Translational vs Rotational Equilibrium Explained

Translational equilibrium means the forces sum to zero, so the centre of mass does not accelerate. Rotational equilibrium means the torques sum to zero, so the angular velocity does not change. The CED says a system can be in one without the other: a ladder whose forces balance can still rotate.

AP Physics: Unit 5 (topics 2.4 Newton's First Law, 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form). The two conditions are defined in different units of AP Physics 1. Translational equilibrium is Topic 2.4, Newton's First Law, in Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. LO 2.4.A asks students to describe the conditions under which a system's velocity remains constant, and carries EK 2.4.A.1 through 2.4.A.5: the net force as the vector sum of all forces, the definition of translational equilibrium with the derived equation sum of F_i = 0, Newton's first law, the statement that forces may be balanced in one dimension but unbalanced in another with the velocity changing only in the direction of the unbalanced force, and the definition of an inertial reference frame. Suggested skills for Topic 2.4 are 1.C, 2.A, 3.B and 3.C. Rotational equilibrium is Topic 5.5 in Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent. LO 5.5.A asks students to describe the conditions under which a system's angular velocity remains constant. EK 5.5.A.1 states that a system may exhibit rotational equilibrium, meaning constant angular velocity, without being in translational equilibrium, and vice versa; its sub-points cover force diagrams, the definition of rotational equilibrium with the relevant equation sum of tau_i = 0, and the rotational analog of Newton's first law. EK 5.5.A.2 adds that unbalanced torques mean the angular velocity must be changing. Suggested skills for Topic 5.5 are 1.C, 2.A, 2.B and 3.B, and the boundary statement says AP Physics 1 does not expect students to simultaneously analyze rotation in multiple planes. EK 5.6.A.3 links the two: to fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently. Neither equilibrium condition is printed on the equation sheet; the sheet prints the two second laws, a_sys = sum F / m_sys and alpha_sys = sum tau / I_sys, from which both follow by setting the left side to zero. AP Physics C: Mechanics carries the same two topics with the same numbering.

Two sums, and neither one implies the other

These are not two names for one idea, and they are not two halves that always arrive together. They are two separate tests you run on the same free-body diagram.

\sum \vec{F}_i = 0 \qquad \text{and} \qquad \sum \tau_i = 0

AP Physics 1 EK 2.4.A.2 defines the first: translational equilibrium is a configuration of forces such that the net force exerted on a system is zero. EK 5.5.A.1.ii defines the second: rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero. Note the parallel wording and the one word that changes.

Then the CED does something unusual and says the two are independent, in a sentence written for exactly this confusion. EK 5.5.A.1: a system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa.

That is the whole page. Four combinations exist and all four are physically ordinary:

Both. A book on a table. A beam on two supports. Nothing changes.
Translational only. Two equal and opposite forces at the two ends of a rod. The centre of mass stays put and the rod spins up. Worked example one.
Rotational only. A ball thrown with no spin. It accelerates through the air while its angular velocity holds at zero. Worked example two.
Neither. A ball rolling down a ramp: it speeds up and it spins up.

A reader who takes only that four-way table away has the answer. The rest of this page is the cases and the marks.

Translational vs rotational equilibrium, side by side

Question you are asking	Translational equilibrium	Rotational equilibrium
The condition	$\sum \vec{F} = 0$	$\sum \tau = 0$
CED definition	EK 2.4.A.2	EK 5.5.A.1.ii
What stays constant	The velocity of the centre of mass	The angular velocity
What is zero	The acceleration $\vec{a}$	The angular acceleration $\alpha$
The first law behind it	EK 2.4.A.3	EK 5.5.A.1.iii
The second law behind it	$\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ , EK 2.5.A.2	$\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}}$ , EK 5.6.A.2
What each term in the sum needs	A magnitude and a direction	A magnitude, a sign, and a chosen axis
Does where the force acts matter	No	Yes, that is the entire content of a torque
Number of equations it gives you	One per axis, so usually two	One per plane, so usually one
Can a force contribute nothing	Only if it is zero	Yes, if its line of action passes through the axis
Unit of each term	$\text{N}$	$\text{N}\cdot\text{m}$
Which unit of AP Physics 1	Unit 2, Topic 2.4	Unit 5, Topic 5.5

The "where the force acts" row is the origin of every difference below it. A force sum does not care where a force is applied; slide it anywhere on the body and $\sum \vec{F}$ is unchanged. A torque sum cares about nothing else. That is why one condition can hold while the other fails: moving a force changes one sum and not the other.

