"The art of structure is where to put the holes."
— Robert Le Ricolais, structural engineer

M5L1 through M5L3 were all about motion — things spinning, rolling, accelerating, conserving angular momentum. Now we arrive at the other half of the module title: Equilibrium. What conditions must hold for a structure to remain perfectly still? The answer turns out to require two independent equations, not one. A net force of zero prevents the object from translating. But a net force of zero is not enough — a pair of equal-and-opposite forces at different points can still produce a net torque that sets the object spinning. Equilibrium demands both ΣF = 0 and Στ = 0. This lesson is about learning to use both conditions simultaneously — and about the single most powerful exam skill in static equilibrium: choosing where to place your pivot point.

The Two Conditions for Static Equilibrium

An object is in static equilibrium if and only if both of these are satisfied simultaneously:

Condition 1: Translational Equilibrium

ΣF = 0

The vector sum of all external forces is zero. Separately: ΣF_x = 0 and ΣF_y = 0. This ensures the center of mass does not accelerate — the object does not slide or fall.

Condition 2: Rotational Equilibrium

Στ = 0

The sum of all torques about any chosen pivot point is zero. This ensures the object does not begin to rotate. The choice of pivot is yours — and choosing wisely eliminates unknowns.

The Pivot Strategy: Place your pivot at the location of an unknown force. That force produces zero torque about its own point of application (moment arm = 0), eliminating it from the Στ = 0 equation entirely. This is the central exam skill of static equilibrium.

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

CC5.4 Apply the conditions of static equilibrium to analyze structures and extended bodies

★ LO5.4.1 State the two conditions for static equilibrium (ΣF = 0 and Στ = 0) and explain why both are necessary

★ LO5.4.2 Construct a free-body diagram for an extended object, identifying all forces and their points of application

★ LO5.4.3 Apply the pivot-point strategy: choose the pivot to eliminate an unknown force from the torque equation

★ LO5.4.4 Locate the center of gravity of a system of masses and explain its role in equilibrium analysis

★ LO5.4.5 Solve equilibrium problems involving beams, ladders, hinges, and cables

★ LO5.4.6 Identify the tipping condition: when an object's center of gravity moves outside its base of support

Required Reading

Read these sections in order. Chapter 12.1 establishes the two conditions; 12.2 develops the problem-solving strategy including pivot selection; 12.3 applies both to a range of real structures.

12.1 Conditions for Static Equilibrium 12.2 Examples of Static Equilibrium 12.3 Stress, Strain, and Elastic Modulus

Optional Reading

Explore More

Physics Classroom: Equilibrium & Statics Khan Academy: Torque and Equilibrium PhET: Balancing Act Simulation PhET: Torque Simulation

Media

Video 1 establishes the two conditions conceptually; Video 2 teaches the pivot-point strategy step by step with beam problems; Video 3 tackles ladder problems and distributed loads including center of gravity.

Video 1: Two Conditions — Why One Is Not Enough

Both ΣF = 0 and Στ = 0 Are Required

It is entirely possible to have zero net force on an object but still have it start rotating. This video builds the intuition for why both conditions are independently necessary.

A pair of equal-and-opposite forces applied at different points: zero net force, but nonzero torque — the object spins
Translational equilibrium vs. rotational equilibrium: conceptual distinction
Writing the two vector equations in component form for 2D problems
Why the torque condition holds about any pivot point when an object is in equilibrium
Setting up a systematic free-body diagram for an extended object: forces, positions, and directions all matter

Time: 9:00

Video 2: The Pivot-Point Strategy

Choose Your Pivot to Eliminate Unknowns

The single most powerful technique in static equilibrium problems: deliberately placing the pivot at a location that kills an unknown force and gives you a solvable equation immediately.

