Friction and limiting equilibrium

Friction is a contact force that opposes the relative motion, or the tendency for relative motion, between two surfaces. Its magnitude is not usually known in advance.

For the standard A level model,

$$\boxed{0\leq F\leq \mu R},$$

where $F$ is the frictional force, $R$ is the normal reaction and $\mu$ is the coefficient of friction. At limiting equilibrium, the object is on the point of slipping and

$$\boxed{F=\mu R}.$$

The word limiting is essential. In ordinary equilibrium, friction takes whatever value is needed, up to the maximum $\mu R$.

Prerequisites

You should be able to:

• draw forces on an isolated body using free body diagrams;
• apply Newton’s laws, including $\sum F=ma$;
• resolve forces parallel and perpendicular to a slope;
• use $W=mg$ for weight and distinguish mass from weight;
• solve linear equations and inequalities.

Unless a question specifies another value, use $g=9.8\ \mathrm{m\,s^{-2}}$.

What friction does

Friction acts:

• parallel to the surfaces in contact;
• opposite actual sliding, if the surfaces are sliding;
• opposite the motion that would occur without friction, if the body is at rest.

The normal reaction $R$ acts perpendicular to the contact surface. It is not always equal to $mg$. An angled pull, an angled push, or other vertical forces can change $R$, which then changes the maximum possible friction $\mu R$.

Friction is not always $\mu R$

If a stationary block needs $3\ \mathrm N$ of friction to remain at rest and $\mu R=8\ \mathrm N$, then $F=3\ \mathrm N$, not $8\ \mathrm N$. Use $F=\mu R$ only when the body is on the point of slipping, or when the model for a sliding body explicitly requires it.

A useful physical picture

Imagine gradually increasing a horizontal pull $P$ on a stationary block. While the block remains at rest,

$$F=P$$

and friction grows with the pull. When $P$ reaches $\mu R$, friction cannot grow further. This is limiting equilibrium. Any larger pull produces a resultant force and the block accelerates.

A reliable solution method

1. Draw a free body diagram.
2. Decide the actual motion or the likely direction of impending motion.
3. Draw friction in the opposite direction.
4. Resolve perpendicular to the surface to find $R$.
5. Resolve parallel to the surface using $F\leq\mu R$, or $F=\mu R$ only when justified.
6. Apply equilibrium, $\sum F=0$, or dynamics, $\sum F=ma$.
7. Check that $F\leq\mu R$, $R\geq0$, and the assumed direction is consistent.

If your assumed direction is wrong, the algebra often produces a negative acceleration or an impossible coefficient. That is useful information, not a reason to discard the sign.

Horizontal surfaces

Worked example 1: friction below its limiting value

A $5\ \mathrm{kg}$ block rests on a rough horizontal floor. The coefficient of friction is $0.4$. A horizontal force of $12\ \mathrm N$ acts on the block. Determine whether it moves and find the frictional force.

There is no vertical acceleration, so

$$R=5g=49\ \mathrm N.$$

The limiting friction is

$$\mu R=0.4(49)=19.6\ \mathrm N.$$

Only $12\ \mathrm N$ is needed to balance the applied force, and $12<19.6$. Therefore the block remains at rest and

$$\boxed{F=12\ \mathrm N}.$$

Setting $F=19.6\ \mathrm N$ would falsely predict a resultant force opposite the pull. Static friction does not create motion.

Worked example 2: acceleration on a rough floor

A $10\ \mathrm{kg}$ crate is pulled horizontally by $50\ \mathrm N$. It is moving, and the model takes friction to have magnitude $\mu R$, where $\mu=0.3$. Find its acceleration.

Vertically,

$$R=10g=98\ \mathrm N.$$

Hence

$$F=\mu R=0.3(98)=29.4\ \mathrm N.$$

Taking the direction of motion as positive,

$$50-29.4=10a.$$

Therefore

$$\boxed{a=2.06\ \mathrm{m\,s^{-2}}}.$$

Self-check 1

A $4\ \mathrm{kg}$ block rests on a rough horizontal plane with $\mu=0.25$. Find the least horizontal force that will put it on the point of moving.

