Two-dimensional equilibrium problems

A ladder leaning against a wall, a beam bolted to a bracket, a signboard hanging from a cable — each is a rigid body sitting perfectly still under a tangle of pushes, pulls, and twists. Two-dimensional equilibrium problems ask you to find the unknown forces, usually support reactions, that keep such a body from moving or rotating. The tool is always the same three scalar equations applied to a correctly drawn free-body diagram. Master this and you have the entry ticket to every later structures topic — trusses, frames and machines, and friction problems such as ladders and wedges all reduce, eventually, to exactly this calculation. This article builds the method, works a full beam-and-cable problem from scratch, then extends a ladder problem into a friction check.

The core insight

A rigid body in a plane is in equilibrium only if the sum of forces in each of two directions and the sum of moments about any point are all simultaneously zero, so a correctly drawn free-body diagram plus $\sum F_x=0,\ \sum F_y=0,\ \sum M_O=0$ solves for exactly three independent unknowns.

The core idea

A rigid body at rest in a plane obeys three independent equilibrium equations:

\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum M_O = 0

where: $F_x$ and $F_y$ are the rectangular components of every external force acting on the body, and $M_O$ is the moment of every force about a chosen point $O$ in the plane, computed as $M_O = Fd$ with $d$ the perpendicular distance from $O$ to the force's line of action. $O$ can be any point — it need not lie on the body — and it is smart to choose it on the line of action of an unknown force, since that force then contributes zero to the moment equation and drops out. You may trade one or both force equations for extra moment equations, provided the final three stay independent: $\sum M_A=0,\ \sum M_B=0,\ \sum F_x=0$ works as long as line $AB$ is not perpendicular to the $x$ -axis. Two shortcuts save algebra. A two-force member carries load at only two points and no couple, so those two forces must be equal, opposite, and collinear along the line joining the points — the member is either in pure tension or pure compression. A three-force member carries exactly three forces and no couple; equilibrium then requires the three forces to be either concurrent (all lines of action meet at one point) or all mutually parallel.

Procedure. Isolate the body with a free-body diagram (FBD), replacing every support by the reaction(s) it can supply; choose convenient axes; write $\sum F_x=0$ , $\sum F_y=0$ , $\sum M_O=0$ ; solve the three simultaneous equations. A single rigid body in 2-D is statically determinate only when the number of independent unknown reaction components equals exactly three — fewer and the body is not fully constrained, more and the extra reactions cannot be found from equilibrium alone.

Figure

Free body diagram of a beam pinned at A with a cable at B and a downward load, showing the three equilibrium equations sum Fx sum Fy and sum moments about A

Derivation ASolving a beam pinned at one end and cable-supported at the other

The goal of this derivation is to turn a free-body diagram of a beam that is pinned at one end and held by an inclined cable at the other, carrying its own weight plus a point load, into closed-form expressions for the cable tension and the two pin-reaction components, using nothing beyond the three scalar equilibrium equations and basic trigonometry, and then to check the result numerically against a beam of length $6\ \text{m}$ , combined vertical load $1300\ \text{N}$ at $3\ \text{m}$ from the pin, and a cable at $40°$ .

Step 1. Draw the FBD and name the unknowns. Beam $AB$ of length $L$ is pinned at $A$ , giving two unknown reaction components $A_x$ (horizontal) and $A_y$ (vertical). At $B$ , a distance $L$ from $A$ , a cable pulls at angle $\theta$ to the beam, with tension $T$ unknown; its components are $T\cos\theta$ (horizontal) and $T\sin\theta$ (vertical). The beam's own weight and any point load combine, for a beam loaded only vertically, into a single downward resultant $W$ acting at distance $a$ from $A$ . There are three unknowns — $T$ , $A_x$ , $A_y$ — matching the three available equations.

Step 2. Take moments about $A$ first, because both $A_x$ and $A_y$ pass through $A$ and drop out, leaving one equation in the one remaining unknown $T$ :

\sum M_A = 0: \quad T\sin\theta\,(L) - W\,a = 0 \ \Rightarrow\ T = \frac{Wa}{L\sin\theta}

Step 3. Substitute the numbers of the reference beam: $W = 1300\ \text{N}$ , $a = 3\ \text{m}$ , $L = 6\ \text{m}$ , $\theta = 40°$ , $\sin40° = 0.643$ .

T = \frac{1300(3)}{6(0.643)} = \frac{3900}{3.858} = 1011\ \text{N}

Step 4. With $T$ known, the two force equations give the pin reaction directly:

\sum F_x = 0: \quad A_x = T\cos\theta = 1011(0.766) = 774\ \text{N}

\sum F_y = 0: \quad A_y = W - T\sin\theta = 1300 - 1011(0.643) = 650\ \text{N}

Step 5. Combine $A_x$ and $A_y$ into a single resultant pin reaction and check the direction: $A = \sqrt{774^2+650^2} = 1011\ \text{N}$ at $\theta_A = \tan^{-1}(650/774) = 40.0°$ above horizontal — the pin reaction turns out equal in magnitude to $T$ and parallel to the cable, a coincidence of this particular geometry that is worth verifying rather than assuming in general. Result: $T \approx 1011\ \text{N}$ , pin reaction $\approx 1011\ \text{N}$ at $40°$ above horizontal — matching the source worked example exactly.