Why any point can serve as the pivot when the body is in static equilibrium
A force produces zero torque about a pivot placed at its own point of application (r = 0)
Step-by-step: uniform beam supported at one end by a hinge, other end by a cable at angle θ
Pivot at the hinge eliminates hinge forces entirely from the torque equation — solve for cable tension immediately
Then use ΣF = 0 to find the hinge reaction forces
Non-uniform beam: locating the weight's effective point of application

Time: 11:30

Video 3: Ladders, Centers of Gravity, and Tipping

Real Structures with Distributed Mass and Multiple Contact Forces

Ladders leaning against walls are the classic multi-force equilibrium problem. This video also addresses how center of gravity differs from center of mass and introduces the tipping condition.

Ladder against a frictionless wall: four forces, two equations, one clever pivot
The role of static friction at the base — why a ladder slides when friction is insufficient
Center of gravity: the single point where gravity's torque equals the distributed weight torque
Center of gravity vs. center of mass: they coincide in uniform gravitational fields; differ when g varies (e.g., very tall structures)
Tipping condition: the object tips when its center of gravity moves outside the base of support
Why wide, low vehicles are stable; why tall, narrow objects are not

Time: 12:00

Deeper Look: Equilibrium from First Principles

Why the Pivot Can Be Anywhere

The Pivot-Independence of Equilibrium

When an object is in static equilibrium, the torque condition Στ = 0 holds about every possible pivot point simultaneously. This is not obvious — here is why it is true.

Let point A and point B be two different pivot choices separated by displacement d. The torque of force F about A is r_A × F, and about B is r_B × F = (r_A + d) × F.

                        ΣτB = ΣτA + d × ΣF
                    

If both equilibrium conditions hold (ΣF = 0 and Στ_A = 0), then Στ_B = 0 + d × 0 = 0 as well. The torque condition is satisfied at every pivot.

Practical consequence: You are free to write the torque equation about whatever point is most convenient. The result will be correct regardless. Choosing the pivot at an unknown force's point of application is not a trick — it is a legitimate and rigorous choice.

How Many Independent Equations?

In 2D, static equilibrium yields three independent scalar equations: ΣF_x = 0, ΣF_y = 0, and Στ = 0. You can solve for at most three unknowns. Systems with more unknowns than equations are statically indeterminate — they require material deformation (elasticity) to solve. A table with four legs is statically indeterminate; a three-legged stool is not.

Stable, Unstable, and Neutral Equilibrium

Not All Equilibrium Is Equal

An object can satisfy ΣF = 0 and Στ = 0 and still be either safe or dangerous, depending on what happens when it is displaced slightly.

Type	What happens on small displacement	Physical criterion	Example
Stable	Returns to original position	CG rises on displacement; potential energy is a minimum	Ball in a bowl; rocking chair on curved feet
Unstable	Moves further from original position	CG falls on displacement; potential energy is a maximum	Pencil balanced on its tip; inverted pendulum
Neutral	Stays in new position	CG height unchanged; potential energy is constant	Ball on a flat surface; cylinder rolling on a flat surface

Key insight: For extended objects, stability is governed by the relationship between the center of gravity and the base of support. An object is stable as long as a vertical line through its CG falls within the base of support. When that line falls outside the base — even slightly — the gravitational torque tips the object over rather than restoring it.

Engineering Application

This is why cranes have massive counterweights, why sports cars have low centers of gravity, and why the Leaning Tower of Pisa (CG still above its base — barely) has not fallen. It is also why loaded 18-wheelers are restricted on curved on-ramps — lateral acceleration shifts the effective CG location.

The Human Body as a Lever System

Muscles, Bones, and Mechanical Advantage

The human musculoskeletal system is a collection of lever systems operating under static equilibrium conditions. Understanding them requires applying Στ = 0 about joint pivot points.

The forearm as a third-class lever:

The elbow joint is the pivot. The bicep muscle attaches very close to the joint — typically about 5 cm from the elbow — while the hand holds a weight 35 cm from the elbow. Applying Στ = 0 about the elbow:

                        Fmuscle × 0.05 m = Wload × 0.35 m

                        Fmuscle = 7 × Wload

The bicep must exert seven times the weight being held. Holding a 10 kg weight requires roughly 700 N of bicep force. This seems inefficient — and it is, from a force perspective — but the short moment arm also means the muscle needs to contract only a tiny amount to produce a large arc of motion at the hand. The body trades force for speed and range of motion.