Answer

At limiting equilibrium,

$$P=F=\mu R=0.25(4g)=\boxed{9.8\ \mathrm N}.$$

Angled pulls and pushes

An upward component of a pull reduces $R$. A downward component of a push increases $R$. Since limiting friction is $\mu R$, the angle changes both the driving force and the resistance.

Worked example 3: pulling at an angle

A $20\ \mathrm{kg}$ box is pulled by a force of magnitude $P$ acting at $30^\circ$ above the horizontal. The coefficient of friction is $0.25$. Find $P$ when the box is on the point of moving.

The impending motion is forwards, so limiting friction acts backwards. Vertically, the box is in equilibrium:

$$R+P\sin30^\circ=20g,$$

so

$$R=196-\frac P2.$$

Horizontally,

$$P\cos30^\circ=F=0.25R.$$

Substitute for $R$:

$$\frac{\sqrt3}{2}P=\frac14\left(196-\frac P2\right).$$

Multiplying by $8$ gives

$$4\sqrt3P=392-P,$$

and therefore

$$\boxed{P=\frac{392}{1+4\sqrt3}=49.4\ \mathrm N\text{ to 3 s.f.}}$$

Check that $R>0$: here $R\approx171\ \mathrm N$, so contact is maintained.

Pulling versus pushing

For the same angle and coefficient, pulling upwards usually requires less force than pushing downwards because pulling reduces the normal reaction and therefore reduces limiting friction.

Rough inclined planes

For a plane inclined at angle $\theta$ to the horizontal, resolve weight into:

$$mg\sin\theta\quad\text{down the plane}, \qquad mg\cos\theta\quad\text{perpendicular into the plane}.$$

If no other force has a perpendicular component,

$$R=mg\cos\theta.$$

Do not decide the direction of friction merely because the plane slopes. Friction opposes the actual or impending motion. A strong force up the plane can make friction act down the plane.

Worked example 4: coefficient from limiting equilibrium

A particle rests on a rough plane inclined at $25^\circ$ to the horizontal. It is on the point of sliding down. Find $\mu$.

Because impending motion is down the plane, friction acts up the plane. Perpendicular to the plane,

$$R=mg\cos25^\circ.$$

Parallel to the plane, equilibrium gives

$$F=mg\sin25^\circ.$$

At limiting equilibrium, $F=\mu R$, so

$$mg\sin25^\circ=\mu mg\cos25^\circ.$$

Cancel $mg$:

$$\boxed{\mu=\tan25^\circ\approx0.466}.$$

This general result, $\mu=\tan\theta$, applies to a body on the point of sliding freely down a plane when weight, reaction and friction are its only forces.

Worked example 5: force needed to move up a slope

A $6\ \mathrm{kg}$ particle lies on a rough plane inclined at $20^\circ$, with $\mu=0.35$. A force $P$ acts up the plane. Find the least $P$ that will cause motion up the plane.

At the threshold of upward motion, friction acts down the plane. Since $P$ has no perpendicular component,

$$R=6g\cos20^\circ.$$

Resolving up the plane at limiting equilibrium,

$$P-6g\sin20^\circ-\mu R=0.$$

Thus

$$\begin{aligned} P &=6g\sin20^\circ+0.35(6g\cos20^\circ)\\ &=39.4\ \mathrm N\quad\text{to 3 s.f.} \end{aligned}$$

so $\boxed{P=39.4\ \mathrm N}$.

Worked example 6: acceleration down a rough slope

A $3\ \mathrm{kg}$ particle slides down a plane inclined at $35^\circ$. The coefficient of friction is $0.2$. Find its acceleration.