The "can a force contribute nothing" row is the practical consequence. A $500$ N force applied straight at the axis adds $500$ N to the force sum and $0$ to the torque sum. There is no equivalent move for the force sum, which is why torque problems have a free choice of pivot and force problems do not.

The equation-count row is why you need both to solve a statics problem. Two unknown support forces cannot come out of $\sum F_y = 0$ alone: that is one equation for two unknowns. The torque condition supplies the second equation, and it is the only place the geometry of the problem enters.

The case that separates them: two forces that cancel and still turn the rod

Take a uniform rod of mass $12$ kg and length $0.80$ m, lying flat and free. Push one end with $15$ N and push the other end with $15$ N in the opposite direction, both perpendicular to the rod.

The force sum is exactly zero. Two forces, equal magnitudes, opposite directions. $\sum \vec{F} = 0$ with nothing left over. The rod is in translational equilibrium, and by EK 2.4.A.3 the velocity of its centre of mass stays constant, which here means the centre of mass does not move at all.

The torque sum is not zero, and it is not close. Both forces push the rod around in the same rotational sense, because they act on opposite sides of the centre and point in opposite directions. Their torques add rather than cancel: $15 \times 0.40 + 15 \times 0.40 = 12.0\ \text{N}\cdot\text{m}$ . The rod spins up at $18.75\ \text{rad/s}^2$ .

This arrangement, two equal and opposite forces with different lines of action, is the cleanest possible demonstration that the two conditions are independent, and worked example one runs the numbers about three different axes to show something else that only happens in this case: when the forces already balance, the net torque is the same about every axis. About the centre, about either end, about a point off the rod entirely, the answer is $12.0\ \text{N}\cdot\text{m}$ each time.

That result is worth holding onto, because it is what makes statics problems solvable. If a system is in translational equilibrium, you may compute torques about whichever point you like, and picking the point where an unknown force acts deletes that unknown from the equation. The how to calculate torque guide sets out that routine; what this page adds is the reason it is allowed, which is the force condition holding first.

The reverse case: rotational equilibrium while accelerating hard

The other direction happens just as often and is easier to miss, because nothing looks like it is being balanced.

Throw a ball with no spin. Once it leaves your hand the only force on it is gravity, and EK 2.6.A.1.iii says the gravitational force on a system can be considered to be exerted on the system's centre of mass. A force acting at the centre of mass has zero lever arm about the centre of mass, so its torque about that axis is zero.

So the ball satisfies $\sum \tau = 0$ about its centre of mass. It is in rotational equilibrium, and EK 5.5.A.1.iii delivers the consequence: its angular velocity is constant. If it left your hand without spinning, it arrives without spinning. If you gave it a spin, that spin rate holds all the way.

Meanwhile $\sum \vec{F} = mg \neq 0$ . The ball is accelerating downward at $9.8\ \text{m/s}^2$ for the whole flight. Worked example two puts numbers on both statements for a $0.42$ kg ball: $4.12$ N of net force, $9.8\ \text{m/s}^2$ of acceleration, and $0$ angular acceleration.

Every projectile in the course is an example. So is a skydiver before the parachute opens, and so is any object in free fall. The pattern to recognise: a single force acting at the centre of mass gives rotational equilibrium automatically and translational equilibrium never.

One caution on the wording. Rotational equilibrium about the centre of mass is not rotational equilibrium about every axis. Take the ball's torque about a point on the ground and gravity has a lever arm, so the sum is not zero there. When the force condition fails, the axis matters, and a claim of rotational equilibrium has to name the axis it is about. That is the mirror image of the result in the previous section.

A ladder whose forces balance and which rotates anyway

The rod is a clean demonstration; a ladder is the version an exam will actually put in front of you.

A uniform ladder leans against a smooth wall with its foot on a rough floor. Four forces act: its weight at the midpoint, the floor's normal force upward, the floor's friction horizontally, and the wall's normal force horizontally at the top. Because the wall is smooth, it pushes only perpendicular to itself.

