Friction is lazy: it supplies just enough force to stop slipping, up to a maximum of \(\mu R\). Decide whether it has reached that maximum, and the problem becomes ordinary resolving.
A rough contact gives a normal reaction \(R\) and a friction force \(F\), with \(F \le \mu R\). While the particle stays at rest, \(F\) is only as large as needed. In limiting equilibrium ("about to slip") \(F = \mu R\), acting against the direction the particle is about to move.
Fig. 1: a particle at rest on a rough plane inclined at 30°. The normal reaction R is perpendicular to the plane; friction F acts up the plane, against the direction the particle would slide
Section 01
Two parts of one contact force.
When two rough surfaces touch, the contact force has a component perpendicular to the surface (the normal reaction \(R\)) and a component along it (friction \(F\)). Friction opposes the motion, or the motion that would happen without it.
Maximum friction is \(\mu R\), where \(\mu\) is the coefficient of friction (a number, no units, usually between 0 and 1).
At rest: \(F \le \mu R\). Friction takes whatever value keeps the particle in equilibrium, if it can.
Limiting equilibrium ("about to slip", "on the point of moving"): \(F = \mu R\) exactly.
Total contact force has magnitude \(\sqrt{F^2 + R^2}\).
Smooth means \(\mu = 0\): no friction at all.
Worked example Does it move?
A box of mass 8 kg rests on rough horizontal ground, \(\mu = 0.4\). A horizontal force of 20 N is applied. Does the box move? Find the friction force.
Vertically: \(R = 8 \times 10 = 80\) N.
Maximum friction \(= \mu R = 0.4 \times 80 = 32\) N.
The pull, 20 N, is less than 32 N, so friction can balance it and the box stays at rest.
Horizontally: \(F = 20\) N.
Answer: The box does not move; the friction force is 20 N (not 32 N).
Examiner trap
Never write \(F = \mu R\) automatically. It is only true when the particle is moving or about to slip. A box at rest with a small push has friction equal to the push.
Section 02
Limiting equilibrium on the level.
In limiting equilibrium you can replace \(F\) by \(\mu R\). Watch the normal reaction: a force pulling upwards at an angle reduces \(R\), and a force pushing downwards increases it.
Worked example Pulled at an angle
A block of mass 5 kg rests on rough horizontal ground. A force of 30 N at \(25^\circ\) above the horizontal acts on it, and the block is about to slip. Find \(\mu\) and the magnitude of the total contact force.
Vertically: \(R + 30\sin 25^\circ = 50\), so \(R = 50 - 12.68 = 37.32\) N.
Contact force \(= \sqrt{27.19^2 + 37.32^2} = 46.2\) N.
Answer: \(\mu = 0.729\) (3 s.f.); contact force 46.2 N.
Your turn
A block of mass 6 kg on rough horizontal ground is pushed by a force of 40 N acting at \(30^\circ\) BELOW the horizontal. The block is about to move. Find \(\mu\).
Using \(R = mg\) when the applied force has a vertical component. For the 5 kg block this gives \(\mu = 27.19 \div 50 = 0.544\), a very common wrong answer.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
When is the friction force equal to \(\mu R\)?
Quick check+10 XP
A box of mass 10 kg rests on rough horizontal ground, \(\mu = 0.3\). A horizontal push of 20 N does not move it. What is the friction force?
Quick check+10 XP
A particle on a rough plane inclined at \(30^\circ\) is about to slip, with no other forces acting. Find \(\mu\).
Quick check+10 XP
The normal reaction on a particle is 12 N and the friction force is 5 N. What is the magnitude of the total contact force?
Section 03
Rough slopes.
On a slope, resolve parallel and perpendicular to the plane. Decide which way the particle would move without friction (or is about to move), and point friction the other way.
Worked example Will it slide?
A particle of mass 3 kg is placed on a rough plane inclined at \(20^\circ\), \(\mu = 0.5\). Show that it stays at rest and find the friction force.
Perpendicular: \(R = 30\cos 20^\circ = 28.19\) N, so \(\mu R = 14.10\) N.
Down the plane, the weight component is \(30\sin 20^\circ = 10.26\) N.
\(10.26 < 14.10\), so friction can hold the particle: it stays at rest.
Parallel: \(F = 10.26\) N up the plane.
Answer: Friction is 10.3 N up the plane, less than the maximum 14.1 N.
Worked example Range of a holding force
A particle of mass 4 kg is on a rough plane inclined at \(35^\circ\), \(\mu = 0.25\). A force \(P\) N parallel to the plane, acting up the plane, keeps it in equilibrium. Find the least and greatest values of \(P\).
\(R = 40\cos 35^\circ = 32.77\) N, so \(\mu R = 8.192\) N; weight component down the plane \(= 40\sin 35^\circ = 22.94\) N.
Least \(P\): the particle is about to slip DOWN, so friction acts up: \(P + 8.192 = 22.94\), \(P = 14.75\).
Greatest \(P\): the particle is about to slip UP, so friction acts down: \(P = 22.94 + 8.192 = 31.13\).
Answer: \(14.8 \le P \le 31.1\)
Worked example A horizontal force on a rough slope
A particle of mass 2 kg is on a rough plane inclined at \(40^\circ\), \(\mu = 0.6\). A horizontal force of \(P\) N pushes it towards the plane, and it is about to slide up the plane. Find \(P\).
A particle rests on a rough plane inclined at angle \(\alpha\), with no other forces acting, and is about to slip. Show that \(\mu = \tan\alpha\), and find \(\mu\) when \(\alpha = 25^\circ\).
\(R = mg\cos\alpha\) and \(F = mg\sin\alpha\). Limiting: \(mg\sin\alpha = \mu mg\cos\alpha\), so \(\mu = \tan\alpha\). For \(25^\circ\), \(\mu = 0.466\).
Examiner trap
Friction changes direction between "about to slip down" and "about to slip up". Using the same sign for both gives one answer twice instead of a range.
Stage 2 · Exam
Exam-style questions
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Exam style+20 XP
A block of mass 5 kg on rough horizontal ground is pulled by a force of 30 N at \(25^\circ\) above the horizontal and is about to slip. Find \(\mu\).
Exam style+20 XP
A particle of mass 4 kg is on a rough plane inclined at \(35^\circ\), \(\mu = 0.25\). Find the least force parallel to the plane that keeps it at rest.
Exam style+20 XP
A particle of mass 2 kg is on a rough plane inclined at \(40^\circ\), \(\mu = 0.6\). A horizontal force \(P\) N pushes it towards the plane and it is about to slide UP the plane. Find \(P\).
Exam style+20 XP
A particle of mass 3 kg rests on a rough plane inclined at \(20^\circ\), \(\mu = 0.5\). Find the frictional force on it.
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Friction under control.
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