Static vs Dynamic Equilibrium: The Difference

Both mean the same thing: the net force is zero, so the acceleration is zero. Static equilibrium is the case where the velocity is zero, dynamic equilibrium where it is constant but not zero. The equations are identical, and the AP CEDs call both translational equilibrium.

AP Physics: Unit 2 (topics 2.4 Newton's First Law, 2.5 Newton's Second Law). Both cases are one topic in AP Physics 1: 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 five pieces of essential knowledge: 2.4.A.1, the net force as the vector sum of all forces exerted on the system; 2.4.A.2, translational equilibrium as a configuration of forces such that the net force exerted on a system is zero, with the derived equation sum of F_i = 0; 2.4.A.3, Newton's first law, that if the net force exerted on a system is zero the velocity of that system will remain constant; 2.4.A.4, 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 2.4.A.5, the definition of an inertial reference frame. Suggested skills are 1.C, 2.A, 3.B and 3.C, and the topic carries no boundary statement. Topic 2.5 supplies the contrast at EK 2.5.A.1 and EK 2.5.A.3. The phrases static equilibrium and dynamic equilibrium appear nowhere in the AP Physics 1 CED or the AP Physics C: Mechanics CED; both use translational equilibrium. The equilibrium condition is not printed on the equation sheet: the Table of Information prints a_sys = sum F / m_sys, and equilibrium is that equation with the left side zero. AP Physics C: Mechanics carries the identical Topic 2.4 inside a Unit 2 weighted at 20 to 25 percent.

The physics is the same, only v differs

Most comparisons on this site resolve into a real difference. This one resolves into a non-difference, and that is the point worth paying for.

AP Physics 1 EK 2.4.A.2: translational equilibrium is a configuration of forces such that the net force exerted on a system is zero. There is no clause about velocity in that definition. Zero net force means zero acceleration, and zero acceleration means the velocity is not changing. It says nothing about what the velocity is.

So the split is a labelling of one condition by an incidental fact about the object:

Static equilibrium: $\sum \vec{F} = 0$ and $\vec{v} = 0$ .
Dynamic equilibrium: $\sum \vec{F} = 0$ and $\vec{v}$ constant, not zero.

Everything you do differs by nothing. Same free-body diagram, same $\sum F_x = 0$ and $\sum F_y = 0$ , same algebra, same answers. A book on a table and a car cruising at $30$ m/s produce identical force equations, and there is no measurement you could make of the forces alone that would tell you which one you are looking at.

EK 2.4.A.3 is why: Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant. Remain, not vanish. Zero is one of the values that velocity is allowed to remain at, and it is not privileged.

The useful reflex, then, is not to sort a problem into static or dynamic. It is to stop asking whether the object is moving and start asking whether its velocity is changing.

Static vs dynamic equilibrium, side by side

Question you are asking	Static equilibrium	Dynamic equilibrium
Net force	$\sum \vec{F} = 0$	$\sum \vec{F} = 0$
Acceleration	$0$	$0$
Velocity	$0$	Constant, not zero
Is the velocity changing	No	No
Free-body diagram	Forces balance	Forces balance, identically
Equations you write	$\sum F_x = 0$ , $\sum F_y = 0$	$\sum F_x = 0$ , $\sum F_y = 0$
Which law applies	Newton's first, EK 2.4.A.3	Newton's first, EK 2.4.A.3
Can you tell from the forces alone	No	No
Depends on the reference frame	Yes	Yes
Term used in the AP CEDs	Neither appears	Neither appears
What the CEDs call it	Translational equilibrium	Translational equilibrium
Everyday case	Book on a table	Skydiver at terminal speed

Read the first six rows. They match on every line. That is unusual for a comparison table and it is the honest result: the physics does not distinguish these cases, and any page that gives them different methods is inventing a difference.

The reference-frame row is the one that shows the split is not even about the object. EK 2.4.A.5 defines an inertial reference frame as one from which an observer would verify Newton's first law, and EK 1.4.A.1 says the choice of reference frame determines the direction and magnitude of quantities measured by an observer in that frame. A passenger asleep on a train is in static equilibrium according to another passenger and in dynamic equilibrium according to someone on the platform. The forces are the same in both descriptions and both observers are right, because the two frames are both inertial and the acceleration is zero in each. A distinction that flips when you walk to the window is not a distinction about the physics.