Worked example

A ladder $AB$ of length $5\ \text{m}$ and weight $250\ \text{N}$ rests with end $A$ on rough ground and end $B$ against a smooth vertical wall, inclined at $60°$ to the horizontal. A person weighing $700\ \text{N}$ stands $2\ \text{m}$ up the ladder, measured along the ladder from $A$ . The wall is smooth, so it can only push horizontally; the ground alone provides friction. Find the normal reaction at the ground, the friction force at the ground, and the normal reaction at the wall required for equilibrium.

Step 1. Draw the FBD. At $A$ : normal force $N_A$ (up) and friction $f$ (horizontal, resisting sliding). At $B$ : normal force $N_B$ (horizontal, since the wall is smooth). Convert the geometry to horizontal distances from $A$ : the ladder's weight acts at its midpoint, $2.5\cos60° = 1.25\ \text{m}$ ; the person acts at $2\cos60° = 1.0\ \text{m}$ ; point $B$ sits $5\cos60° = 2.5\ \text{m}$ horizontal and $5\sin60° = 4.33\ \text{m}$ vertical from $A$ .

Step 2. Apply $\sum F_y = 0$ , which involves only vertical forces: $N_A - 250 - 700 = 0 \Rightarrow N_A = 950\ \text{N}$ . Apply $\sum F_x = 0$ : $f - N_B = 0 \Rightarrow f = N_B$ , so the two horizontal unknowns are tied together and a moment equation is needed to pin down their value.

Step 3. Take moments about $A$ , which eliminates both $N_A$ and $f$ since they pass through $A$ :

\sum M_A = 0: \quad N_B(4.33) - 250(1.25) - 700(1.0) = 0

N_B = \frac{312.5+700}{4.33} = \frac{1012.5}{4.33} = 233.8\ \text{N}

So $f = 233.8\ \text{N}$ , $N_A = 950\ \text{N}$ , and $N_B = 233.8\ \text{N}$ . The ground pushes up with $950\ \text{N}$ , friction supplies $233.8\ \text{N}$ horizontally, and the wall pushes back with $233.8\ \text{N}$ .

Competitive stretch

Whether the ladder actually stays put depends on whether the ground can supply that much friction. If the coefficient of static friction at the ground is $\mu_s = 0.3$ , the maximum friction available is $\mu_s N_A = 0.3(950) = 285\ \text{N}$ . Since the required friction $f = 233.8\ \text{N}$ is less than $285\ \text{N}$ , the ladder is safely in equilibrium — but if $\mu_s$ were as low as $0.24$ , the maximum available friction would drop to $228\ \text{N}$ , just short of the $233.8\ \text{N}$ needed, and the ladder would begin to slip. This is exactly the check the source article flags as the natural next step into Coulomb friction.

Common traps

Forgetting a support reaction component. A pin has two unknowns; a roller has one; a fixed support has three (two forces plus a moment). Miscounting makes the equilibrium equations over- or under-determined.
Choosing a moment center carelessly. Placing it on an unknown force's line of action is a legitimate way to eliminate that unknown, but the resulting equation is still only one of three independent ones — it does not give you a free fourth equation.
Sign errors in moment arms. Always measure the perpendicular distance from the moment center to the force's line of action, never distance measured along the body itself.
Assuming a three-force member is automatically concurrent. The shortcut only applies when there are exactly three forces and no applied couple; check both conditions before using it.

Ask the tutor: If you want to practice choosing the best moment center for a tricky FBD, or work through another beam, ladder, or bracket problem step by step, ask the AI tutor.

How do I know which point to choose for the moment equation?

Pick a point that lies on the line of action of an unknown force you want to eliminate — that force then has zero perpendicular distance to the point, so it drops out of the moment equation entirely. In the beam example, taking moments about the pin $A$ removes both unknown pin components $A_x$ and $A_y$ in one step, leaving a single equation in the single unknown $T$ . There is no requirement that the point lie on the body itself; any point in the plane is valid, so choose whichever one kills the most unknowns.

Can I use two moment equations instead of one force equation and one moment equation?

Yes, as long as the three equations you end up with are independent. A common valid combination is $\sum M_A = 0$ , $\sum M_B = 0$ , $\sum F_x = 0$ , where $A$ and $B$ are two different points on the body. The one condition to check is that the line $AB$ is not perpendicular to the $x$ -axis used in the force equation — if it were, the two moment equations would end up encoding the same information as each other plus the force equation, and you would not actually have three independent constraints.

Why does the problem have to be statically determinate before I even start solving?

Because the three equilibrium equations for a single rigid body in a plane can only ever solve for three independent unknowns. If the supports supply more than three independent reaction components, the extra ones cannot be found from equilibrium alone — the structure is statically indeterminate and needs additional methods involving material deformation, which is beyond this topic. If the supports supply fewer than three, the body is not fully constrained and can move, so there is no equilibrium to solve for. Counting reaction unknowns (pin = 2, roller = 1, fixed support = 3) before writing any equation saves you from setting up an unsolvable system.

Two-dimensional equilibrium problems

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads