Topic 5.3 Torque: define torque as the product of force and lever arm, compute it as $\tau = rF\sin\theta$ and as a cross product, and combine torques about an axis.

What this topic is asking

The College Board (Topic 5.3) wants you to define torque as the rotational effect of a force, to compute it from the lever arm as $\tau = rF\sin\theta$ and as a cross product $\vec{\tau} = \vec{r}\times\vec{F}$, and to combine torques about an axis. Torque is to rotation what force is to translation: it is the cause of angular acceleration, and getting its magnitude and sense right is the foundation of the rest of the unit.

Defining torque

Torque depends on three things: how big the force is, how far from the axis it acts, and the angle. A force applied far from the axis (large $r$) and perpendicular to the position vector ($\theta = 90^\circ$, so $\sin\theta = 1$) gives the maximum torque, which is why door handles sit at the edge of the door and wrenches have long handles. A force pointing straight toward or away from the axis ($\theta = 0$ or $180^\circ$) produces no torque, because only the perpendicular component turns the object.

The lever arm

A useful way to compute torque is the lever arm, the perpendicular distance from the axis to the line of action of the force (the line along which the force points, extended both ways). Then $\tau = r_\perp F$ with $r_\perp = r\sin\theta$. This picture makes it obvious that sliding the force along its line of action does not change the torque, only the perpendicular distance to that line matters. The two views, "force times lever arm" and "perpendicular component of force times distance", give the same $rF\sin\theta$ and you can use whichever is more convenient.

Torque as a cross product

The full vector definition of torque is the cross product:

The magnitude recovers $rF\sin\theta$, and the direction is given by the right-hand rule: point the fingers from $\vec{r}$ toward $\vec{F}$ and the thumb gives the torque direction, perpendicular to both. For rotation in a plane, this means a counterclockwise torque points out of the page and a clockwise torque points into it. On the exam you usually need only the sign (counterclockwise positive, clockwise negative), but the cross-product framing is the rigorous statement and matters for angular momentum.

Combining torques

Several forces produce several torques about an axis, and they add (as signed quantities or vectors) to give the net torque:

Adopt a sign convention, take counterclockwise as positive, and add the torques with their signs. A balanced object has zero net torque (the equilibrium condition of the next topic); an unbalanced net torque produces angular acceleration through the rotational form of Newton's second law. Computing each force's torque about the chosen axis and summing is the core skill for both statics and dynamics of rotation.

Try this

Q1. A $20$ N force is applied perpendicular to a $0.30$ m lever. Calculate the torque. [2 points]

• Cue. $\tau = rF\sin 90^\circ = (0.30)(20) = 6.0$ N m.

Q2. Explain why pushing a door near its hinge requires more force than pushing near its edge to open it. [2 points]

• Cue. Torque $\tau = rF$; near the hinge $r$ is small, so a larger force is needed for the same opening torque.