12.1 Conditions for Translational and Rotational Equilibrium

The Crane That Does Nothing

[Video: A construction crane on a city skyline holds a concrete slab suspended thirty meters above the street. The camera zooms in on the steel cable --- taut, motionless. It pans down to the counterweight at the back of the crane arm, then to the massive bolts anchoring the crane's base to the foundation. Overlay graphics appear: force arrows at the cable attachment point, at the pivot, at the counterweight, at the base. Every arrow is enormous --- tens of thousands of newtons. A torque diagram appears around the pivot point, showing clockwise and counterclockwise contributions. The final frame shows the net force readout: 0 N. The net torque readout: 0 N m. Text: "Nothing moves. Everything pushes."]

Look at those numbers. The cable pulls with a force that could crush a car. The counterweight pushes down with the weight of a small truck. The base bolts resist forces that would rip apart most materials. And the result of all of this? Nothing happens.

The crane does not accelerate. It does not rotate. It just sits there, silently, under enormous stress.

This is not a trivial situation. "Nothing moves" does not mean "nothing is going on." It means that every single force and every single torque has been precisely balanced so that the net effect is zero. Change one cable tension by a few percent, shift the load a meter to the left, and the whole structure could begin to rotate or translate --- with catastrophic results.

So here is the question: what exactly must be true about the forces and torques on a rigid body for it to remain in equilibrium?

You already know pieces of the answer. In Chapter 5, you learned that $\sum \vec{F} = \vec{0}$ means no translational acceleration. In Chapter 10, you learned that $\sum \vec{\tau} = \vec{0}$ means no angular acceleration. But you studied those conditions separately. In this section, we put them together --- and discover that both must hold simultaneously, and that satisfying one does not guarantee the other.

A Beam, Two Supports, and a Prediction

Before we formalize anything, let's build some intuition.

Picture a uniform beam --- say, a 4-meter wooden plank with a weight of 200 N --- resting horizontally on two sawhorses. The sawhorses are positioned at the two ends of the beam.

By symmetry, each sawhorse supports half the beam's weight: 100 N each.

Now imagine you slide sawhorse B (the right one) inward, toward the center of the beam.

Prediction --- commit before reading on:

As you move sawhorse B closer to the center of the beam, does the force that sawhorse A (the left one) must exert:

(a) Increase? (b) Decrease? (c) Stay the same?

Pick one and give a brief reason. Don't hedge --- commit to an answer.

Take a moment. This is not a trick question, but it does require careful thought.

[Interactive: Predict-Then-Reveal. Student selects (a), (b), or (c) and types a brief justification. After submitting, the explanation below unlocks.]

Here is what happens. The beam's weight acts at its center, 2 m from sawhorse A. If sawhorse B is at the far end (4 m from A), the torques about A give:

$$N_B \cdot 4 = 200 \cdot 2 \implies N_B = 100 \; \text{N}$$

Now slide sawhorse B to the 3-meter mark. The torque equation about A becomes:

$$N_B \cdot 3 = 200 \cdot 2 \implies N_B = 133 \; \text{N}$$

Force balance ($N_A + N_B = 200$ N) gives $N_A = 67$ N.

The force on sawhorse A decreased. Moving B closer to the center forced B to carry more of the load, because B's shorter lever arm means it needs a larger force to produce the same torque about A.

If you got this wrong, good --- this is exactly the kind of situation where quick intuition fails and careful torque analysis is essential. If you got it right, notice that you could not have reached the answer from force balance alone. You needed the torque condition.

Exploring Two Conditions at Once

Let's play with this more systematically.

[Interactive: Beam Equilibrium Explorer. A horizontal beam is shown on screen, with two triangular supports whose positions can be dragged left and right along the beam. One or more point loads can be placed anywhere on the beam by clicking. The beam's own weight is shown as a downward arrow at its center of mass.

As the student adjusts support positions and load placements, two panels update in real time:

Force panel: Shows $\sum F_y$ with a color-coded bar. Green when $\sum F_y = 0$, red otherwise. Displays numerical values of all vertical forces.
Torque panel: Shows $\sum \tau$ (about a student-selectable pivot point) with a similar color-coded bar. Green when $\sum \tau = 0$, red otherwise. Displays the individual torque contributions with their signs.