The last two rows are checkable and worth stating plainly. The phrases "static equilibrium" and "dynamic equilibrium" appear nowhere in the AP Physics 1 Course and Exam Description, and nowhere in the AP Physics C: Mechanics one. Both use "translational equilibrium", defined once at EK 2.4.A.2 and used again in Topic 5.5. The two terms are standard physics vocabulary and worth knowing, because textbooks and teachers use them; they are just not exam vocabulary, so a question will describe the situation rather than name it.

The case that separates them: one object passing through both

The sharpest demonstration is not two objects but one, watched through a change.

A $65$ kg person stands on a bathroom scale in a lift. Worked example one runs four phases:

Lift at rest. Scale reads $637$ N. Static equilibrium.
Lift accelerating upward at $1.2\ \text{m/s}^2$ . Scale reads $715$ N. Not equilibrium at all.
Lift moving upward at a constant $2.0$ m/s. Scale reads $637$ N. Dynamic equilibrium.
Lift slowing at $1.2\ \text{m/s}^2$ while still moving up. Scale reads $559$ N. Not equilibrium.

Phases one and three give the same number. The person is stationary in one and travelling at $2.0$ m/s in the other, and the scale cannot tell the difference, because the scale reports a force and the forces are identical.

The two phases that differ are the accelerating ones, and they are the ones nobody calls equilibrium. So the boundary that actually matters in this problem runs between phases 1 and 3 on one side and phases 2 and 4 on the other. It does not run between static and dynamic.

This also settles the everyday intuition that motion needs a force to sustain it. Between phase 2 and phase 4 the lift is moving upward the entire time and the net force is zero for most of it. Motion does not need a net force; a change in motion does.

The trap: zero velocity is not the same as equilibrium

The two-way street here catches more marks than the definition does. Reading "equilibrium" as "at rest" is one error. Reading "at rest" as "equilibrium" is the worse one, because it looks safe.

Throw a ball straight up. At the very top its velocity is exactly zero. It is not in equilibrium, static or otherwise. Gravity is still the only force on it, the net force is its full weight, and it is accelerating downward at $9.8\ \text{m/s}^2$ the whole time it is up there. Worked example two puts $2.94$ N on that for a $0.30$ kg ball.

The difference between the ball at the top and a book on a table is not the velocity, which is zero in both. It is that the book has a second force, the normal force, and the ball does not.

Other versions of the same trap:

A block momentarily at rest on the way up a rough incline. Velocity zero, net force down the slope, accelerating.
A mass at the extreme of a spring oscillation. Velocity zero, restoring force maximum, acceleration maximum. EK 2.8.A.3 puts the spring force always directed toward the equilibrium position of the object and spring system, and at the extreme that force is at its largest.
A ball at the instant it reverses during a bounce. Velocity zero, and the contact force is nowhere near balancing out.

The rule that survives all of these: equilibrium is a statement about the sum of the forces, and you can only settle it by adding up the forces. The velocity is an output, not a test. Static equilibrium is the case where that output happens to be zero, and "happens to be" is doing real work in that sentence.

One axis at a time, and partial equilibrium

There is a third case that neither label covers and the CED does.

EK 2.4.A.4: forces may be balanced in one dimension but unbalanced in another. The system's velocity will change only in the direction of the unbalanced force.

A projectile is the standard example and it is in the course from Unit 1. Horizontally the net force is zero, so the horizontal velocity is constant: that axis is in equilibrium for the whole flight. Vertically the net force is the weight, so the vertical velocity changes at $9.8\ \text{m/s}^2$ . Calling the projectile "in equilibrium" or "not in equilibrium" is a poor description of it either way; the accurate statement names the axis.

The same thing happens on an incline, which is why the axes are conventionally tilted to lie along and across the slope. Across the slope the forces balance and stay balanced. Along it they usually do not. Splitting the problem that way turns one awkward two-dimensional question into one trivial equation and one real one, and the inclined plane problems guide runs that routine.

