System of Particles and Rotational Motion · Lesson 7 of 15
Torque and the Turning Effect of Forces
“Calculate moments of force and choose useful origins and perpendicular lever arms.”
• Define torque as r × F. • Use force components and perpendicular moment arms. • Identify zero-torque configurations. • Distinguish torque from work despite matching dimensions.
Force alone does not specify turning effect
Push a door near its hinge and then at its outer edge using the same force. The outer push usually turns it more effectively. The turning effect depends on the point of application, the force direction and the reference point or axis. Torque, also called moment of force, combines these features. A force can accelerate a body’s CM while also changing its rotation.
The force’s line of action is the entire straight line along which it acts. The lever arm is perpendicular to that line, not necessarily the length of a rod. Resolve the force into a radial component parallel to r and a perpendicular component. The radial part has no moment; only the perpendicular part contributes. Torque is zero if force vanishes, its application point is the origin, or its line of action passes through the origin.
Problem
A 20 N force acts 0.30 m from a hinge at 30° to the radius.
- 1.The perpendicular force is 20 sin30° = 10 N. Torque magnitude is 0.30(10) = 3.0 N m.
- 2.Equivalently, the lever arm is 0.30 sin30° = 0.15 m, and torque is 20(0.15) = 3.0 N m.
- 3.A perpendicular 20 N force at the same point would give 6.0 N m, the largest possible magnitude for that force and radius.
Direction, components and reference point
Use the right-hand rule for r × F. In a plane, choose an out-of-plane positive axis; anticlockwise moments are then positive and clockwise moments negative. Add torques algebraically only when referring to the same axis. Reversing F alone reverses torque. Reversing both r and F preserves it. A different origin usually changes r and therefore changes the torque.
Problem
Let r = (1, −1, 1) m and F = (7, 3, −5) N.
- 1.The x component is (−1)(−5) − (1)(3) = 2 N m.
- 2.The y component is (1)(7) − (1)(−5) = 12 N m.
- 3.The z component is (1)(3) − (−1)(7) = 10 N m. Thus τ = (2,12,10) N m.
Torque has dimensions ML²T⁻², the same as energy, and SI unit newton metre. Nevertheless, torque is an axial vector describing turning effect, while work is a scalar describing energy transfer. We normally retain N m for torque rather than labelling it joules. Matching dimensions do not make physical quantities interchangeable.
Quiz
A force along a rod through its pivot gives which torque?
Its line of action passes through the pivot, giving zero lever arm.
The moment arm is which distance?
The force times this shortest distance gives the torque magnitude.
Reversing the force direction alone does what?
Cross products are linear in force.
Torque and energy share what?
Their dimensions match but torque is a vector and energy a scalar.
At fixed force and radius, maximum torque occurs at what angle?
The sine factor is 1 for a perpendicular force.
Key Takeaways
• Torque combines force with its location and direction. • Use a perpendicular moment arm or perpendicular force component. • Specify the reference origin or axis and a sign convention. • Equal dimensions do not turn torque into energy.