Equilibrium - statics questions

Conditions for the Equilibrium of Three Non-Parallel Forces

If we say that an object is under the influence of forces which are in equilibrium, we mean that the object is not accelerating - there is no net force acting.The object may still be travelling - but at a constant velocity - but in most questions the object will be stationary.

The following pointers will help you to solve problems that involve a body acted on by three co-planar forces.

The lines of action of the three forces must all pass through the same point.

ii)

The principle of moments: the sum of all the clock-wise moments about any point must have the same magnitude as the sum of all the anti-clockwise moments about the same point.

iii)

a) The sum of all the forces acting vertically upwards must have the same magnitude as the sum of all the forces acting vertically downwards
b) The sum of all the forces acting horizontally to the right must have the same magnitude as the sum of all the forces acting horizontally to the left.

So - if you resolve all of the forces the 'ups' will equal the 'downs' and the 'rights' will equal the 'lefts'.

Example questions

Q1. The diagram on the right represents a uniform ladder leaning against a smooth vertical wall. Draw in the force arrow that is acting from the ground to keep the ladder in equilibrium.

The weight arrow is vertically downwards from the centre of the ladder.

The wall (being smooth) exerts a force normally outwards (at 90^o) on the ladder.

When you extend these two, you find that they cross. As the ladder is in equilibrium (stable) the force from ground on the base of the ladder must be directed upwards towards the crossing point.

The force from the ground is the result of two forces - the normal force from the ground and friction acting along the ground.

The three forces will form a right angled triangle. Therefore you can use Pythagoras and trig. to work out the value of the force from the ground if you are given the angle the ladder is positioned at.

Q2. The diagram below represents a uniform beam which is in equilibrium. The block resting on the beam has a mass of 2.00 kg. A rope is holding the system in equilibrium. Calculate the magnitude of the tension T in the rope.

W = 2.00 × 9·81 = 19.6 N

Resolving the tension to give the verical force we get T cos 40^o

That is then the force making the clockwise moment.

So by the principle of moments we get:

clockwise moments = anticlockwise moments

(T cos 40^o) × 25 = 19.6 × 50

T = 51·2 N

Q3. The object in the diagram below has a mass of 40kg. Calculate the magnitude of the tensions (T₁ and T₂) in the supporting wires.

The forces acting on the mass are in equilibrium horizontally.

Therefore, considering magnitudes only, we can write

T₁cos 60^o = T₂cos 10^o

The forces acting on the mass are also in equilibrium vertically.

Therefore, considering magnitudes only, we can write

T₁cos 30^o + T₂cos 80^o = mg

These two equations can be solved simultaneously to find T₁ and T₂.

See free body diagrams for how to simplify questions you have to tackle.