PhysicsCore26 min read

Friction, Slopes and Connected Particles

What holds a block still, and what happens when two are tied together

This topic appears in:

A Level Mathematics (9709) · ch.4
01

Friction adjusts, up to a limit

Definition

Coefficient of friction — μ, the constant of proportionality between the maximum friction and the normal reaction: F ⩽ μR, with equality only at the point of slipping.

Friction is unlike the other forces in a mechanics question, because its size is not fixed. Push a heavy box gently and it does not move: friction has matched your push exactly. Push harder and it still does not move, because friction has increased to match again. Push hard enough and it finally slides.

So friction supplies whatever is needed to prevent motion — but only up to a maximum. That maximum is proportional to how hard the surfaces are pressed together, which is measured by the normal reaction R.

This gives the single most important consequence in the topic: F = μR is true only when the object is on the point of sliding, or already sliding. At any other time friction is less than μR, and writing F = μR for a block sitting comfortably still is simply wrong.

F ⩽ μRin generalF = μRon the point of slipping, or slidingon a horizontal surface:R = mgon a slope of angle θ:R = mg cos θthe inequality is the whole idea — equality is a special case
F
the frictional forcealways opposes the motion, or the attempted motion
μ
coefficient of frictiondepends on the two surfaces, not on mass or area
R
the normal reactionperpendicular to the surface, and not always equal to mg
02

On a slope, the weight splits in two

A slope is where most marks in this topic are won or lost, and the reason is always the same: the weight still points straight down, but the motion is along the slope. So the weight must be resolved into a component along the slope and a component perpendicular to it.

The component mg sin θ acts down the slope and is what tries to move the block. The component mg cos θ presses into the surface, so the normal reaction matches it: R = mg cos θ. Notice that R is now smaller than the weight, and shrinks further as the slope steepens — which means the available friction shrinks at exactly the moment more of it is needed.

On The forces, watch how mg splits and how R shrinks as the slope steepens. Then switch to Will it slip? and raise θ: the driving bar grows while the friction bar shrinks, and where they cross is tan⁻¹μ — an angle that does not depend on the mass at all.

Which component takes the sine?

The component along the slope is mg sin θ and the one into it is mg cos θ. If you are unsure, test with θ = 0: a flat surface should have nothing pulling along it and the full weight pressing down. Since sin 0 = 0 and cos 0 = 1, that works only with sine along and cosine into — a two-second check that settles it every time.

03

The angle of friction

Tilt a slope slowly until the block just begins to move. At that instant friction is at its maximum, so the two expressions are equal.

Setting mg sin θ = μ mg cos θ and cancelling mg from both sides leaves tan θ = μ. The mass has vanished completely — a heavy block and a light one made of the same material slip at exactly the same angle, which is a genuinely surprising result and a favourite exam question.

at the point of slipping:mg sin θ = μ mg cos θtan θ = μθ = tan⁻¹ μthe mass cancels — the angle is the same for any blockthis angle is called the angle of friction
θ
the angle of frictionthe steepest slope the block can rest on
μ
coefficient of frictionequals the tangent of that angle
04

Two particles, one string

Connected particles look harder than they are, because the string enforces a simple fact: both objects move together. They have the same speed and the same acceleration at every instant, since the string is inextensible.

The tension is the same throughout a light string passing over a smooth pulley. "Light" means the string has no mass to accelerate, and "smooth" means the pulley does not rub — remove either assumption and the tensions on the two sides would differ.

The method is then mechanical: write F = ma for each particle separately, taking the positive direction as the direction that particle actually moves. Two equations, two unknowns — the acceleration and the tension.

Worked example

Particles of mass 5 kg and 3 kg hang either side of a smooth pulley on a light inextensible string. Find the acceleration of the system and the tension in the string. Take g = 9.8 m s⁻².

  1. The 5 kg mass descends and the 3 kg mass rises, both with acceleration a.The string is inextensible, so the accelerations have the same magnitude. Deciding the direction of motion first keeps the signs consistent.
  2. For the 5 kg mass, taking down as positive: 5g − T = 5a.Weight acts down, tension up. Its weight exceeds the other so it is the one that falls.
  3. For the 3 kg mass, taking up as positive: T − 3g = 3a.For this particle the positive direction is upwards, because that is the way it moves.
  4. Add the two equations: 5g − 3g = 8a, so 2g = 8a and a = 2.45 m s⁻².Adding eliminates T immediately — the standard reason for writing the equations this way round.
  5. Substitute back: T = 3g + 3a = 29.4 + 7.35 = 36.75 N.Either equation works; using the simpler one reduces the chance of error.
  6. Check: T lies between the two weights, 29.4 N and 49 N.The tension must be less than the falling weight and more than the rising one, otherwise nothing would accelerate. A quick sanity check that catches sign errors.

a = 2.45 m s⁻², T = 36.8 N

Why the tension sits between the two weights

If T equalled the heavier weight, that particle would be in equilibrium and nothing would move. If T equalled the lighter weight, the same would be true on the other side. So the tension must lie strictly between them — and checking that is the fastest way to spot a sign error in the two equations.