Other body levers:

Neck: The head (~5 kg) pivots at the atlanto-occipital joint; neck extensor muscles are close to the pivot, so they must exert very large forces to hold the head erect — a significant source of neck strain during forward head posture.
Foot/calf: Standing on tiptoe pivots at the ball of the foot; the Achilles tendon (close to the pivot) must support the entire body weight acting at the ankle.
Lower back: The spine pivots at the sacrum; back muscles attach very close to the pivot, so bending forward at 60° requires back muscles to generate forces up to 10 × body weight — a major source of disc compression injury.

Why "Lift with Your Legs"

When lifting with a bent back, the torque about the lumbar spine from the load plus the upper body weight is enormous — and the short erector spinae moment arm means the muscle force required is extreme. Lifting with a straight back and bending at the knees moves the load closer to the spine's axis of rotation, dramatically reducing both the torque and the required muscle force.

Practice and Apply — Conceptual

The Pivot-Point Strategy

Put the Static Equilibrium Solution Steps in Order

Arrange these steps into the correct sequence for solving a static equilibrium problem using the pivot strategy. Drag and drop to reorder.

Draw a complete free-body diagram showing all forces and their exact points of application on the extended object
Set up a coordinate system and establish a positive direction for torques (usually CCW = positive)
Choose the pivot point at the location of an unknown force you want to eliminate from the torque equation
Write Στ = 0 about the chosen pivot, calculating the torque from each remaining force as ±r⊥F
Solve the torque equation for the first unknown (typically a cable tension or support force)
Write ΣF_x = 0 and ΣF_y = 0 to solve for remaining unknown force components
Check: verify that the torque condition is also satisfied about a different pivot point as a consistency check

Equilibrium or Not?

Sort Each Situation into the Correct Category

Each scenario describes forces acting on an extended rigid body. Decide whether it is in translational equilibrium, rotational equilibrium, both, or neither. Sort into the correct bucket.

Situations to Sort

Two equal forces pulling opposite ends of a rod in opposite directions along the rod's axis
A book sitting motionless on a table with gravity and normal force acting at the same point
A steering wheel with two hands applying equal forces on opposite sides, both pushing clockwise
A uniform beam balanced on a fulcrum at its center with equal weights at each end
A door being held open: the hinge pushes right, your hand pushes left at the outer edge with equal magnitude
A spinning top precessing at constant rate with gravity and the normal force acting at the tip

Both ΣF = 0 and Στ = 0

ΣF = 0 but Στ ≠ 0

Στ = 0 but ΣF ≠ 0

Neither condition satisfied

Common Errors in Equilibrium

Think It Through — Then Flip

Each card presents a common student error or misconception about static equilibrium. Decide what is wrong before flipping to see the correction.

"If ΣF = 0, the object must be in equilibrium."

Incorrect — Translational Equilibrium Only

ΣF = 0 prevents translation but says nothing about rotation. A couple — two equal-and-opposite forces applied at different points — has zero net force but produces a net torque. The object can begin to rotate even though its center of mass does not accelerate. Full equilibrium requires both conditions satisfied.

"You must place the pivot at the geometric center of the object."

Incorrect — The Pivot Is Your Free Choice

When an object is in static equilibrium, Στ = 0 holds about every point. You may place the pivot anywhere that is mathematically convenient — and the strategically smart choice is at the location of an unknown force you want to eliminate. Placing the pivot at the geometric center is usually the worst choice because it keeps all unknowns in the equation.

"For a uniform beam, I can place gravity's force anywhere along the beam."

Incorrect — Gravity Acts at the Center of Gravity

For torque calculations, the entire weight of a uniform beam acts as a single downward force at the beam's center of gravity — which coincides with the geometric center for a uniform beam. Placing W at the wrong position changes the torque calculation and gives a wrong answer. For non-uniform beams, you must locate the CG correctly first.