Now look at what $\sum \vec{F} = 0$ actually pins down. Vertically it fixes the floor's normal force at the ladder's weight. Horizontally it says the friction force equals the wall's push. It does not say what either of those two horizontal forces is. Any pair of matching horizontal forces satisfies it: $10$ N against $10$ N, $30$ N against $30$ N, $200$ N against $200$ N.

Only one of those values leaves the ladder standing. Worked example three takes an $18$ kg, $5.0$ m ladder at $60^\circ$ and finds that the wall must push with $50.9$ N. Set the pair at $30$ N instead and the force sum is still exactly zero while the net torque about the foot is $-90.6\ \text{N}\cdot\text{m}$ , and the ladder starts to swing down. Perfect force balance, and it falls.

This is the case worth memorising, because it shows what the torque condition contributes that the force condition cannot: the torque equation is the only one in the problem that knows how long the ladder is or what angle it leans at. Strip it out and the geometry disappears, and a ladder at $80^\circ$ becomes indistinguishable from one at $20^\circ$ .

It also shows the direction of the dependency. The torque condition then hands back a requirement on the floor: friction must supply $50.9$ N against a normal force of $176.4$ N, so the coefficient of static friction must be at least $0.289$ . Below that the ladder slips, which is a translational failure caused by a rotational calculation. Static friction takes whatever value is required up to its maximum, so the two conditions end up talking to each other through it.

Where the confusion costs a mark

Answering a two-condition question with one condition. A beam on two supports has two unknowns and $\sum F_y = 0$ is one equation. Writing only the force equation produces an underdetermined system, and the usual guess, splitting the load evenly, is right only when the load sits at the midpoint.
Concluding that balanced forces mean no rotation. The rod in worked example one has perfectly balanced forces and an angular acceleration of $18.75\ \text{rad/s}^2$ . EK 5.5.A.1 exists to stop this inference.
Concluding that balanced torques mean no acceleration. A thrown ball has zero net torque about its centre of mass and is accelerating at $9.8\ \text{m/s}^2$ .
Reading either condition as "nothing is moving". Both are constant-rate conditions, not zero-rate ones. EK 2.4.A.3 gives constant velocity; EK 5.5.A.1.iii gives constant angular velocity. A wheel spinning at a steady rate is in rotational equilibrium.
Summing torques about different axes in one equation. Every term in $\sum \tau = 0$ must be taken about the same axis. Mixing axes produces a number with no meaning.
Changing the axis part-way through and expecting the same net torque. That works only when the force condition already holds. If $\sum \vec{F} \neq 0$ , the net torque genuinely differs from axis to axis, and "the net torque is zero" is then a claim about one specific axis.
Adding a force's magnitude into the torque sum. The two sums have different units. A term in newtons in a $\sum \tau$ equation is a lost mark before any arithmetic.
Forgetting the body's own weight in the torque sum. It acts at the centre of mass by EK 2.6.A.1.iii, and it contributes zero torque only when the axis is at the centre of mass. In the ladder it contributes $220.5\ \text{N}\cdot\text{m}$ about the foot.
Using a free-body diagram for a torque problem. EK 5.3.B.1.ii asks for a force diagram, which differs from a free-body diagram in exactly the way that matters here: it depicts the location at which those forces are exerted relative to the axis of rotation. Collapsing the forces to a point erases the torques.

When you need both, and why static problems feel like torque problems

Most exam problems that use the phrase equilibrium want both conditions at once, on an object that is not moving at all. That gives three usable equations in a plane: $\sum F_x = 0$ , $\sum F_y = 0$ and $\sum \tau = 0$ .

The standard order of operations is worth stating because it is not the order the conditions are taught in.

Draw the forces at the points where they act, not at the centre.
Choose an axis through the location of an unknown force. That force gets a lever arm of zero and vanishes.
Solve $\sum \tau = 0$ for whichever unknown is left.
Feed that back into $\sum F_x = 0$ and $\sum F_y = 0$ for the rest.

The torque equation goes first because step 2 makes it the only equation with a single unknown in it. That is why these problems feel like torque problems even though half the answer comes from the force conditions. The step-by-step method is in how to calculate torque and the free-body work is in how to draw a free body diagram; neither is repeated here.