The method, every time

  1. Draw both particles separately, with every force on each.
  2. Same acceleration magnitude for both, because the string does not stretch.
  3. Same tension throughout, because the string is light and the pulley smooth.
  4. Write F = ma for each, taking positive in the direction that particle moves.
  5. Add the equations to eliminate T and find a; substitute back for T.
  6. If one particle is on a table, its weight is balanced by R and does not appear in its equation of motion — only tension and friction do.

Practice questions

5 questions · 18 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 7 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
A block rests on a rough horizontal table and is not moving. Explain why it is wrong to write F = μR for the friction acting on it.
Model answer

F = μR gives the maximum friction available, which occurs only when the block is on the point of slipping or already sliding. A stationary block that is not on the point of moving experiences only as much friction as is needed to balance the other forces, so F < μR.

Examiner tip. The mark is for the word maximum, and for stating the general relation as an inequality.

SQ2[3 marks]
A block is on the point of sliding down a rough slope inclined at angle θ. Show that tan θ = μ.
Model answer

Resolving perpendicular to the slope: R = mg cos θ. On the point of sliding, friction is at its maximum and acts up the slope, so resolving along the slope gives mg sin θ = μR = μ mg cos θ. Dividing both sides by mg cos θ gives tan θ = μ.

Examiner tip. Three marks: resolving perpendicular, resolving along, and the division. The mass cancelling is the point of the result.

SQ3[2 marks]
Two blocks of the same material, one of mass 2 kg and one of mass 20 kg, rest on identical adjustable slopes. Which slips first as the slopes are slowly raised? Justify your answer.
Model answer

Neither — they slip at the same angle. At the point of slipping tan θ = μ, and the mass cancels out of that derivation entirely. Since both blocks are of the same material on the same surface, μ is identical, so the critical angle is identical.

Examiner tip. The answer is counter-intuitive and the mark is for the justification, not the assertion. Point at the cancellation of mg.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A box of mass 12 kg rests on a rough horizontal floor with μ = 0.35. A horizontal force of 50 N is applied. Determine whether the box moves, and if so find its acceleration. Take g = 9.8 m s⁻².
Full working
  1. R = mg = 12 × 9.8 = 117.6 NHorizontal surface, so the normal reaction equals the weight.[1]
  2. Maximum friction = μR = 0.35 × 117.6 = 41.16 NThis is the largest force friction can supply.[1]
  3. 50 N > 41.16 N, so the box moves.The comparison is the answer to the first part and must be stated.[1]
  4. Resultant = 50 − 41.16 = 8.84 N, so a = 8.84/12 = 0.737 m s⁻²Once sliding, friction is exactly μR, so it can be subtracted directly.[1]

It moves, with a = 0.737 m s⁻²

Exam questions

1 · 7 marks

Multi-part questions with a full mark scheme.

Q1[7 marks]
Particles A of mass 4 kg and B of mass 6 kg are connected by a light inextensible string over a smooth pulley at the edge of a table. A lies on the rough horizontal table with μ = 0.2; B hangs freely. The system is released from rest. Take g = 9.8 m s⁻².
(a) Find the acceleration of the system.
(b) Find the tension in the string.
(c) State one assumption made in modelling the string, and explain its effect.
Mark scheme
  1. (a) For A on the table: R = 4g = 39.2 N, so friction = μR = 0.2 × 39.2 = 7.84 NA slides, so friction is at its maximum and opposes the motion.[1]
  2. For A, along the table: T − 7.84 = 4aA's weight is balanced by R and does not enter the horizontal equation.[1]
  3. For B, taking down as positive: 6g − T = 6a, that is 58.8 − T = 6aB falls, so down is the positive direction for B.[1]
  4. Adding: 58.8 − 7.84 = 10aAdding eliminates T, as always with connected particles.[1]
  5. a = 50.96/10 = 5.10 m s⁻²Less than g, as it must be, since friction and A's inertia both resist.[1]
  6. (b) T = 4a + 7.84 = 20.38 + 7.84 = 28.2 NSubstituting into A's equation. T is less than B's weight of 58.8 N, as expected.[1]
  7. (c) The string is light, so its mass need not be accelerated and the tension is the same throughout; and inextensible, so both particles share one acceleration.Either assumption earns the mark provided its consequence is explained.[1]

(a) 5.10 m s⁻²; (b) 28.2 N; (c) light and inextensible — equal tension, equal acceleration