The practical form of EK 2.4.A.4 in an exam: write $\sum F_x = 0$ and $\sum F_y = 0$ as two independent claims and check each separately. A body at constant velocity satisfies both. A body in circular motion at constant speed satisfies neither in the usual axes, since its velocity is changing direction. Constant speed is not constant velocity, and only constant velocity is equilibrium.

Where the confusion costs a mark

Adding a force to keep something moving. A crate sliding at constant velocity has zero net force. Drawing a leftover forward force to account for the motion is a classic free-body diagram error, and it comes straight from reading equilibrium as rest.
Assuming an object at rest is in equilibrium. A ball at the top of its flight is at rest and has a net force of its full weight. Check the forces, not the velocity.
Assuming an object in motion is not in equilibrium. A skydiver at terminal speed is falling fast and is in equilibrium exactly, because the drag has grown to match the weight. So is a puck sliding on frictionless ice.
Treating constant speed in a circle as equilibrium. The speed is constant and the velocity is not, so the acceleration is nonzero and the net force points to the centre. This one is worth naming because the phrase "moving at a constant rate" sounds like the dynamic case and is not.
Switching methods for a moving object. There is no separate dynamic-equilibrium method. Same diagram, same two equations.
Concluding that balanced forces mean nothing is happening. Balanced forces also permit steady rotation to be spinning up, if the torques do not balance. Which sum you are setting to zero is a separate question, covered in translational vs rotational equilibrium.
Using the terms in a free-response answer as though they were course vocabulary. They are not in the CED. "The net force is zero" is unambiguous and always scores; "it is in static equilibrium" invites the reader to work out what you meant.
Forgetting that the frame has to be inertial. EK 2.4.A.5 requires it, and the boundary statement for Topic 1.4 says that unless otherwise stated the frame of reference of any problem may be assumed to be inertial. In an accelerating lift, the person is not in equilibrium in the ground frame, and the exam works in the ground frame.

The diagram work is in how to draw a free body diagram and the summing in how to find net force; neither routine is repeated here.

When the distinction does earn its keep

Having spent the page saying the two are the same, it is worth being exact about the one place the difference is real, because a page that says "no difference at all" is also wrong.

The difference is in what else you are allowed to assume. If an object is genuinely at rest and stays at rest, then two extra things follow for free:

Static friction is available, and it is the larger coefficient. EK 2.7.B.3 says the coefficient of static friction is typically greater than the coefficient of kinetic friction for a given pair of surfaces. A stationary block on a slope is held by static friction, which adopts whatever value prevents slipping up to $\mu_s F_N$ . A block sliding at constant velocity down the same slope is held by kinetic friction, fixed at $\mu_k F_N$ , which is why the constant-velocity angle is the shallower of the two. That is a genuine numerical difference between a static and a dynamic problem on the same surfaces, and it belongs to static vs kinetic friction.
The rotational condition usually comes with it. A rigid body sitting still is not turning either, so a problem that says "at rest" is normally handing you $\sum \tau = 0$ as well as $\sum \vec{F} = 0$ . A body in dynamic equilibrium may be spinning up while its forces balance.

Both of those are things the word "static" tells you about the situation. Neither is a difference in the equilibrium condition itself.

There is also a plain reading advantage. When a problem says an object is at rest, it has told you a lot: no kinetic friction, no drag, no rotation, and every velocity in the problem is zero. When it says constant velocity, it has told you the net force is zero and nothing else. Read "at rest" as extra information rather than as a different physics.

What the CED asks, and how the exam frames it

All of this lives in one topic. Topic 2.4, Newton's First Law, in Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the AP Physics 1 multiple-choice section over about 22 to 27 class periods.

LO 2.4.A asks you to describe the conditions under which a system's velocity remains constant. That phrasing is worth reading twice: the objective is about the velocity remaining constant, not about it being zero. Its five pieces of essential knowledge are the whole content of this comparison.

EK 2.4.A.1: the net force on a system is the vector sum of all forces exerted on the system.
EK 2.4.A.2: translational equilibrium is a configuration of forces such that the net force exerted on a system is zero. Derived equation: $\sum \vec{F}_i = 0$ .
EK 2.4.A.3: Newton's first law states that if the net force exerted on a system is zero, the velocity of that system will remain constant.
EK 2.4.A.4: forces may be balanced in one dimension but unbalanced in another; the system's velocity will change only in the direction of the unbalanced force.
EK 2.4.A.5: an inertial reference frame is one from which an observer would verify Newton's first law of motion.