A status banner at the top reads "TRANSLATIONAL EQUILIBRIUM" (green check or red X) and "ROTATIONAL EQUILIBRIUM" (green check or red X) independently.

Guided prompts appear in sequence:

"Place both supports symmetrically. Is the beam in equilibrium?"
"Now add a load near one end. Are both conditions still satisfied?"
"Can you find a configuration where $\sum F = 0$ but $\sum \tau \neq 0$? What would happen to the beam physically?"
"Can you find a configuration where $\sum \tau = 0$ but $\sum F \neq 0$? What would that look like?"

For prompt 3, a hint is available: "Try replacing the two supports with two equal upward forces placed symmetrically, but with the beam's weight shifted off-center." For prompt 4, a hint reads: "Imagine two forces of equal magnitude pushing in the same direction (both up), and one force pushing down but placed exactly at the pivot. The torques might cancel, but the net force would not."]

This interactive is worth spending several minutes on. The central discovery is this: the force condition and the torque condition are independent. You can satisfy one without the other. And when only one is satisfied, the beam is not in equilibrium --- it either accelerates linearly, rotates, or both.

Pause and think: What did you find for prompt 3 --- a situation where forces balance but torques do not? What would happen to that beam in real life? It would not fly off in any direction (no net force), but it would start spinning. The two balanced forces act as a couple --- a pair of forces that produces pure rotation without translation.

The Two Conditions: Why Both Are Necessary

Let's now state the conditions formally and see why each one, on its own, is not enough.

Condition 1: Translational Equilibrium

$$\sum \vec{F} = \vec{0}$$

This is Newton's second law with $\vec{a} = \vec{0}$. If the net force on an object is zero, its center of mass does not accelerate. It either stays at rest or moves at constant velocity in a straight line.

You first encountered this in Section 5.6. It handles translation --- motion of the center of mass through space.

Condition 2: Rotational Equilibrium

$$\sum \vec{\tau} = \vec{0}$$

This is Newton's second law for rotation with $\alpha = 0$. If the net torque about any point is zero, the object has no angular acceleration. It either stays non-rotating or rotates at a constant angular velocity.

You first encountered this in Section 10.3. It handles rotation --- spinning about an axis.

Why both?

Because translation and rotation are independent types of motion for a rigid body. A rigid body can translate without rotating (a hockey puck sliding), rotate without translating (a spinning top), or do both at once (a rolling wheel). Stopping one does not automatically stop the other.

Here are two contrasting examples that make this concrete:

Example A: Forces balance, but torques do not.

Imagine you grip opposite edges of a book lying on a table and push with equal forces in opposite directions --- your right hand pushes the far edge to the left, your left hand pushes the near edge to the right. Both forces are 5 N.

The net force is zero: $5 \, \text{N (left)} + 5 \, \text{N (right)} = 0$. The book's center of mass does not accelerate. But the torques from these two forces are both clockwise (or both counterclockwise, depending on perspective). The net torque is decidedly not zero. The book spins.

This pair of equal and opposite forces separated by a distance is called a couple. A couple produces pure rotation with zero net force.

Example B: Torques balance, but forces do not.

Imagine a seesaw with a pivot at the center. Two children of the same weight sit at equal distances from the pivot. The torques balance: the clockwise torque from one child equals the counterclockwise torque from the other. $\sum \tau = 0$.

But now consider all the forces. The two children push down, and the pivot pushes up. If the pivot's upward force does not equal the total weight of both children, then $\sum F \neq 0$, and the entire seesaw accelerates downward (or upward).

For complete equilibrium, the pivot must supply an upward force equal to the combined weight of both children. You need both conditions.

The Three Scalar Equations in 2D

Most problems in this chapter take place in two dimensions. In 2D, the vector conditions $\sum \vec{F} = \vec{0}$ and $\sum \vec{\tau} = \vec{0}$ reduce to three scalar equations:

$$\sum F_x = 0$$

$$\sum F_y = 0$$

$$\sum \tau = 0$$

Here $\sum \tau$ is the net torque computed about some chosen point (you choose the pivot --- more on that in Section 12.3), and the sign convention is your standard one: counterclockwise positive, clockwise negative (or vice versa, as long as you are consistent).