"A hinge always exerts only a vertical force on a beam."

Incorrect — A Hinge Exerts Forces in Both Directions

A hinge is a pin joint that can exert a force in any direction — both horizontal and vertical components. This is why hinge problems have two unknown force components (F_hx and F_hy), not one. The pivot strategy is especially valuable here: placing the pivot at the hinge eliminates both unknown components from the torque equation in one move.

Practice and Apply — Computational

Quick Reference: Static Equilibrium

                        Condition 1: ΣFx = 0 and ΣFy = 0

                        prevents translation

                        Condition 2: Στ = 0 (any pivot)

                        prevents rotation

                        Torque: τ = r⊥ × F = rF sinθ

                        r⊥ = perpendicular distance from pivot to line of action

                        CG location: xcg = Σ(mixi) / Σmi

                        same as center of mass in uniform g

Problem 1 — Uniform Beam with a Cable

A uniform beam of mass 20 kg and length 4.0 m is attached to a wall by a hinge at its left end. A cable attached to the right end makes an angle of 30° above the horizontal and supports the beam horizontally. Find (a) the tension in the cable and (b) the horizontal and vertical components of the hinge force.

Setup: The beam is horizontal. Forces: weight W = mg = (20)(9.8) = 196 N downward at the center (x = 2.0 m from hinge); cable tension T upward-and-right at 30° at x = 4.0 m; hinge forces F_hx and F_hy at x = 0.

Pivot strategy: Place pivot at the hinge (x = 0). This eliminates F_hx and F_hy from the torque equation — both act at zero moment arm.

(a) Στ = 0 about the hinge:

CCW positive. T·sin30° acts upward at x = 4.0 m (CCW torque). W acts downward at x = 2.0 m (CW torque).

T sin30° × 4.0 − W × 2.0 = 0

T(0.500)(4.0) = 196 × 2.0

2.0T = 392 ⇒ T = 196 N

(b) ΣF_x = 0:

F_hx + T cos30° = 0 ⇒ F_hx = −T cos30° = −196(0.866) = −170 N (hinge pulls beam toward wall)

ΣF_y = 0:

F_hy + T sin30° − W = 0 ⇒ F_hy = 196 − 196(0.500) = 196 − 98 = 98 N upward

Notice the pivot strategy delivered T directly from one clean equation. If we had placed the pivot at the center, all three unknowns would have appeared simultaneously.

Problem 2 — Non-Uniform Beam with Two Supports

A non-uniform beam of mass 15 kg and length 5.0 m is supported at two points: support A at x = 0 and support B at x = 5.0 m. The beam's center of gravity is located 2.0 m from end A. A 40 kg crate sits 1.5 m from end A. Find the upward forces F_A and F_B exerted by each support.

Identify all forces: F_A up at x = 0; F_B up at x = 5.0 m; beam weight W_beam = 147 N down at x = 2.0 m; crate weight W_crate = (40)(9.8) = 392 N down at x = 1.5 m.

Pivot at A (x = 0) to eliminate F_A:

Στ_A = 0:

F_B(5.0) − W_beam(2.0) − W_crate(1.5) = 0

5.0 F_B = 147(2.0) + 392(1.5) = 294 + 588 = 882

F_B = 176.4 N

ΣF_y = 0:

F_A + F_B − W_beam − W_crate = 0

F_A = 147 + 392 − 176.4 = 362.6 N

Check by pivoting at B:

F_A(5.0) − W_beam(3.0) − W_crate(3.5) = 362.6(5.0) − 147(3.0) − 392(3.5) = 1813 − 441 − 1372 = 0 ✓

The non-uniform beam's center of gravity is given explicitly here. In practice you would locate it by finding where the beam balances, or by calculating it from the mass distribution.

Problem 3 — Ladder Against a Frictionless Wall

A uniform 8.0 m ladder of mass 12 kg leans against a smooth (frictionless) vertical wall, making a 60° angle with the ground. The floor exerts both normal and friction forces on the base. Find the forces on the ladder and the minimum coefficient of static friction needed to prevent slipping.

