Locate the force before calculating its torque
A force diagram for rotation must retain where each force acts. Replacing an extended body by a dot loses the distance and orientation needed to calculate turning effects. Mark the rotation axis, the point of application and the force’s line of action. The position vector runs outward from the chosen axis to the application point.
Learning outcome: Compare 0°, 30°, 90° and 150° at r=0.4 m. Sketch the force at the application point, extend its line of action and mark the perpendicular lever arm. Verify one calculation using Fℓ and another using rF sinφ.
Starting recap
Review signed graph slopes and areas, extended-body force diagrams, Newton’s laws and center-of-mass coordinates. State the rotation axis.
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Try
A 10 N force acts 0.4 m to the right of a pivot at 30° above the outward radius. Its upward component is 5 N, its lever arm is 0.4 sin30°=0.2 m, and its counterclockwise torque is 2 N m. At 150° the horizontal component reverses but the upward component and torque remain the same.
Open the experimentPredict
Can equal force magnitudes at the same radius produce different torques? Can a nonzero force produce none?
Experiment
Compare 0°, 30°, 90° and 150° at r=0.4 m. Sketch the force at the application point, extend its line of action and mark the perpendicular lever arm. Verify one calculation using Fℓ and another using rF sinφ.
Distinguish radius from perpendicular lever arm
Worked teaching model · synthetic values · A 10 N force acts at (r,0), at angle phi counterclockwise from the positive horizontal radius. Pivot is fixed at the origin. The plotted line of action and coordinates are original geometry, not a load-bearing design.
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Force line and perpendicular lever arm
Pivot at (0,0) m; application at (0.4,0) m. Force components (8.6603,5) N. Lever arm 0.2 m.
Application locations and directions are meaningful; force arrows are schematic unless explicitly labeled otherwise. Exact quantities and units follow in the ledger.
| Quantity | Value | Unit |
|---|---|---|
| Force application x | 0.4 | m |
| Force application y | 0 | m |
| Force x component | 8.66025 | N |
| Force y component | 5 | N |
| Perpendicular lever arm | 0.2 | m |
| Counterclockwise torque | 2 | N m |
Accessible plot coordinates
| time (s) | angular velocity (rad/s) |
|---|
Only the force component perpendicular to the radius contributes to the turning effect.
τ=rF sinφ=Fℓ. The angle is between outward radius and force, not between the force and an arbitrary horizontal line in every diagram. Radial force can be nonzero while its torque about this pivot is zero.
Only the force component perpendicular to the radius contributes to the turning effect.
τ=rF sinφ=Fℓ. The angle is between outward radius and force, not between the force and an arbitrary horizontal line in every diagram. Radial force can be nonzero while its torque about this pivot is zero.
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If a force acts through the axis but the application point is far away, why does using the application distance alone give a wrong turning effect?
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