Three equations means you can solve for at most three unknowns. This is a critical constraint. If your free-body diagram has more than three unknown quantities (force magnitudes, directions, positions), the problem is either statically indeterminate (requires additional information, such as material properties) or you need to identify additional geometric constraints.

For now, the important point is structural: static equilibrium in 2D gives you exactly three independent equations to work with. Every problem in this chapter comes down to setting up these three equations carefully and solving them.

The force equations handle the translational side. The torque equation handles the rotational side. You need all three.

Connecting to What You Already Know

This section is not introducing new physics. It is combining physics you already have.

What you learned	Where you learned it	How it appears here
Free-body diagrams	Section 5.1	Every equilibrium problem starts with a careful FBD
Translational equilibrium: $\sum \vec{F} = \vec{0}$	Section 5.6	One of the two required conditions
Rotational equilibrium: $\sum \vec{\tau} = \vec{0}$	Section 10.3	The other required condition
Torque depends on lever arm	Section 10.3	Torque calculations require distances from the pivot
Choice of axis (pivot point)	Section 10.7	You are free to choose any point as the torque reference --- and a smart choice simplifies the algebra
Center of mass	Section 8.6	Gravity acts at the center of mass of a uniform body

The new idea in this chapter is not any single condition. It is the simultaneous demand that both conditions hold. And the new skill is learning to set up problems where both conditions are active --- where you cannot solve for the unknowns without using force balance and torque balance together.

Think of it this way: Chapter 5 gave you the brakes for translation. Chapter 10 gave you the brakes for rotation. This chapter asks: what happens when you apply both brakes at once?

A Quick Example: Checking Both Conditions

Let's see the two conditions in action with a concrete setup.

Setup: A uniform 6-meter beam of weight $W = 300$ N rests horizontally on two supports. Support A is at the left end ($x = 0$). Support B is at $x = 4$ m. The beam extends 2 m past support B on the right side.

Question: Find the support forces $N_A$ and $N_B$. Is the beam in equilibrium?

Step 1: Free-body diagram.

Forces on the beam: - $N_A$ upward at $x = 0$ - $N_B$ upward at $x = 4$ m - $W = 300$ N downward at $x = 3$ m (center of the 6-meter beam)

Step 2: Torque balance. Choose support A ($x = 0$) as the pivot (this eliminates $N_A$ from the torque equation, since its lever arm is zero):

$$\sum \tau_A = 0: \quad N_B \cdot 4 - W \cdot 3 = 0$$

$$N_B = \frac{300 \cdot 3}{4} = 225 \; \text{N}$$

Step 3: Force balance.

$$\sum F_y = 0: \quad N_A + N_B - W = 0$$

$$N_A = 300 - 225 = 75 \; \text{N}$$

Step 4: Interpret. Support B carries more of the weight because the beam's center of mass is closer to B. Support A carries less. The beam extends past B, creating a slight overhang, but the center of mass is still between the two supports, so the beam does not tip.

Notice that we needed both equations. The torque equation alone gave us $N_B$. The force equation then gave us $N_A$. Neither equation alone determines both unknowns.

Practice

Layer 1: Concrete

A uniform horizontal beam has a length of 5 m and weighs 400 N. It rests on two supports: support A at $x = 0$ (the left end) and support B at $x = 5$ m (the right end). A 600 N load is placed at $x = 2$ m.

(a) Draw a free-body diagram of the beam showing all forces.

(b) Write the torque-balance equation about support A and solve for $N_B$.

(c) Write the force-balance equation and solve for $N_A$.

(d) Verify your answers by computing $\sum \tau$ about support B. Does it equal zero?

