Static equilibrium and free body diagrams

The idea

A body is in equilibrium when it isn’t accelerating, and in static equilibrium when it stays at rest. For a rigid body that needs two things at once:

• No net force, so it doesn’t start translating
• No net moment, so it doesn’t start rotating

In 2D that gives three equations:

$\boxed{\sum F_x=0\sum F_y=0\sum M_O=0}$

Three equations means at most three unknowns per body. Those unknowns are usually support reactions or the force in a link, which is exactly what you need to size a pin, a bearing or a motor.

Newton’s three laws (the foundation)

1. A body at rest, or moving at constant velocity in a straight line, stays that way unless a net force acts on it.
2. $\sum \mathbf{F}=m\mathbf{a}$. Equilibrium is the special case $\mathbf{a}=0$.
3. Action and reaction are equal, opposite and collinear. A pin pushes on a beam exactly as hard as the beam pushes on the pin.

A car at constant speed is in equilibrium: the traction force exactly equals the drag. While it’s speeding up, it isn’t, and the difference is $ma$ (see Friction and Newton’s second law).

Supports: what each one can push back with

A support resists the motions it prevents. Replace each support with the reactions it can apply:

Support	Stops	Reactions	Unknowns
Cable or rope	Moving away along the cable	Tension, pulling along the cable	1
Roller or smooth surface	Moving into the surface	A normal force, perpendicular to the surface	1
Smooth pin (revolute joint)	Translation in any direction	$A_x$ and $A_y$; no moment, since it can still rotate	2
Fixed support (welded or clamped)	Translation and rotation	$A_x$, $A_y$ and a moment $M_A$	3

If you guess a reaction’s direction wrong, nothing breaks: the answer just comes out negative, meaning it acts the other way.

Two-force members: a link pinned at both ends with no load in between can only carry force along its length, in tension or compression. That turns two unknowns ($A_x$, $A_y$) into one (the force along the link). Connecting rods and hydraulic cylinders are usually two-force members.

The free body diagram (FBD)

You can’t write the equilibrium equations reliably without one. An FBD shows the body on its own with every external force on it:

1. Isolate the body: draw it with nothing attached.
2. Draw axes.
3. Add the applied loads, including the weight $mg$ at the centre of mass.
4. Replace each support with its reactions from the table above.
5. Add dimensions and angles.

Worked example: a beam on a pin and a roller

A 2.4 m beam is pinned at A and rests on a roller at B. It carries 500 N at 0.8 m from A and 300 N at 1.8 m from A. Ignore the beam’s weight. Find the reactions.

Answer

Unknowns: $A_x$, $A_y$ (pin) and $B_y$ (roller): three, so it’s solvable. Moments about A (this kills $A_x$ and $A_y$, so only one unknown is left): $\sum M_A=0$: $B_y(2.4)-500(0.8)-300(1.8)=0$ → $B_y=940/2.4=$ 391.7 N ↑ $\sum F_y=0$: $A_y+391.7-500-300=0$ → $A_y=408.3$ N ↑ $\sum F_x=0$: no horizontal loads, so $A_x=0$ Check: moments about B: $-A_y(2.4)+500(1.6)+300(0.6)=-980+800+180=0$ ✓

The trick that makes these quick: take moments about the point with the most unknowns (usually a pin). They drop out, and you solve for one unknown straight away.

Worked example: a hanging load (concurrent forces)

A 25 kg sign hangs from two cables meeting at a ring. The left cable is at 30° above the horizontal and the right one at 60°. Find both tensions.

Answer

The forces all meet at the ring, so moments give nothing; use the two force equations. $W=25\times 9.81=245.3$ N. $\sum F_x=0$: $T_2\cos 60°=T_1\cos 30°$ → $T_1=0.577T_2$ $\sum F_y=0$: $T_1\sin 30°+T_2\sin 60°=245.3$ → $0.289T_2+0.866T_2=245.3$ $T_2=212$ N, $T_1=123$ N. The steeper cable carries more of the load.

Common mistakes

• Leaving a reaction off the FBD, usually the horizontal reaction at a pin.
• Taking moments about a point where nothing cancels, which leaves you with simultaneous equations you didn’t need.
• Putting the weight in the wrong place. It acts at the centre of mass, which for a uniform beam is the middle.
• Mixing up which forces belong to which body when a machine has several parts. Draw a separate FBD for each part, with equal and opposite forces at the shared pins.

Check yourself

1. A 2 m cantilever is fixed into a wall at A and carries 150 N at its free end. Find the reactions at A.

$A_x=0$, $A_y=150$ N ↑, and a moment $M_A=150\times 2=300$ N·m anticlockwise (it resists the load trying to turn the beam clockwise).

2. A uniform 40 kg plank, 3 m long, rests on supports at its two ends. A 70 kg person stands 1 m from the left end. Find both support forces.

Weights: 392 N at 1.5 m and 687 N at 1 m. Moments about the left end: $R_R(3)=392(1.5)+687(1)=1275$ → $R_R=425$ N. Then $R_L=392+687-425=$ 654 N.

3. How many unknowns does a beam with a fixed support at one end and a roller at the other have, and why is that a problem?

3 + 1 = 4 unknowns, but only 3 equations. It’s statically indeterminate: you’d need the beam’s stiffness to solve it, which is beyond this module.

Next: what happens when the forces don’t balance, in Friction and Newton’s second law.