Forces: Normal from wall N_w (horizontal, pointing away from wall) at the top; Normal from floor N_f (upward) at the base; Friction from floor f (horizontal, toward wall) at the base; Weight W = (12)(9.8) = 117.6 N downward at the center (4.0 m along ladder from base).

Pivot at the base — eliminates N_f and f (both act there):

Στ_base = 0:

N_w acts at the top of the ladder. Its moment arm = 8.0 sin60° = 6.93 m (horizontal distance to base).

W acts downward at the ladder's midpoint. Its moment arm = 4.0 cos60° = 2.0 m (horizontal distance from base).

N_w(6.93) − W(2.0) = 0

N_w = 117.6 × 2.0 / 6.93 = 33.95 N

ΣF_x = 0: f − N_w = 0 ⇒ f = N_w = 33.95 N (friction points toward wall)

ΣF_y = 0: N_f − W = 0 ⇒ N_f = 117.6 N

Minimum coefficient of static friction:

μ_s,min = f / N_f = 33.95 / 117.6 = 0.289

If the floor's coefficient of static friction is less than 0.289, the ladder will slide. As the ladder angle decreases (becomes more horizontal), the moment arm of W increases while that of N_w decreases — so a shallower ladder requires a larger friction force and is harder to keep in place.

Problem 4 — Cantilever Bracket (Hinge + Horizontal Cable)

A 3.0 m horizontal shelf bracket is attached to a wall by a hinge at its left end and a horizontal cable at 1.8 m from the wall (60% of its length). The bracket itself has mass 5.0 kg; a 25 kg load hangs from the free right end. Find the tension in the cable and the hinge reaction.

Forces: Hinge at x = 0 (components F_hx, F_hy); horizontal cable tension T (pointing toward wall, so T acts in the −x direction on the bracket) at x = 1.8 m — wait, a horizontal cable along a horizontal bracket would produce no vertical support. Let us re-specify: the cable is attached to the wall above the hinge and runs diagonally to the bracket at x = 1.8 m, making angle θ = 50° above horizontal.

Setup (cable at 50° at x = 1.8 m):

Bracket weight: W_b = 49 N at x = 1.5 m (center)

Load: W_L = 245 N at x = 3.0 m

Cable tension T at 50° at x = 1.8 m; vertical component = T sin50°

Pivot at hinge (x = 0):

Στ = 0: T sin50°(1.8) − W_b(1.5) − W_L(3.0) = 0

T(0.766)(1.8) = 49(1.5) + 245(3.0) = 73.5 + 735 = 808.5

1.379 T = 808.5 ⇒ T = 586.3 N

ΣF_x = 0: F_hx − T cos50° = 0 ⇒ F_hx = 586.3(0.643) = 377 N (hinge pulls bracket toward wall)

ΣF_y = 0: F_hy + T sin50° − W_b − W_L = 0

F_hy = 294 − 586.3(0.766) = 294 − 449.1 = −155 N (hinge pushes bracket downward)

The negative F_hy tells us the hinge must actually push down on the bracket — the cable is pulling the right end of the bracket upward so strongly that the left end would lift off the hinge without a downward pin force. This kind of sign surprise is why solving formally and trusting the math matters.

Problem 5 — Center of Gravity and Tipping

A filing cabinet is 0.50 m wide and 1.30 m tall. It has mass 18 kg distributed uniformly. A fully loaded top drawer (6.0 kg) is pulled out 0.35 m, shifting its effective center of mass to 0.35 m in front of the cabinet's front face. (a) Find the combined center of gravity of the system. (b) Determine whether the cabinet will tip, assuming it rests on a flat floor with the base extending from x = 0 to x = 0.50 m. (c) What is the maximum drawer extension before tipping?

Coordinate system: x = 0 at the back of the cabinet base; x = 0.50 m at the front face. Cabinet CG is at x = 0.25 m (geometric center). Drawer pulled 0.35 m out from the front face: drawer CG is at x = 0.50 + 0.35 = 0.85 m from the back.

(a) Combined CG:

x_cg = (m_cab·x_cab + m_drawer·x_drawer) / (m_cab + m_drawer)

x_cg = (18 × 0.25 + 6.0 × 0.85) / (18 + 6.0)

x_cg = (4.50 + 5.10) / 24.0 = 9.60 / 24.0 = 0.40 m from back

(b) Tipping check: The base extends from x = 0 to x = 0.50 m. The combined CG at x = 0.40 m lies within the base. The cabinet does not tip with this drawer extension.

(c) Maximum extension before tipping: Tipping begins when x_cg = 0.50 m (CG over front edge).

(18 × 0.25 + 6.0 × x_d) / 24.0 = 0.50

4.50 + 6.0 x_d = 12.0

x_d = 7.50 / 6.0 = 1.25 m from back of cabinet

Maximum extension from front face = 1.25 − 0.50 = 0.75 m

This is why filing cabinet instructions warn against opening multiple fully loaded drawers simultaneously — two drawers each pulled halfway can shift the CG outside the base just as easily as one drawer pulled all the way.

Problem 6 — Human Forearm as a Lever

A person holds a 5.0 kg textbook in their hand with the forearm horizontal. The bicep muscle attaches 4.0 cm from the elbow joint; the hand (with book) is 36 cm from the elbow. The forearm itself has mass 1.8 kg with its center of gravity 16 cm from the elbow. (a) Find the force exerted by the bicep muscle. (b) Find the reaction force at the elbow joint.

System: Forearm + hand + book in static equilibrium. Pivot at elbow joint.

Forces:

Bicep force F_B upward at d_B = 0.040 m from elbow
Forearm weight W_fa = (1.8)(9.8) = 17.64 N down at d_fa = 0.16 m from elbow
Book weight W_book = (5.0)(9.8) = 49.0 N down at d_hand = 0.36 m from elbow
Joint reaction force F_J (upward or downward?) at elbow, moment arm = 0

(a) Στ = 0 about elbow joint (F_J has zero moment arm — eliminated):

F_B(0.040) − W_fa(0.16) − W_book(0.36) = 0

0.040 F_B = 17.64(0.16) + 49.0(0.36) = 2.822 + 17.64 = 20.46

F_B = 511.5 N ≈ 512 N

For comparison, the 5 kg book weighs only 49 N — the bicep must exert about 10.5× the book's weight.

(b) ΣF_y = 0:

F_B + F_J − W_fa − W_book = 0

F_J = W_fa + W_book − F_B = 17.64 + 49.0 − 511.5 = −444.9 N

The negative sign means the joint force acts downward on the forearm — the humerus is pulling the forearm down (compressing the elbow joint) while the bicep pulls it up. The joint must withstand a compressive force of about 445 N just to hold a 5 kg book.

This is why elbow and shoulder injuries are so common in athletes: the forces in the joint itself are many times larger than the forces being exerted on external objects.

Ready to Move On?

Self-Check Before M5L5

M5L5 (Elasticity) builds directly on static equilibrium: when real materials deform under applied forces, the equilibrium equations still apply — but now you also relate force to deformation through material properties. Make sure you can do all of these before moving on:

State both equilibrium conditions and identify a situation that satisfies one but not the other.
A uniform 6.0 m beam (30 kg) is hinged at the left end and supported by a vertical rope at the 4.0 m mark. A 50 kg sign hangs from the right end. Find the rope tension without using the hinge force at all.
Explain in one sentence why placing the pivot at a hinge is almost always the best first move in a beam problem.
A box is 0.60 m wide and has its CG at 0.25 m from its left base edge. Is it stable, at the tipping point, or already tipped? What if the CG were at 0.35 m?
Why does the bicep muscle need to exert forces much larger than the weight it supports? What does this trade-off gain for the body?

In M5L5, you will extend equilibrium into the regime where materials stretch, compress, and shear under load. The same force analysis from this lesson provides the internal forces that produce the deformations you will quantify with Young's modulus, the shear modulus, and the bulk modulus.