Check your answer

**(a)** Forces on the beam: - $N_A$ upward at $x = 0$ - $N_B$ upward at $x = 5$ m - Weight of beam: 400 N downward at $x = 2.5$ m (center of a uniform beam) - Applied load: 600 N downward at $x = 2$ m **(b)** Torques about support A (counterclockwise positive): $$\sum \tau_A = N_B \cdot 5 - 400 \cdot 2.5 - 600 \cdot 2 = 0$$ $$5 N_B = 1000 + 1200 = 2200$$ $$N_B = 440 \; \text{N}$$ **(c)** Force balance: $$N_A + N_B = 400 + 600 = 1000 \; \text{N}$$ $$N_A = 1000 - 440 = 560 \; \text{N}$$ **(d)** Check by computing torques about support B: $$\sum \tau_B = -N_A \cdot 5 + 400 \cdot 2.5 + 600 \cdot 3 = -2800 + 1000 + 1800 = 0 \; \checkmark$$ The torque about B is indeed zero, confirming our answers. Notice that $N_A > N_B$ because the 600 N load is closer to A than to B, so A carries more of that load.

Layer 2: Pattern

For each of the following setups, determine whether the system is in (i) translational equilibrium, (ii) rotational equilibrium, or (iii) both.

(a) A wheel rolls across a flat floor at constant velocity without slipping. No net horizontal force acts on it.

(b) A rod is pinned at its center. You push the left end up with 10 N and the right end up with 10 N. The rod has negligible weight.

(c) A rod is pinned at its center. You push the left end up with 10 N and the right end down with 10 N. The rod has negligible weight.

(d) A uniform beam rests on a single support at its center of mass. No other forces act (besides gravity and the normal force from the support).

Check your answer

**(a)** Both translational and rotational equilibrium. The wheel moves at constant velocity (no translational acceleration) and rolls without slipping at constant speed (no angular acceleration). $\sum F = 0$ and $\sum \tau = 0$. Note: equilibrium does not require the object to be at rest --- constant velocity counts. **(b)** Not in translational equilibrium. The two 10 N forces are both upward, so $\sum F_y = 20$ N (upward). The pin must exert 20 N downward for force balance. If the pin does exert that force, then $\sum F = 0$. Now check torques about the center: both forces are upward at equal distances from the center, so one produces a clockwise torque and the other counterclockwise. They cancel: $\sum \tau = 0$. So if the pin supplies the correct reaction, this system *is* in both equilibria. If the pin cannot supply the reaction force, it is not in translational equilibrium. **(c)** Translational equilibrium is satisfied (the two forces cancel: 10 N up + 10 N down = 0), but rotational equilibrium is *not*. Both forces produce torques in the same rotational direction about the center --- one pushes the left end up and the other pushes the right end down, both creating clockwise rotation. This is a couple. The rod spins. **(d)** Both. Gravity pulls down at the center of mass, the support pushes up at the same point. $\sum F = 0$ and $\sum \tau = 0$ (both forces act at the same point, so neither produces a torque about that point). The beam is in complete equilibrium.

Layer 3: Structure

Why do you need both equilibrium conditions? Answer this question by constructing two specific examples:

(a) Describe a physical setup where $\sum \vec{F} = \vec{0}$ but $\sum \vec{\tau} \neq \vec{0}$. Explain what motion results.

(b) Describe a physical setup where $\sum \vec{\tau} = \vec{0}$ (about some point) but $\sum \vec{F} \neq \vec{0}$. Explain what motion results.

(c) In your own words, explain why satisfying one condition does not automatically guarantee the other.

Check your answer

**(a)** A couple: two equal-magnitude, opposite-direction forces applied at different points on a rigid body. For instance, push the top of a steering wheel to the right with 5 N and the bottom to the left with 5 N. The forces cancel ($\sum F = 0$), so the center of mass does not accelerate. But both forces create torques in the same direction about the center, so $\sum \tau \neq 0$. The result: the wheel spins in place (pure rotation, no translation). **(b)** Consider a uniform rod of weight $W$ lying on a table, supported at its center. Now apply an additional 10 N force downward at the center. Torques about the center: the weight acts at the center (zero lever arm, zero torque), the support force acts at the center (zero torque), and the applied 10 N also acts at the center (zero torque). So $\sum \tau = 0$ about that point. But $\sum F_y = N - W - 10 \neq 0$ unless the normal force adjusts. If the table cannot provide enough normal force (say the rod is glued to a surface and you pull), then $\sum F \neq 0$ and the rod accelerates linearly downward (or upward) without rotating. A simpler example: a single force applied at the pivot point of a pinned rod. The torque about the pin is zero (zero lever arm), but the net force is not zero. The pin's reaction forces must handle this, or the entire system accelerates. **(c)** The force condition governs the motion of the center of mass. The torque condition governs rotation about the center of mass (or any point). These are independent aspects of a rigid body's motion. A rigid body has six degrees of freedom in 3D (three translational, three rotational) --- or three in 2D ($x$, $y$, and $\theta$). The force equations constrain the translational degrees, and the torque equation constrains the rotational degree. Constraining one set does nothing to constrain the other.

