Connect torque to force and lever arm

Force location changes its turning effect

Imagine pressing downward on a beam that can rotate about a pin at its center. A downward force on the right tends to turn the beam clockwise. The same downward force on the left tends to turn it counterclockwise. A force applied through the pin has no turning effect about that pin, even though the force itself is not zero.

The lab compares initial horizontal configurations of a uniform 4 m beam. The beam has mass 2 kg and the ideal pin is at its center. One fixed external force of 10 N points downward 1 m left of the pin. You choose another downward force and its application distance on the right. These are prescribed external forces, not extra masses attached to the beam.

The diagram marks application locations on a fixed distance scale. It is not a drawing of scaled force arrows. The ledger gives force magnitudes, signs and torques. Moving the adjustable marker changes where the force acts; it does not change the beam length or its mass distribution.

Define torque about a particular pivot

Torque describes the turning effect of a force about a chosen point or axis. For a force perpendicular to a lever, its magnitude is force multiplied by perpendicular distance from the pivot to the force’s line of action. The unit is newton-metre, N·m.

Here positions x are measured along the horizontal beam from the pin, with right positive. Vertical force Fy is positive upward. The signed torque is τ = x Fy. We choose counterclockwise torque as positive. The symbol τ, read ‘tau,’ means torque in N·m, x is metres and Fy is newtons.

The fixed left force has x = −1 m and Fy = −10 N. Its torque is (−1) × (−10) = +10 N·m, counterclockwise. An adjustable 10 N downward force at x = +1 m has torque (+1) × (−10) = −10 N·m, clockwise. These torques cancel even though both applied forces point downward.

Read OpenStax torque for the general lever-arm definition. The widget restricts applied loads to vertical forces on a horizontal beam so that the geometry can be checked directly before introducing angled forces.

Hold force fixed and change distance

Keep adjustable force at 10 N. At x = 0, its torque is zero. At x = 0.5 m, torque is −5 N·m. At x = 1 m, torque is −10 N·m. At x = 2 m, torque is −20 N·m. Doubling the perpendicular distance doubles torque magnitude when force is unchanged.

Capture x = 0, 1 and 2 m. The adjustable force stays 10 N, but net torque including the fixed left load changes from +10 to 0 to −10 N·m. A force reading alone therefore cannot establish the rotational effect of a load. You also need its line of action and the pivot.

A positive or negative torque indicates the direction of angular acceleration for this fixed-axis beam, not a rotation angle already reached. The picture remains horizontal because the lab compares starting states. It does not integrate the beam’s later motion or calculate how fast it rotates.

Trade force against lever arm

To oppose the fixed +10 N·m, require adjustable torque −10 N·m. If the right-hand distance is d and downward force magnitude is F, then Fd = 10 N·m. At d = 0.5 m, F must be 20 N. At d = 1 m, F must be 10 N. At d = 2 m, F must be 5 N.

All three settings give the same adjustable torque, but their force magnitudes differ. Save two of these balanced cases and compare the pivot reaction. A longer lever arm reduces the required applied force for this torque; it does not mean energy or work is created. The present model has no finite rotation or work calculation.

At d = 1.5 m, exact balance requires 10/1.5 = 6⅔ N. The control selects whole newtons, so neither 6 N nor 7 N gives exact balance. That is a resolution limit of the teaching controls, not proof that physical equilibrium is impossible. At d = 0, by contrast, any finite force through the pin has zero torque about it and cannot cancel the fixed left torque.

Use perpendicular distance for angled forces

Distance along an object is not always the lever arm. For a force of magnitude F applied a distance r from the pivot, at angle θ to the position vector, torque magnitude is rF sin θ. Equivalently, use the shortest distance from the pivot to the force’s extended line of action and multiply by F.

At a right angle, sin θ = 1 and the full distance counts. If the force points directly along the beam through the pivot, the lever arm is zero and there is no torque about that pivot. A large force can still have zero torque when its line of action passes through the axis.

Although torque and energy both contain units of force times distance, they are different quantities. Torque describes a turning effect; work requires force through displacement, or torque through angular displacement. Reporting a torque as an amount of energy would hide that distinction.

Transfer beyond the vertical controls

A 12 N force is applied 0.5 m from a pivot at 30 degrees to the position vector. Predict torque magnitude. Then compare with the same force perpendicular to the lever. Determine the sign only after specifying whether the force tends to turn clockwise or counterclockwise.

Check the torque transfer

The angled torque magnitude is 0.5 × 12 × sin 30° = 3 N·m. The perpendicular case gives 0.5 × 12 = 6 N·m. The angle has halved the perpendicular force component. Magnitude alone does not specify direction; with counterclockwise positive, assign +3 N·m for a counterclockwise tendency and −3 N·m for clockwise.