The pairing with the other equilibrium distinction. Whether the object is at rest or moving at a constant rate is a separate question from which of the two conditions you are applying, and confusing the two questions is common. That axis is covered in static vs dynamic equilibrium. Here is the compact version: this page asks which sum you are setting to zero; that page asks what value of $v$ the answer permits.

A note on scope. The Topic 5.5 boundary statement says AP Physics 1 does not expect students to simultaneously analyze rotation in multiple planes, so the torque condition on this exam is one equation, not three, and signs replace vectors.

What the CED asks, and how the exam frames it

The two conditions are defined three units apart in AP Physics 1, which is part of why they are learned separately.

Translational equilibrium: Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. It sits in Topic 2.4, Newton's First Law, at LO 2.4.A: describe the conditions under which a system's velocity remains constant. EK 2.4.A.1 gives the net force as the vector sum of all forces exerted on the system; EK 2.4.A.2 defines translational equilibrium and carries $\sum \vec{F}_i = 0$ as a derived equation; EK 2.4.A.3 states Newton's first law; EK 2.4.A.4 adds that forces may be balanced in one dimension but unbalanced in another, with the system's velocity changing only in the direction of the unbalanced force; EK 2.4.A.5 defines an inertial reference frame. Suggested skills for Topic 2.4 are 1.C, 2.A, 3.B and 3.C.

Rotational equilibrium: Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent. Topic 5.5, Rotational Equilibrium and Newton's First Law in Rotational Form, has one learning objective, LO 5.5.A: describe the conditions under which a system's angular velocity remains constant. Read that next to LO 2.4.A and the parallel is exact, with one word changed. EK 5.5.A.1 is the independence statement, and its sub-points cover force diagrams (5.5.A.1.i), the definition with $\sum \tau_i = 0$ as the relevant equation (5.5.A.1.ii), and the rotational analog of Newton's first law (5.5.A.1.iii). EK 5.5.A.2 gives the corollary: if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing. Suggested skills for Topic 5.5 are 1.C, 2.A, 2.B and 3.B. The boundary statement rules out simultaneous analysis of rotation in multiple planes.

EK 5.6.A.3 is the sentence that ties the two together and is worth quoting whole: to fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently. Two analyses, run separately, on the same object.

On the equation sheet, neither condition is printed as an equation. $\sum \vec{F}_i = 0$ is a derived equation in the CED and $\sum \tau_i = 0$ is a relevant equation; what the AP Physics 1 Table of Information prints is the two second laws, $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$ and $\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}}$ , from which both equilibrium conditions follow by setting the left side to zero.

The CED framing is at Topic 2.4 and Topic 5.5, and the torque and rotational motion practice set has problems that need both conditions at once.

A couple: zero net force, twelve newton metres of torque

A uniform rod of mass $12$ kg and length $0.80$ m lies at rest on a frictionless horizontal surface. A force of $15$ N is applied at one end, perpendicular to the rod, and a force of $15$ N is applied at the other end, perpendicular to the rod and in the opposite direction. Take counterclockwise as positive. Show that the rod is in translational equilibrium, find the net torque about the centre, about the left end, and about a point $2.0$ m off the left end along the rod's line, and find the angular acceleration. The rod's rotational inertia about a perpendicular axis through its centre is $\frac{1}{12}ML^2$ .

