←
Lesson 5 · The moment of a force
A force with a lever
score 0

A force with a lever

A force applied off a pivot does not just push — it TURNS. That turning effect is the moment (engineers say moment for structures, torque for shafts; the arithmetic is identical). τ=rFsin⁡θ\tau = r F \sin\theta, read aloud tau equals r F sine theta: τ\tau is the moment in newton-metres, rr is the lever arm in metres — pivot to the point where the force acts: FF is the force in newtons, and θ\theta is the angle between the arm and the force.

The sine is the whole subtlety. Pull square to the bar and θ=90∘\theta = 90^\circ, sin⁡θ=1\sin\theta = 1, and every newton turns the bolt. Pull at an angle and only the perpendicular share does any turning; the rest runs along the bar and merely tries to pull it out of your hands. Pull straight along the bar, θ=0\theta = 0, and you get nothing at all — which is precisely why nobody tightens a bolt by pulling the wrench toward themselves.

Two habits worth building. First, the arm always beats the muscle: doubling rr doubles the moment for the same pull, and that is the entire argument for a longer breaker bar. Second, watch the arm's units — a bar quoted in centimetres dropped straight into the formula gives you a moment a hundred times too proud, and it looks perfectly reasonable on the page.

Fθτr

τ=rFsin⁡θ\tau = r F \sin\theta

  • τ\tau= Torque (torque)
  • rr= Lever arm length (length)
  • FF= Applied force (force)
  • θ\theta= Angle between arm and force (angle)
Torque solver →
Fθτr

τ=rFsin⁡θ\tau = r F \sin\theta

  • τ\tau= Torque (torque)
  • rr= Lever arm length (length)
  • FF= Force (force)
  • θ\theta= Angle (angle)
Torque with a Lever Arm (τ = rF sin θ) solver →