Suggested skills for Topic 2.4 are 1.C, 2.A, 3.B and 3.C. The topic carries no boundary statement. Topic 2.5, Newton's Second Law, supplies the contrast at EK 2.5.A.1: unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero, and EK 2.5.A.3 adds that the velocity of a system's centre of mass will only change if a nonzero net external force is exerted on that system.

On the equation sheet, $\sum \vec{F}_i = 0$ is not printed. The AP Physics 1 Table of Information prints the second law, $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$ , and the equilibrium condition is that equation with the left side set to zero. There is nothing on the sheet that distinguishes a static problem from a dynamic one, which is the same conclusion this page reached from the definitions.

AP Physics C: Mechanics carries the identical Topic 2.4 with the same numbering, inside a Unit 2 weighted at 20 to 25 percent of that exam. The CED framing is at Topic 2.4 and Topic 2.5, and the forces practice set has problems of both kinds, deliberately not labelled.

Four phases in a lift, and the two that read the same

A $65$ kg person stands on a bathroom scale in a lift. Find the scale reading when the lift is at rest, when it accelerates upward at $1.2\ \text{m/s}^2$ , when it travels upward at a constant $2.0$ m/s, and when it slows at $1.2\ \text{m/s}^2$ while still moving upward. State for each phase whether the person is in equilibrium. Take upward as positive and use $g = 9.8\ \text{m/s}^2$ .

Set up once. Two forces act on the person: the weight $mg$ downward and the normal force $F_N$ from the scale upward. The scale reads $F_N$ . Newton's second law along the vertical axis: $F_N - mg = ma$ , so $F_N = m(g + a)$ with the sign of $a$ carrying the direction. The weight is $mg = (65)(9.8) = 637\ \text{N}$ in every phase.
Phase 1, at rest. $a = 0$ , so $F_N = (65)(9.8 + 0) = 637\ \text{N}$ . The forces balance, $\sum F = 0$ , and the velocity is zero. Static equilibrium.
Phase 2, accelerating upward at $1.2\ \text{m/s}^2$ . $F_N = (65)(9.8 + 1.2) = (65)(11.0) = 715\ \text{N}$ . The forces do not balance: $715 - 637 = 78\ \text{N}$ upward, which is $(65)(1.2)$ as required. Not equilibrium.
Phase 3, constant $2.0$ m/s upward. Constant velocity means $a = 0$ , so $F_N = (65)(9.8 + 0) = 637\ \text{N}$ . The forces balance. Dynamic equilibrium.
Phase 4, slowing at $1.2\ \text{m/s}^2$ while still rising. The velocity is upward and decreasing, so the acceleration is downward: $a = -1.2\ \text{m/s}^2$ . $F_N = (65)(9.8 - 1.2) = (65)(8.6) = 559\ \text{N}$ . Not equilibrium.
Compare phases 1 and 3. $637$ N and $637$ N. The person is stationary in one and travelling at $2.0$ m/s in the other, and every force in the problem is identical. No measurement of forces can separate a static case from a dynamic one.
Compare phases 2 and 4. $715$ N and $559$ N, either side of $637$ N. These are the phases where something is actually different, and neither is called equilibrium.
A sanity check on the whole sequence: the lift starts at rest, speeds up, cruises, slows, and stops. It is in equilibrium at the start, in the middle and at the end, and out of equilibrium only during the two changes. The scale reading over time is $637$ , $715$ , $637$ , $559$ , $637$ N.

$637$ N at rest, $715$ N accelerating upward, $637$ N at constant velocity, $559$ N slowing. Phases 1 and 3 are equilibrium and give identical readings despite one being at rest and the other moving at $2.0$ m/s. Phases 2 and 4 are the only ones where the forces do not balance.

Zero velocity without equilibrium: a ball at the top of its flight

A $0.30$ kg ball is thrown straight upward. At the highest point of its flight its velocity is zero. Find the net force on it and its acceleration at that instant, and compare with a $0.30$ kg book resting on a table. Take upward as positive, ignore air resistance and use $g = 9.8\ \text{m/s}^2$ .

