Statics studies the conditions for equilibrium (rest) of bodies. Designing bridges and buildings, a crane lifting a load without toppling, levers and scissors, the strings of a hanging lamp: all are statics problems. In each one you must answer two questions: will the body slide, and will it turn? Each question has its own condition of equilibrium. The topic is the special case of Newton’s laws when the acceleration is zero.
§1. Basics of statics
For a point particle there is just one condition of equilibrium: the forces add up to zero. Real bodies can also rotate: a door opens easily when you push the handle and hardly at all when you push near the hinges. So the effect of a force depends on where it is applied. Statics usually uses the model of a rigid body, one whose shape does not change under forces. A force on a rigid body can be slid along its line of action without changing the result. For a body to be in equilibrium, two conditions must hold together:
the force condition, so that there is no translation;
the moment condition, so that there is no rotation.
§2. The first condition of equilibrium
ΣF⃗ = 0: the vector sum of all forces is zero
In practice, the components along each axis add up to zero: ΣFₓ = 0, ΣF_y = 0. Choosing an axis along one of the unknown forces simplifies the work.
Worked example
A 2 kg lamp hangs from two strings, each at 60° to the vertical. Find the tension in each string.
The horizontal components are equal and opposite. Vertically: 2T·cos 60° = mg.
T = mg/(2·cos 60°) = 20/(2·0.5) = 20 N.
The tension grows sharply with the angle: at 85° it would be 5.7 times mg, which is why a rope pulled taut snaps easily.
§3. Moment of a force and the second condition
The lever armd is the perpendicular distance from the axis to the line of action of the force.
Moment (torque): M = F·d, unit N·m
The moment measures the turning effect of a force. If a force passes through the axis, its arm and moment are zero. Choose a sign convention: clockwise moments one sign, anticlockwise the other.
Second condition: ΣM = 0, i.e. ΣM_clockwise = ΣM_anticlockwise
If a body is in equilibrium, the moments add to zero about any axis. So it is convenient to take the axis at the point where an unknown force acts; that force then drops out of the equation.
Worked example
A uniform plank 2 m long with mass 10 kg rests on supports at its ends; a 40 kg load is placed 0.5 m from the left end. Find the support forces.
Forces: the plank’s weight 100 N (at the middle, 1 m from the left), the load 400 N (0.5 m from the left), reactions N₁ (left) and N₂ (right).
Moments about the left support: N₂·2 = 100·1 + 400·0.5 = 300 → N₂ = 150 N.
First condition: N₁ = 500 − 150 = 350 N.
§4. Levers and simple machines
A lever is a rigid body that can turn about a pivot. Its condition of balance:
F₁d₁ = F₂d₂
Whatever factor we gain in force, we lose in distance. The golden rule of mechanics: no simple machine gains work (with friction it loses some).
Fixed pulley: an equal-arm lever; it changes the direction of the force without gaining force.
Movable pulley: halves the force, but twice as much rope must be pulled.
Block and tackle: a system of pulleys; the force is reduced by the number of rope sections supporting the load.
Inclined plane: without friction F = mg·h/l.
Winches, screws, wedges, scissors, pliers are forms of the lever or the inclined plane.
Efficiency of a real machine: η = A_useful/A_input = mgh/(Fl).
Worked example
A 100 kg load is pulled steadily up a slope 5 m long and 1 m high with a force of 250 N. What is the efficiency?
Useful work: A_u = mgh = 1000·1 = 1000 J.
Input work: A_in = F·l = 250·5 = 1250 J.
η = 1000/1250 = 0.8 = 80 % (without friction 200 N would be enough).
§5. Centre of gravity and types of equilibrium
The centre of gravity is the point where the resultant of the weights of all parts of a body acts. For uniform symmetric bodies it is at the centre of symmetry: the middle of a rod, the centre of a disc. For point masses:
x_c = (m₁x₁ + m₂x₂ + ...)/(m₁ + m₂ + ...)
A body hung from its centre of gravity balances in any position. Types of equilibrium:
Stable: displaced slightly, the body returns (its centre of gravity rises): a hanging pendulum, a ball in a hollow.
Unstable: displaced slightly, the body falls (its centre of gravity drops): a pencil on its point, a ball on top of a hump.
Neutral: equilibrium in any position (the height of the centre of gravity does not change): a ball on a level table.
A body standing on a base does not topple as long as the vertical through its centre of gravity stays within the base. So for stability the centre of gravity should be low and the base wide (racing cars, tower foundations).
Worked example
A 3 kg mass is fixed to one end of a uniform rod 1 m long with mass 2 kg. How far from that end is the centre of gravity of the system?
The rod’s mass acts at its middle (x = 0.5 m), the load at the end (x = 0).
x_c = (2·0.5 + 3·0)/(2 + 3) = 1/5 = 0.2 m.
§6. Method for statics problems
Isolate the body in equilibrium and draw all the forces on it: weight (at the centre of gravity), support reactions, tensions, friction.
Write the first condition in components: ΣFₓ = 0, ΣF_y = 0.
Choose an axis, preferably where several unknown forces act.
Find each force’s lever arm (the perpendicular from the axis to the line of action) and write ΣM = 0.
Solve the equations; if a reaction comes out negative, it acts opposite to the direction you assumed.
If there are more unknowns than equations (a four-legged table, for example), statics alone cannot solve the problem; the deformation of the body must be taken into account.
Check yourself
A lever has a short arm of 20 cm and a long arm of 1 m. What force lifts a 500 N load?
A 40 N force acts at right angles on a 25 cm spanner. Moment of the force?
What force lifts a 60 kg load with a movable pulley (light pulley, no friction)?
Balls of 1 kg and 3 kg are at the ends of a light rod 40 cm long. How far from the 1 kg ball is the centre of gravity?
Force needed to pull a 100 kg load up a frictionless slope 5 m long and 1 m high?
What is the moment of a force whose line of action passes through the axis?
What kind of equilibrium is a pencil balanced on its point?
A uniform 4 kg rod 1 m long is pivoted at its middle; a 2 kg mass hangs 0.1 m from one end. What mass must hang from the other end for balance?
Common mistakes
Taking the lever arm as the distance from the axis to the point of application: the perpendicular distance is needed.
Writing only one condition of equilibrium: the body must neither slide nor turn.
Placing the weight of a uniform rod at its end instead of its middle.
Thinking a simple machine gains work as well as force.
Ignoring the signs of moments (the sense of rotation).