Why does = at the point of slipping?

Resolving gives mg = mg, and dividing gives = — independent of mass.

3.03MechanicsStretch

Equilibrium

A picture hanging from two strings, a ladder leaning against a wall, a plank resting on two supports — none of them is moving, and that fact alone is enough to calculate the forces on them. An object is in $\textbf{equilibrium}$ when the resultant force is zero, and for an object with size, when the forces also have no overall turning effect.

50 min Forces and Newton’s Laws

The big picture

Equilibrium problems are dynamics problems with $a = 0$ , so every tool from Newton’s laws still applies: force diagrams, resolving, friction. For a particle, zero resultant force means the components in any two perpendicular directions each sum to zero, and three forces in equilibrium can be drawn as a closed triangle. For a rigid body, which can rotate, there is a second condition: the total $\textbf{moment}$ about any point must be zero. Taking moments about a well-chosen point — usually one where an unknown force acts — removes that force from the equation, which is the single most useful trick in statics. Limiting equilibrium, where friction is at its maximum, and tipping, where a reaction becomes zero, are the boundary cases OCR asks about most.

What you'll be able to do

Apply the conditions for equilibrium of a particle by resolving forces
Solve problems involving limiting equilibrium with friction
Calculate moments and apply the conditions for equilibrium of a rigid body
Solve problems involving beams, supports and tipping

Equilibrium of a particle

A particle is in equilibrium if the resultant force on it is zero. Resolving in two perpendicular directions, the components in each direction must sum to zero.

Three forces in equilibrium form a closed triangle when drawn tip to tail, so Lami’s theorem or the sine rule can also be used — but resolving always works.

Example 1A lamp of weight 20 N hangs from two strings making angles of

30^{\circ}

and

60^{\circ}

with the horizontal ceiling. Find the tensions.

1Horizontally:

T_{1}\cos 30^{\circ} = T_{2}\cos 60^{\circ}

2Vertically:

T_{1}\sin 30^{\circ} + T_{2}\sin 60^{\circ} = 20

3From the first,

T_{1} = \dfrac{T_{2}}{\sqrt3}

. Substituting:

\dfrac{T_{2}}{2\sqrt3} + \dfrac{\sqrt3T_{2}}{2} = 20

, so

\dfrac{2T_{2}}{\sqrt3} = 20

Answer

T_{2} = 10\sqrt3 \approx 17.3

N (the steeper string) and

T_{1} = 10

Tip — The steeper string carries more of the weight, so it has the larger tension — a quick sense check on your answer.

Limiting equilibrium

An object on the point of moving is in $\textbf{limiting equilibrium}$ : still balanced, but with friction at its maximum, $F = \mu R$ .

For a particle on a rough slope with no other forces, limiting equilibrium gives $mg\sin\alpha = \mu mg\cos\alpha$ , so $\mu = \tan\alpha$ . The slope angle at which it starts to slip measures the coefficient of friction.

When a force could move an object either up or down a slope, consider both limiting cases: friction acts down the slope when it is about to move up, and up the slope when it is about to move down.

\mu = \tan\alpha

Particle about to slip on a slope under its weight alone.

Moments

The $\textbf{moment}$ of a force about a point is the force multiplied by the perpendicular distance from the point to the line of action of the force. It is measured in newton-metres (N m) and has a sense, clockwise or anticlockwise.

A $\textbf{rigid body}$ is in equilibrium if (1) the resultant force is zero and (2) the sum of moments about $\textit{any}$ point is zero.

A $\textbf{uniform}$ rod or beam has its weight acting at its midpoint.

\text{moment} = F \times d_{\perp}

Perpendicular distance to the line of action.

Example 1A uniform plank

AB

of length 4 m and mass 20 kg rests on supports at

C

, 0.5 m from

A

, and

D

, 1 m from

B

. A person of mass 60 kg stands 2.5 m from

A

. Find the reactions at the supports.