Friction acts up the plane. Perpendicular to the plane,

$$R=3g\cos35^\circ.$$

Taking down the plane as positive,

$$3g\sin35^\circ-0.2(3g\cos35^\circ)=3a.$$

Cancel $3$:

$$a=g(\sin35^\circ-0.2\cos35^\circ).$$

Therefore

$$\boxed{a=4.02\ \mathrm{m\,s^{-2}}\text{ to 3 s.f.}}$$

The mass cancels because both the component of weight and the modelled friction are proportional to mass.

Equilibrium over a range of forces

When a force can vary while a body remains at rest, there may be two limiting cases. At one end of the range the body is about to move down, so friction acts up. At the other it is about to move up, so friction acts down.

Worked example 7: finding the complete range

A $5\ \mathrm{kg}$ particle is held at rest on a rough plane inclined at $30^\circ$ by a force $P$ acting up the plane. The coefficient of friction is $0.2$. Find the range of values of $P$ for equilibrium.

First,

$$R=5g\cos30^\circ, \qquad \mu R=0.2(5g\cos30^\circ)=8.49\ \mathrm N.$$

Resolving up the plane, equilibrium requires

$$P+F-5g\sin30^\circ=0,$$

where signed friction satisfies $|F|\leq8.49$. Hence

$$|5g\sin30^\circ-P|\leq8.49.$$

Since $5g\sin30^\circ=24.5$,

$$|24.5-P|\leq8.49.$$

Therefore

$$\boxed{16.0\ \mathrm N\leq P\leq33.0\ \mathrm N}$$

to $3$ significant figures.

At the lower endpoint, the particle is about to slide down and friction acts up. At the upper endpoint, it is about to move up and friction acts down.

Self-check 2

A particle is on a rough plane with $\mu=0.4$. The plane is inclined at angle $\theta$, and the particle is on the point of slipping down under its own weight. Find $\theta$.

Answer

At limiting equilibrium,

$$mg\sin\theta=\mu mg\cos\theta,$$

so $\tan\theta=0.4$. Therefore

$$\boxed{\theta=\tan^{-1}(0.4)=21.8^\circ\text{ to 3 s.f.}}$$

Self-check 3

A $2\ \mathrm{kg}$ block is pushed by a $15\ \mathrm N$ force at $40^\circ$ below the horizontal across a rough horizontal floor. The modelled friction is $\mu R$, where $\mu=0.3$. Find the acceleration.

Answer

The downward component of the push increases the reaction:

$$R=2g+15\sin40^\circ=29.24\ \mathrm N.$$

Thus $F=0.3R=8.77\ \mathrm N$. Horizontally,

$$15\cos40^\circ-8.77=2a,$$

giving

$$\boxed{a=1.36\ \mathrm{m\,s^{-2}}\text{ to 3 s.f.}}$$

Common misconceptions

• Writing $F=\mu R$ immediately: first ask whether the body is limiting or sliding under that model. Otherwise use $F\leq\mu R$.
• Assuming $R=mg$: resolve perpendicular to the surface. Other forces may have perpendicular components.
• Always drawing friction down a slope: friction opposes motion or impending motion, not the slope itself.
• Using $mg\cos\theta$ down the plane: the component down a plane inclined at $\theta$ is $mg\sin\theta$.
• Treating $\mu$ as a force: $\mu$ is dimensionless. The frictional force $F$ is measured in newtons.
• Finding only one endpoint of an equilibrium range: test impending motion in both directions.

Final checklist

Before accepting an answer, ask:

1. Is friction parallel to the contact surface and in the correct direction?
2. Did I find $R$ by resolving perpendicular to the surface?
3. Is $F=\mu R$ genuinely justified?
4. Does equilibrium require $F\leq\mu R$?
5. Are force units in newtons and acceleration units in $\mathrm{m\,s^{-2}}$?

Next, combine friction with tension and common acceleration in connected particles and pulleys, then apply component equations in dynamics in a plane. For later equilibrium problems involving turning effects, continue to moments.