Equilibrium

Static and dynamic equilibrium, the condition that the net force is zero, solving equilibrium problems in one and two dimensions with components and with a closed triangle of forces, and why tension rises sharply as cables flatten.

Hang a heavy picture from one string and it holds fine. Pull the string tight and nearly horizontal across a wide frame, and the same picture can snap it. A washing line can't be pulled perfectly straight, however hard you try. Both of these follow from one condition: in equilibrium, the forces add to zero.

An object is in equilibrium when the net force on it is zero. By Newton's first law, it then has constant velocity. There are two kinds:

Static equilibrium: at rest and staying at rest (a bridge, a hanging sign, a parked car on a hill).
Dynamic equilibrium: moving at constant velocity (a car cruising on a level road, a skydiver at terminal velocity).

Either way, the condition is the same, and in two dimensions it splits into two equations:

$$\Sigma F_x = 0 \qquad \Sigma F_y = 0$$

Equilibrium along one line

With forces along a line, add them with signs and set the total to zero.

Worked example: terminal velocity

A 72 kg skydiver falls at a steady 55 m s⁻¹. What is the air resistance on the skydiver?

Constant velocity means dynamic equilibrium, so $\Sigma F = 0$.

constant velocity

$F_{air} - mg = 0$, so $F_{air} = 72 \times 9.8 = 706$ N upward, about $7.1 \times 10^2$ N.

up is positive

The speed doesn't appear in the calculation. Whatever the terminal velocity, the drag at that speed equals the weight.

note

Equilibrium in two dimensions

For forces at angles, resolve every force into components along two perpendicular axes and write one equation for each axis. Two equations let you find two unknowns, usually two tensions.

A traffic light hanging from two cables. The junction is in equilibrium, so the three forces drawn tip to tail make a closed triangle.

Worked example: a traffic light on two cables

A 12.0 kg traffic light hangs from two cables. The left cable makes 25° with the horizontal and the right cable makes 40°. Find the tension in each cable.

$mg = 12.0 \times 9.8 = 117.6$ N.

weight

$T_1\cos25° = T_2\cos40°$, so $T_1 = T_2 \dfrac{\cos40°}{\cos25°} = 0.8452\,T_2$.

horizontal

$T_1\sin25° + T_2\sin40° = 117.6$, so $0.8452\,T_2(0.4226) + 0.6428\,T_2 = 117.6$, giving $1.000\,T_2 = 117.6$.

vertical

$T_2 = 118$ N and $T_1 = 0.8452 \times 117.6 = 99.4$ N.

solve

The steeper cable carries more of the load, which makes sense: it points more nearly upward.

check

You can also solve these with a triangle of forces. Draw the weight, then the two tensions, tip to tail. Because the resultant is zero, the arrows must close into a triangle, and the sine rule gives the tensions. For a symmetric hang, with each cable at angle $\theta$ to the horizontal:

$$2T\sin\theta = mg \quad\Rightarrow\quad T = \frac{mg}{2\sin\theta}$$

As $\theta$ gets small, $\sin\theta$ gets small and $T$ grows without limit. Only the vertical components hold the load up, so a nearly flat cable needs a huge tension. That is why a washing line always sags: making it perfectly straight would need an infinite tension.

Worked example: why cables sag

A 20 kg sign hangs from the midpoint of a cable. Compare the tension when each half of the cable makes 30° with the horizontal and when it makes 5°.

$T = \dfrac{20 \times 9.8}{2\sin30°} = \dfrac{196}{1.0} = 196$ N, the same as the sign's weight.

30°

$T = \dfrac{196}{2\sin5°} = \dfrac{196}{0.1743} = 1.12 \times 10^3$ N, nearly six times the weight.

5°

A method that always works

Pick the object or junction that is in equilibrium and draw its free-body diagram.
Choose axes. Often horizontal and vertical; on a slope, along and perpendicular to it.
Resolve every force that isn't along an axis.
Write $\Sigma F_x = 0$ and $\Sigma F_y = 0$.
Solve the simultaneous equations, then check the answer is physically sensible.

InteractiveTension in a hanging sign

Try this

Start at 20 kg with both angles at 30°: each cable holds 196 N, exactly the sign's weight.
Lower both angles to 5°. The tension climbs to 1,124 N and both cables turn red, well past their 500 N rating.
Set θ₁ = 30° and θ₂ = 60°. Now T₁ = 98 N and T₂ = 170 N: the steeper cable carries more.
Watch the triangle of forces. However you set the angles, it always closes, because the net force on the junction is zero.

Equilibrium in structures

Engineers design every bridge, crane and building so that each joint is in equilibrium under its worst expected load, with a safety margin. The Sydney Harbour Bridge's arch carries the deck in compression, and its huge granite-faced pylons were built mainly for the look and to calm public nerves. The forces in the arch itself go into massive bearings at each end, each made to carry about 20,000 tonnes. Suspension and cable-stayed bridges, like the Anzac Bridge, do the opposite: their cables are in tension, and the steeper the cable, the less tension it needs for the same load.

Test yourself

Which object is NOT in equilibrium?
A 5.0 kg lamp hangs from two cables, each at 45° to the horizontal. The tension in each is
A load hangs from the middle of a rope. As the rope is pulled tighter so it sags less, the tension
A car of mass 1400 kg cruises at a constant 90 km/h on a level road. The total resistance force on it is 520 N. The driving force from the road on the tyres is

Label the forces on the junction above the traffic light.

Classify each object.

An object is in equilibrium when the on it is zero. A skydiver at terminal velocity is in equilibrium. In two dimensions, the forces drawn tip to tail form a triangle. For a symmetric hang, $T = mg/(2$$)$, so as a cable becomes flatter the tension .

5 marks

A 30.0 kg sign is held by two cables attached at the same point on the sign. Cable A makes 35° with the horizontal and cable B makes 55° with the horizontal, on the other side. Calculate the tension in each cable, and explain which cable is more likely to break if both are identical.

Your answer