1Weights: plank

20g

at 2 m from

A

; person

60g

at 2.5 m from

A

D

is 3 m from

A

2Moments about

C

R_{D} \times 2.5 = 20g \times 1.5 + 60g \times 2 = 150g

, so

R_{D} = 60g = 588

3Resolving vertically:

R_{C} + R_{D} = 80g

, so

R_{C} = 20g = 196

Answer

R_{C} = 196

R_{D} = 588

Why this works

Taking moments about $C$ eliminated $R_{C}$ entirely, leaving one unknown. Always take moments about the point where the most awkward unknown force acts.

Tipping

As a load moves towards the end of a beam, the reaction at the far support decreases. The beam is $\textbf{on the point of tipping}$ about the nearer support when the other reaction becomes zero.

To find the tipping position, set that reaction to zero and take moments about the pivot support.

Example 1For the plank above, how far towards

B

can the person walk before the plank tips?

1On the point of tipping about

D

R_{C} = 0

2Moments about

D

: the plank’s weight acts 1 m on the

A

side of

D

; the person stands

x

m beyond

D

20g \times 1 = 60g \times x

, so

x = \tfrac13

Answer

\tfrac13

m beyond

D

, i.e.

3\tfrac13

m from

A

Tip — A tipping question is solved by "reaction $= 0$ at the support that lifts off, then moments about the other support". Write both statements.

Think like an examiner

State the equilibrium conditions you are using: resolving, and moments about a named point.
Use perpendicular distances for moments, and state clockwise or anticlockwise.
Justify $F = \mu R$ with "limiting equilibrium".
For tipping, state that the reaction at the other support is zero.

Common misconceptions

Using the distance along a slanting rod instead of the perpendicular distance.

Moment uses the perpendicular distance to the line of action.

Taking moments about a point but still including the force acting there.

A force through the pivot point has zero moment about it.

Placing a uniform beam’s weight at one end.

For a uniform beam, weight acts at the midpoint.

Assuming friction is at its limit in any equilibrium problem.

Only in limiting equilibrium; otherwise

F < \mu R

Equilibrium

\sum F_{x} = 0,\ \sum F_{y} = 0

Particle equilibrium.

\sum M = 0

Rigid body: no resultant moment.

M = F d_{\perp}

Moment of a force.

F = \mu R

Limiting equilibrium.

Stretch yourself

A particle of mass 5 kg rests on a rough slope inclined at $30^{\circ}$ , with $\mu = 0.4$ . A horizontal force $P$ acts on it, pushing it towards the slope. Find the smallest value of $P$ that prevents the particle sliding down the slope.

Hint — On the point of sliding down, friction acts up the slope at its maximum. Resolve along and perpendicular to the slope; $P$ has components in both directions.

Reveal solution

SolutionResolve perpendicular to the slope:

R = 5g\cos 30^{\circ} + P\sin 30^{\circ} = 42.44 + 0.5P

. Along the slope (up positive), with friction

\mu R

up the slope in limiting equilibrium:

P\cos 30^{\circ} + 0.4R = 5g\sin 30^{\circ}

. So

0.8660P + 0.4(42.44 + 0.5P) = 24.5

, i.e.

0.8660P + 16.98 + 0.2P = 24.5

, giving

1.0660P = 7.52

and

P = 7.06

N. Check:

R = 45.97

, friction

= 18.39

, up-slope total

= 6.11 + 18.39 = 24.50

✓.

\blacksquare

Questions students ask

What is the difference between particle and rigid body equilibrium?

A particle can only translate, so only forces must balance. A rigid body can also rotate, so moments must balance too.

Can I take moments about any point?

Yes — for a body in equilibrium the moments balance about every point. Choose one that eliminates an unknown.

What does limiting equilibrium mean?

The object is still at rest but on the point of moving, so friction has reached its maximum value

\mu R

Why does

\mu = \tan\alpha

at the point of slipping?

Resolving gives

mg\sin\alpha = \mu mg\cos\alpha

, and dividing gives

\mu = \tan\alpha

— independent of mass.

Key takeaways

Equilibrium means zero resultant force; resolve in two directions.
Limiting equilibrium: friction $= \mu R$ ; on a slope alone, $\mu = \tan\alpha$ .
Moment $=$ force $\times$ perpendicular distance.
Rigid bodies need both forces and moments to balance.
Tipping occurs when a support reaction falls to zero.