The force condition. Take the upward force at the left end as $+15$ N and the downward force at the right end as $-15$ N in the transverse direction. $\sum F = +15 - 15 = 0$ . There is no force along the rod. So $\sum \vec{F} = 0$ exactly, and by EK 2.4.A.3 the centre of mass keeps whatever velocity it had, which is zero.
Torque about the centre. Each force acts $0.40$ m from the centre and is perpendicular to the rod, so $\sin\theta = 1$ in $\tau = rF\sin\theta$ . The left force pushes the left end up and the right force pushes the right end down, and both of those turn the rod the same way. $\tau_{\text{net}} = (0.40)(15) + (0.40)(15) = 6.0 + 6.0 = 12.0\ \text{N}\cdot\text{m}$ . The torques add, even though the forces cancelled.
Torque about the left end. The left force now acts at the axis, so $r = 0$ and it contributes nothing. The right force acts the full $0.80$ m away: $\tau_{\text{net}} = 0 + (0.80)(15) = 12.0\ \text{N}\cdot\text{m}$ . Same answer, from a completely different pair of terms.
Torque about a point $2.0$ m off the left end, so $2.0$ m from one force and $2.80$ m from the other, with the two forces now on the same side of the axis and pointing opposite ways: $\tau_{\text{net}} = (2.80)(15) - (2.00)(15) = 42.0 - 30.0 = 12.0\ \text{N}\cdot\text{m}$ . Three axes, one answer. That is the general result for a system already in translational equilibrium.
Rotational inertia: $I = \frac{1}{12}(12)(0.80)^2 = \frac{1}{12}(12)(0.64) = 0.64\ \text{kg}\cdot\text{m}^2$ .
Angular acceleration, from EK 5.6.A.2: $\alpha = \frac{\tau_{\text{net}}}{I} = \frac{12.0}{0.64} = 18.75\ \text{rad/s}^2$ . Not small, and not zero.
The verdict on each condition. Translational equilibrium: satisfied, $a = 0$ , the centre of mass never moves. Rotational equilibrium: failed, $\alpha = 18.75\ \text{rad/s}^2$ , so by EK 5.5.A.2 the angular velocity must be changing. The rod spins in place.
A physical check: after $1.0$ s the rod is turning at $18.75$ rad/s, close to three revolutions per second, while its centre of mass is exactly where it started. Any account of this rod that used only the force condition would have predicted nothing happening at all.

$\sum \vec{F} = 0$ , so the centre of mass never moves. The net torque is $12.0\ \text{N}\cdot\text{m}$ about the centre, about the left end and about a point $2.0$ m off the end, the same value about every axis because the forces already balance. With $I = 0.64\ \text{kg}\cdot\text{m}^2$ the rod spins up at $18.75\ \text{rad/s}^2$ . Translational equilibrium holds and rotational equilibrium fails.

A thrown ball: zero net torque, four newtons of net force

A $0.42$ kg ball is thrown and is in flight with air resistance negligible. It was released with no spin. Find the net force on it, its acceleration, the net torque about its own centre of mass, and its angular acceleration. Then find the net torque about a point on the ground $3.0$ m horizontally from the ball at one instant. Take downward as positive and use $g = 9.8\ \text{m/s}^2$ .

The forces. Gravity only, because the exam conventions state that air resistance is assumed to be negligible unless otherwise stated. $\sum F = mg = (0.42)(9.8) = 4.116\ \text{N}$ downward.
Translational verdict. $\sum \vec{F} \neq 0$ , so the ball is not in translational equilibrium. $a = \frac{4.116}{0.42} = 9.8\ \text{m/s}^2$ downward, as it must be.
Torque about the centre of mass. EK 2.6.A.1.iii says the gravitational force on a system can be considered to be exerted on the system's centre of mass. The lever arm from the centre of mass to a force acting at the centre of mass is zero, so $\tau = r_{\perp}F = (0)(4.116) = 0$ .
Rotational verdict. $\sum \tau = 0$ about the centre of mass, so the ball is in rotational equilibrium, and $\alpha = \frac{0}{I} = 0$ whatever the ball's rotational inertia is. By EK 5.5.A.1.iii its angular velocity is constant, and it was released at zero, so the ball arrives without spin.
Now change the axis. About a point on the ground $3.0$ m horizontally away, gravity has a lever arm of $3.0$ m, since the weight is vertical and the horizontal separation is the perpendicular distance from that point to the force's line of action. $\tau = (3.0)(4.116) = 12.3\ \text{N}\cdot\text{m}$ .
Read that carefully. The same ball, the same instant, the same single force: zero net torque about one axis and $12.3\ \text{N}\cdot\text{m}$ about another. This is allowed precisely because the ball is not in translational equilibrium, so the axis-independence result from worked example one does not apply.
The consequence for how you state the answer. "The ball is in rotational equilibrium" is only a complete claim once you say about which axis. About its centre of mass, yes, for the whole flight. About a fixed point on the ground, no.

Net force $4.12$ N downward and $a = 9.8\ \text{m/s}^2$ , so no translational equilibrium. Net torque zero about the centre of mass and $\alpha = 0$ , so rotational equilibrium holds there and the ball keeps whatever spin it was thrown with. About a ground point $3.0$ m away the net torque is $12.3\ \text{N}\cdot\text{m}$ , which is why the axis has to be named.

A ladder in perfect force balance that falls over

A uniform ladder of mass $18$ kg and length $5.0$ m leans at $60^\circ$ to the horizontal against a smooth vertical wall, with its foot on a rough floor. First suppose the wall pushes horizontally with $30$ N and the floor's friction is $30$ N. Show that the force condition is satisfied and find the net torque about the foot. Then find the wall force that actually gives equilibrium, and the smallest coefficient of static friction the floor must have. Take counterclockwise as positive and use $g = 9.8\ \text{m/s}^2$ .

Set up the four forces. Weight $mg = (18)(9.8) = 176.4\ \text{N}$ downward at the midpoint; the floor's normal force $F_N$ upward at the foot; the floor's friction $f$ horizontally at the foot; the wall's normal force $F_w$ horizontally at the top, pointing away from the wall. The wall is smooth, so it exerts no vertical force.
Geometry. The top of the ladder is at a height $h = (5.0)\sin 60^\circ = (5.0)(0.86603) = 4.330\ \text{m}$ . The midpoint is a horizontal distance $d = (2.5)\cos 60^\circ = (2.5)(0.5) = 1.250\ \text{m}$ from the foot.
The force condition with $F_w = f = 30$ N. Vertically: $F_N - 176.4 = 0$ , so $F_N = 176.4\ \text{N}$ . Horizontally: $F_w - f = 30 - 30 = 0$ . Both components are exactly zero. The ladder is in translational equilibrium.
The torque condition, about the foot. $F_N$ and $f$ both act at the foot, so both have $r = 0$ and neither contributes. The weight acts $1.250$ m horizontally from the foot and points down: $\tau = -(1.250)(176.4) = -220.5\ \text{N}\cdot\text{m}$ , turning the ladder down the wall. The wall force acts $4.330$ m up and points away from the wall: $\tau = +(4.330)(30) = +129.9\ \text{N}\cdot\text{m}$ .
Net torque: $129.9 - 220.5 = -90.6\ \text{N}\cdot\text{m}$ . Not zero. The forces balance and the ladder rotates. With a rotational inertia of $\frac{1}{3}ML^2 = \frac{1}{3}(18)(25) = 150\ \text{kg}\cdot\text{m}^2$ about the foot, that is an initial angular acceleration of $\frac{-90.6}{150} = -0.604\ \text{rad/s}^2$ .
What the wall force has to be. Set $\sum \tau = 0$ about the foot: $(4.330)F_w = 220.5$ , so $F_w = \frac{220.5}{4.330} = 50.92\ \text{N}$ . This is the only value that works, and the force condition alone could never have found it, because $\sum \vec{F} = 0$ is satisfied for every value of $F_w$ as long as $f$ matches it.
Back to the force condition for the rest. Horizontally, $f = F_w = 50.92\ \text{N}$ . Vertically, $F_N = 176.4\ \text{N}$ , unchanged.
The friction requirement. Static friction adopts whatever value is needed up to its maximum, $\mu_s F_N$ , so the floor needs $\mu_s \geq \frac{50.92}{176.4} = 0.2887$ . A smooth-tiled floor with $\mu_s$ below that cannot hold the ladder, and it slips.
Symbolic check. The condition reduces to $\mu_s \geq \frac{1}{2\tan\theta}$ , and $\frac{1}{2\tan 60^\circ} = \frac{1}{2(1.7321)} = 0.2887$ , matching. The mass cancels out, which is why a heavier ladder does not help, and the requirement gets harsher as the ladder is laid flatter.

With $F_w = f = 30$ N the force sum is exactly zero and the net torque about the foot is $-90.6\ \text{N}\cdot\text{m}$ , giving an angular acceleration of $-0.604\ \text{rad/s}^2$ : balanced forces, and the ladder swings down. True equilibrium needs $F_w = 50.9$ N, matched by $50.9$ N of friction against a $176.4$ N normal force, so $\mu_s \geq 0.289$ . The torque condition is the only one in the problem that knows the ladder's length or angle.

Frequently asked questions

What is the difference between translational and rotational equilibrium?

Translational equilibrium is the condition that the forces on a system sum to zero, so the velocity of its centre of mass stays constant. Rotational equilibrium is the condition that the torques on it sum to zero about a chosen axis, so its angular velocity stays constant. The AP Physics 1 CED defines the first at essential knowledge 2.4.A.2 and the second at 5.5.A.1.ii, using deliberately parallel wording: a configuration of forces such that the net force is zero, and a configuration of torques such that the net torque is zero. The force sum does not care where each force is applied and the torque sum cares about nothing else, which is why one can hold while the other fails.

Can an object be in rotational equilibrium but not translational equilibrium?

Yes, and the AP Physics 1 CED states it explicitly at essential knowledge 5.5.A.1: a system may exhibit rotational equilibrium, meaning constant angular velocity, without being in translational equilibrium, and vice versa. A thrown ball is the everyday case. Gravity is considered to act at the centre of mass, so its torque about the centre of mass is zero and the ball's spin rate never changes, while the net force on it is its full weight and it accelerates downward at 9.8 metres per second squared throughout. The reverse case is a rod pushed by two equal and opposite forces at its two ends: the forces cancel exactly and the torques add, so the centre of mass stays put while the rod spins up.

Can an object with zero net force still rotate?

Yes. Zero net force fixes the acceleration of the centre of mass at zero and says nothing at all about rotation, because moving a force to a different point on the body changes its torque without changing the force sum by anything. Two equal and opposite 15 newton forces applied to opposite ends of a 0.80 metre rod give a force sum of exactly zero and a net torque of 12 newton metres, spinning a 12 kilogram rod up at 18.75 radians per second squared while its centre of mass does not move. A ladder is the practical version: its forces can balance perfectly while the net torque about its foot is nonzero, and it swings down.

Does the choice of pivot point change the net torque?

It depends on whether the forces already balance. If the system is in translational equilibrium, the net torque is the same about every axis, which is why statics problems let you put the pivot wherever it is convenient, usually at an unknown force so that force drops out. If the net force is not zero, the net torque genuinely differs from axis to axis. A thrown ball has zero net torque about its own centre of mass and a nonzero net torque about a point on the ground, at the same instant. So a claim that a system is in rotational equilibrium is incomplete until you name the axis, unless you have already established that the forces sum to zero.

Do you need both conditions to solve a statics problem?

Almost always, because the force conditions alone do not supply enough equations. A beam resting on two supports has two unknown support forces and only one useful force equation, the vertical one, so the system is underdetermined until the torque condition adds a second. The torque equation is also the only equation in the problem that contains any geometry, so without it a 5 metre beam and a 50 centimetre one would give the same answer. The usual order is to write the torque equation first, about an axis chosen at one of the unknown forces so that force contributes nothing, solve for the other unknown, and then use the force equations for the rest.

Is rotational equilibrium the same as not rotating?

No. It means the angular velocity is constant, which includes the case of a constant nonzero spin. Essential knowledge 5.5.A.1.iii gives the rotational analog of Newton's first law: a system will have a constant angular velocity only if the net torque exerted on it is zero. A merry-go-round turning at a steady rate satisfies that just as fully as a bridge that never turns. The corollary at 5.5.A.2 is the other half: if the torques on a rigid system are not balanced, its angular velocity must be changing. Reading the term as no rotation is the same mistake as reading translational equilibrium as at rest.

What is the equation for rotational equilibrium?

The sum of the torques about a chosen axis equals zero. The AP Physics 1 CED lists it as a relevant equation under essential knowledge 5.5.A.1.ii, alongside the sum of the forces equalling zero, which appears as a derived equation under 2.4.A.2. Neither is printed as a separate line on the equation sheet. What the Table of Information prints is the two second laws, the acceleration of a system as the net force over the system's mass and the angular acceleration as the net torque over the rotational inertia, and both equilibrium conditions come from setting the left-hand side of one of those to zero. Every term in the torque sum must be taken about the same axis and carry its own sign.