this case) to the person’s hand. If this distance is large, the torque is higher. Only the component of the force perpendicular to the lever arm contributes to the torque. Therefore, pushing the wrench perpendicular to the lever arm is more advantageous. If multiple people apply force to rotate the same wrench, then the torque applied by each person simply adds up to generate the resultant torque.
Torque problems can be solved by following the step-by-step strategy listed below:
This text is adapted from Openstax, University Physics Volume 1, Section 10.6: Torque.
Consider a meter scale pivoted at its center. Three forces of 3.0 N, 5.0 N, and 2.0 N are acting on it at distances of 0.25 meters, 0 meters, and 0.5 meters, respectively.
The angles made by the forces with the scale are 60°, 90°, and 30°, respectively.
Recall that the magnitude of the torque is expressed as the product of the lever arm and the sine component of the force.
As the values for the applied force, lever arm, and the angles are known, the torque due to each force can be calculated.
Therefore, the magnitude of torque τ1 is 0.650 Nm. Since the rotation of the scale is clockwise, due to F1, the torque τ1 is negative.
The force F2 is acting at the pivot point. Therefore, the torque τ2 is zero. The magnitude of the torque τ3 is equal to 0.5 Nm. Since the scale rotates counterclockwise, the torque is positive.
Therefore, the net torque on the scale is −0.15 Nm, directed into the plane of the paper.
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