Layer 4: Debug

A student analyzes the following situation:

A uniform 4-meter beam of weight 200 N is supported at both ends. A 300 N load hangs from the midpoint. The student writes:

$$\sum F_y = N_A + N_B - 200 - 300 = 0 \implies N_A + N_B = 500 \; \text{N}$$

The student then says: "The forces balance, so the beam is in equilibrium. By symmetry, $N_A = N_B = 250$ N."

(a) Is the student's force-balance equation correct?

(b) Is the student's conclusion that the beam is in equilibrium justified by only checking force balance?

(c) Now suppose the 300 N load is moved from the midpoint to a position 1 m from support A. The student again applies "symmetry" and claims $N_A = N_B = 250$ N. Show, using the torque condition, that the student's answer is wrong. Find the correct values.

Check your answer

**(a)** Yes, the force-balance equation is correct. The total upward force must equal the total downward force for translational equilibrium. **(b)** No. Force balance alone does not guarantee equilibrium. The student also needs to verify $\sum \tau = 0$. In this particular case (both supports at the ends, both the beam's weight and the load at the midpoint), symmetry *does* make the torques balance, so the conclusion happens to be correct --- but the reasoning is incomplete. The student got lucky. **(c)** With the 300 N load at $x = 1$ m (1 m from support A at $x = 0$, with support B at $x = 4$ m): Torques about A: $$\sum \tau_A = N_B \cdot 4 - 200 \cdot 2 - 300 \cdot 1 = 0$$ $$4 N_B = 400 + 300 = 700 \implies N_B = 175 \; \text{N}$$ Force balance: $$N_A = 500 - 175 = 325 \; \text{N}$$ The correct answers are $N_A = 325$ N and $N_B = 175$ N --- not 250 N each. The load is closer to A, so A bears more of it. The student's error was assuming symmetry when the load placement is asymmetric. Only the torque condition reveals this asymmetry. This is exactly why checking only $\sum F = 0$ is not enough. The force equation tells you the *total* support force. The torque equation tells you how it is *distributed* between the supports. You need both.

Reflection

Think back over this section.

What does "equilibrium" actually require beyond "nothing is moving"?

Consider this: a car driving at a constant 60 km/h on a straight, flat road is in translational equilibrium --- $\sum F = 0$, and the center of mass has zero acceleration. A wheel spinning at constant angular velocity is in rotational equilibrium --- $\sum \tau = 0$, and the angular acceleration is zero.

Equilibrium is not about being motionless. It is about having zero acceleration --- both translational and rotational. And for a rigid body, you must check both independently, because forces and torques govern different aspects of the body's motion.

In your own words: why is it dangerous to check only one equilibrium condition and declare a structure "safe"?

Looking Ahead

You now know the two conditions that define rigid-body equilibrium, and you have seen that both are necessary. But knowing the conditions and using them effectively are different things.

In the next section, we turn to the role of the center of mass and support geometry. Where the center of mass sits relative to the supports determines whether an object balances or tips --- and this geometric reasoning will become your primary tool for analyzing stability. You have already seen hints of this: the beam's center of mass determined how the weight distributed between the supports. In Section 12.2, we make that connection precise.

The problems will also get more interesting. Beams on two supports are a starting point, but the real payoff comes when you analyze ladders, cranes, bridges, and suspended signs --- systems where geometry, friction, and strategic pivot choice all matter at once. The two conditions you learned here are the foundation for all of it.