List the forces on the ball at the top. Gravity, and nothing else: it has left the hand and air resistance is negligible by the standing exam convention. There is no upward force at all. The idea that something must be holding it there is what makes this instant confusing.
Net force: $\sum F = -mg = -(0.30)(9.8) = -2.94\ \text{N}$ , that is $2.94$ N downward. Not zero, so the ball is not in equilibrium.
Acceleration: $a = \frac{-2.94}{0.30} = -9.8\ \text{m/s}^2$ . The same value it had on the way up and will have on the way down. Only the velocity turned around.
Now the book on the table. Two forces: weight $mg = 2.94\ \text{N}$ down and the normal force from the table. $\sum F = F_N - 2.94 = 0$ gives $F_N = 2.94\ \text{N}$ up, and $a = 0$ . Static equilibrium.
Line the two up. Velocity: $0$ and $0$ . Mass: $0.30$ kg and $0.30$ kg. Weight: $2.94$ N and $2.94$ N. Number of forces: one and two. Net force: $2.94$ N and $0$ . Acceleration: $9.8\ \text{m/s}^2$ and $0$ .
The velocity row matched and the verdict did not. Everything that decides the question is in the count of forces, and the velocity contributed nothing. That is why the test for equilibrium has to be run on the force sum.
A useful consequence to carry into a graph question: on a velocity-time graph for the ball, the top of the flight is where the curve crosses zero, and the slope there is $-9.8\ \text{m/s}^2$ exactly as it is everywhere else. The graph is a straight line through that point with no kink and no flat spot.

At the top the ball has a net force of $2.94$ N downward and an acceleration of $9.8\ \text{m/s}^2$ downward, so it is not in equilibrium despite having zero velocity. The book has the same weight, an equal normal force, zero net force and zero acceleration. Same velocity, opposite verdicts, and the deciding difference is the second force.

A hanging sign, at rest and then at constant velocity

A $12$ kg sign hangs from two cables that each make $35^\circ$ with the horizontal, arranged symmetrically. Find the tension in each cable. Then find the tension when the whole assembly is mounted in a truck travelling at a constant $25$ m/s along a straight level road, and state what changes when the truck brakes. Use $g = 9.8\ \text{m/s}^2$ and take upward and forward as positive.

Weight: $mg = (12)(9.8) = 117.6\ \text{N}$ downward.
Set up the two equations. By symmetry both cables have the same tension $T$ . Horizontally the two horizontal components point in opposite directions and cancel, so $\sum F_x = 0$ is satisfied automatically and tells you nothing. Vertically, each cable contributes $T\sin 35^\circ$ upward.
Solve the vertical equation. $2T\sin 35^\circ - 117.6 = 0$ , so $T = \frac{117.6}{2(0.57358)} = \frac{117.6}{1.14715} = 102.5\ \text{N}$ .
A check on the shape of that answer. Each tension is larger than half the weight, $58.8$ N, because only the vertical component of each cable does any lifting. The shallower the cables, the larger the tension: at $10^\circ$ the same sign would need $338.6$ N per cable.
Now put it in a truck at a constant $25$ m/s. Constant velocity in a straight line means $a = 0$ , so $\sum \vec{F} = 0$ , so the equations are the two you have already written. $T = 102.5\ \text{N}$ , unchanged. The sign hangs at exactly the same angle and the cables carry exactly the same load at $25$ m/s as at rest.
Nothing in the working referred to the velocity. The only quantity that entered was $a = 0$ , and it entered the same way in both cases. That is the whole content of the static and dynamic distinction reduced to an arithmetic observation.
Now the truck brakes. There is a backward acceleration, so $\sum \vec{F} \neq 0$ and the assembly is not in equilibrium in either sense. The sign swings forward until the cables can supply a horizontal component, the two tensions stop being equal, and $\sum F_x = 0$ is no longer available. The problem has changed shape, and it changed at the moment the acceleration became nonzero, not at the moment the sign started moving.
A frame check, since it is the objection most often raised here: to a passenger in the cruising truck the sign is at rest and to someone on the roadside it is doing $25$ m/s. Both frames are inertial while the speed is constant, both give $a = 0$ , and both compute $T = 102.5$ N. The labels static and dynamic swap between the two observers and the physics does not.

