Topic 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form: apply the condition of zero net torque for rotational equilibrium, alongside zero net force, to analyze balanced rigid bodies.

What this topic is asking

The College Board (Topic 5.5) wants you to apply rotational equilibrium: a rigid body is in equilibrium only when the net torque is zero as well as the net force. This extends Newton's first law to rotation: with zero net torque, an object's rotational state (at rest or spinning steadily) does not change. You use the two conditions together to analyze balanced beams, signs and ladders.

The two conditions for equilibrium

A point particle needs only $\sum F = 0$ to be in equilibrium, but an extended rigid body needs $\sum \tau = 0$ as well, because a balanced set of forces can still produce a net turning effect. A seesaw with equal forces at unequal distances has zero net force but a net torque, so it rotates.

Newton's first law in rotational form

This mirrors the translational first law (Topic 2.3): no net force means no change in velocity; no net torque means no change in angular velocity. The two laws together describe what "balanced" means for an extended body.

Solving equilibrium problems

The practical method is to choose the axis cleverly. Because $\sum \tau = 0$ holds about any axis for a body in equilibrium, you pick the axis to pass through an unknown force you want to eliminate (a force through the axis has zero lever arm and so contributes no torque). This often turns a problem with two unknown support forces into a single torque equation with one unknown, which you solve directly; then $\sum F = 0$ gives the remaining force. The routine is: draw a free-body diagram showing every force and where it acts, treat the object's weight as acting at its center of mass, choose an axis, write the torque equation ($\sum \tau = 0$, balancing clockwise against counterclockwise), and write the force equation ($\sum F = 0$). This is the rotational counterpart of the equilibrium analysis from Unit 2, now requiring the extra torque condition because the body is extended rather than a point. Balanced beams, diving boards, hanging signs supported by a cable, and ladders leaning against walls are all the same problem: identify the forces and their lever arms, set the net torque to zero about a smart axis, and solve. Choosing the axis to kill an unknown is the single most useful trick, and it is heavily rewarded on the exam.

Try this

Q1. A $30$ N weight hangs $0.80$ m from a pivot. Calculate the torque it produces about the pivot. [2 points]

• Cue. $\tau = rF = (0.80)(30) = 24$ N$\cdot$m.

Q2. A rigid body has zero net force but a net torque of $5$ N$\cdot$m. State whether it is in equilibrium and what it does. [1 point]

• Cue. It is not in equilibrium; the net torque makes it angularly accelerate (start to rotate).