Section 5.2: Torque
Torque is the rotational equivalent of force. It measures how effectively a force causes an object to rotate about an axis. Torque depends not only on the magnitude of the force but also on the distance from the axis of rotation and the angle at which the force is applied.
\( \tau \) = torque (N·m)
\( r \) = lever arm (distance from pivot) (m)
\( F \) = applied force (N)
\( \theta \) = angle between force and lever arm
If the force is applied perpendicular to the lever arm (\( \theta = 90^\circ \)), torque is maximized. If the force is applied along the lever arm (\( \theta = 0^\circ \)), torque is zero.
Example: Opening a Door
A person applies a force of 40 N perpendicular to a door at a distance of 0.8 m from the hinges. Find the torque applied on the door.
Since the force is perpendicular (\( \theta = 90^\circ \)): \( \tau = rF = 0.8 \times 40 = 32 \, \text{N·m} \). The torque applied is 32 N·m.
Example: Wrench at an Angle
A 50 N force is applied at the end of a 0.3 m wrench at an angle of 60° to the wrench. Find the torque about the pivot.
\( \tau = rF \sin \theta = 0.3 \times 50 \times \sin 60^\circ \). \( \tau = 15 \times 0.866 \approx 13 \, \text{N·m} \). The torque is approximately 13 N·m.
Practice Problems
- A force of 25 N is applied perpendicular to a lever arm of length 0.6 m. Find the torque.
- A 100 N force is applied at 30° to a lever of length 0.5 m. Calculate the torque.
- What is the torque when a 60 N force acts along a 1.2 m lever arm?
- A child pushes a swing with a force of 20 N at a distance of 1.5 m from the pivot, at an angle of 45°. Find the torque.
- Why does using a longer wrench make loosening a bolt easier?