Each cable carries $102.5$ N, at rest and at a constant $25$ m/s alike, because the only fact used was that the acceleration is zero. Braking makes the acceleration nonzero, and only then do the answers change. Two observers in different inertial frames would label the cruising case differently and compute the same tension.

Frequently asked questions

What is the difference between static and dynamic equilibrium?

Only the value of the velocity. Both mean the net force on the object is zero and therefore its acceleration is zero. In static equilibrium the velocity is zero; in dynamic equilibrium it is constant but not zero. The free-body diagram, the equations and the algebra are identical in the two cases, so no measurement of the forces could tell them apart. The AP Physics 1 CED defines the shared condition at essential knowledge 2.4.A.2 as translational equilibrium, a configuration of forces such that the net force exerted on a system is zero, and says nothing about the velocity being zero.

Is an object moving at constant velocity in equilibrium?

Yes, fully. Newton's first law, at essential knowledge 2.4.A.3, states that if the net force exerted on a system is zero, the velocity of that system will remain constant, and the reverse reading is just as valid: constant velocity means the forces balance. A car cruising at a steady 30 metres per second on a straight road, a puck sliding on frictionless ice and a skydiver at terminal speed are all in equilibrium. Motion does not require a net force to sustain it. What requires a net force is a change in motion, which is why drawing a leftover forward force on a crate that slides at constant speed is a defect.

Can an object at rest not be in equilibrium?

Yes, and this is the more expensive version of the confusion. A ball thrown straight up has zero velocity at the top of its flight and is not in equilibrium: gravity is the only force on it, so a 0.30 kilogram ball has a net force of 2.94 newtons downward and an acceleration of 9.8 metres per second squared downward at that instant. A block momentarily at rest partway up a rough incline is another case, as is a mass at the extreme of a spring oscillation, where the restoring force is at its largest. Equilibrium is a statement about the sum of the forces, so it has to be settled by adding up the forces, never by looking at the velocity.

Do the AP Physics CEDs use the terms static and dynamic equilibrium?

No. Neither phrase appears in the AP Physics 1 Course and Exam Description or in the AP Physics C: Mechanics one. Both use translational equilibrium, defined at essential knowledge 2.4.A.2 as a configuration of forces such that the net force exerted on a system is zero, and both also use rotational equilibrium for the torque condition at 5.5.A.1.ii. Static and dynamic equilibrium are standard physics vocabulary that many textbooks and teachers use, so they are worth recognising, but an exam question will describe the situation instead. In a free-response answer, writing that the net force is zero is unambiguous and always scores.

Is the free-body diagram different for static and dynamic equilibrium?

No. In both cases the forces must sum to zero, so the arrows balance in exactly the same way and the equations you write from them are the same two: the sum of the horizontal components is zero and the sum of the vertical components is zero. A 12 kilogram sign hanging from two cables at 35 degrees needs 102.5 newtons in each cable whether it is bolted to a wall or riding in a truck at a constant 25 metres per second. The diagram only changes when the acceleration becomes nonzero, at which point the situation is not equilibrium of either kind.

Is an object moving in a circle at constant speed in dynamic equilibrium?

No. Constant speed is not constant velocity, because the direction is changing continuously, so the acceleration is not zero and the forces cannot sum to zero. An object in uniform circular motion has a net force directed toward the centre of its path, and that force is what keeps changing the direction of the velocity. The phrase moving at a constant rate makes this sound like the dynamic equilibrium case, and it is not, which is why the test is always whether the velocity vector is changing rather than whether the speedometer reading is.

How do you tell whether a problem is a static or a dynamic equilibrium problem?

You do not need to, because the method is the same either way: draw the forces, resolve them onto axes, and set each component sum to zero. What the words at rest do tell you is extra information that is often useful. A stationary object is held by static friction rather than kinetic friction, and static friction has the larger coefficient, so a block resting on a slope and a block sliding down it at constant speed do not need the same angle. A stationary rigid body is usually also not rotating, so the problem is handing you the zero-net-torque condition as well. Read at rest as a package of extra facts, not as a different physics.