What is M8.6: Friction?

Friction acts along the surface, opposing the motion or the tendency to move, and satisfies F≤ R where R is the normal reaction. Find R first by resolving perpendicular to the surface; on a slope R=mgcos, and an applied force with a vertical component changes R.

How many marks are M8.6 questions worth?

M8.6 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has M8.6 appeared in?

We have not yet indexed a verified past-paper appearance for M8.6 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M8.6

Understand and use the F ≤ μR model for friction; coefficient of friction; motion of a body on a rough surface; limiting friction and statics.

Draft — not yet indexed

Friction

Worked answers and methods for M8.6 on Edexcel A-level Maths 9MA0.

Definition

Specification M8.6: understand and use the $F\leq\mu R$ model for friction; coefficient of friction; motion of a body on a rough surface; limiting friction and statics.

Explanation

Friction acts along the surface, opposing the motion or the tendency to move, and satisfies $F\leq\mu R$ where $R$ is the normal reaction. Find $R$ first by resolving perpendicular to the surface; on a slope $R=mg\cos\alpha$ , and an applied force with a vertical component changes $R$ .
Friction equals $\mu R$ only when the body is moving or is in limiting equilibrium, on the point of moving; in ordinary equilibrium friction takes whatever value the other forces need, and the question is whether that value exceeds $\mu R$ .
To decide whether a body on a slope stays at rest, compare the component of weight down the slope, $mg\sin\alpha$ , with the maximum friction $\mu mg\cos\alpha$ and state the conclusion.
For a body in motion write $F=\mu R$ and $F=ma$ along the surface, with friction opposite to the velocity, which reverses when the body slides back down.
A range of values for a force comes from the two limiting cases.

On a rough horizontal surface, friction opposes impending or actual motion and satisfies

F\leq\mu R

Worked example

A block of mass

6

kg is moving on a rough horizontal surface under the action of a horizontal force of magnitude

20

N acting in the direction of motion. The coefficient of friction between the block and the surface is

0.25

. Find the acceleration of the block.

(4)

(Total for Question 1 is 4 marks)

Working

1.Resolve vertically: $R=6g=58.8$ N.
2.The block is moving, so friction is limiting: $F=\mu R=0.25\times58.8=14.7$ N, acting against the motion.
3.Horizontally, $F=ma$ : $20-14.7=6a$ , so $a=\dfrac{5.3}{6}=0.883$ m s $^{-2}$ .

Answer: $0.88$ m s $^{-2}$ (2 s.f.)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	$R=6g=58.8$ N	B1
	$F=\mu R$ used: $0.25\times$ their $R$	M1
	Equation of motion with the correct terms: $20-F=6a$	M1
	$a=0.88$ m s $^{-2}$ or $0.883$	A1
Notes
M1: Condone sign errors
A1: awrt 0.88

(4 marks)

Common mistakes

Don't reverse the direction of friction so that it assists the impending relative motion.
Don't set friction equal to the limiting value in every situation, even when equilibrium requires a smaller force.

Mark-scheme wording to use

“(2/7) × 4.9 oe seen … M1 (2/7) × their (a) (must be a numerical value) … 1.4, 1.40 or (1/7)g … A1” — Pearson 9MA0, June 2024 Paper 32, Q1(b)
“Equation of motion down the plane … M1 … mg sin α − F = ma or mg sin α − F = −ma … A1 … (F =) μ × their R … M1 … A1* Given answer correctly obtained, with at least one further line of working with both trig ratios substituted as fractions.” — Pearson 9MA0, June 2024 Paper 32, Q3(b)
“B1 for R = 3gcos(20) and hence maximum friction 0.9gcos(20). M1 for comparing or subtracting 3gsin(20) and 0.9gcos(20). A1* for the justified conclusion that the box starts to move down the plane” — Pearson 9MA0, Specimen Paper 32, Q3(c)

What the examiner wants

$R$ before $F$ , and $\mu R$ only when friction is limiting. The June 2024 Paper 32 report says most found the limiting friction by multiplying their reaction by $\frac27$ and then realised the applied force had the same value, though some, seeing the word “magnitude”, squared or square-rooted a correct answer and lost the accuracy mark.

Exam tip

Decide whether friction is limiting; otherwise use the force balance to find its actual value within the inequality.

Worked practice

Tier 1 · Easy

A block is in equilibrium on a rough horizontal surface. The normal reaction is

80\,\text{N}

and the coefficient of friction is

0.30

. A horizontal force of

10\,\text{N}

acts on the block. Find the friction force and the greatest possible friction force.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Friction $=10\,\text{N}$ opposite to the applied force Greatest possible friction $=24\,\text{N}$	3
Notes
Equilibrium requires friction to balance the $10\,\text{N}$ applied force, so $F=10\,\text{N}$ . The limiting value is $\mu R=0.30(80)=24\,\text{N}$ ; the actual friction is smaller because the block is not in limiting equilibrium.

(3 marks)

Tier 2 · Standard

4\,\text{kg}

block rests on a rough horizontal surface with coefficient of friction

0.30

. A horizontal force

P

is applied. Using

g=9.8\,\text{m s}^{-2}

, find the greatest value of

P

for equilibrium and the friction force when

P=10\,\text{N}

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Greatest $P=11.76\,\text{N}$ , or $11.8\,\text{N}$ to 3 significant figures When $P=10\,\text{N}$ , friction is $10\,\text{N}$ opposite to $P$ .	4
Notes
The normal reaction is $R=4g=39.2\,\text{N}$ , so the limiting friction is $\mu R=0.30(39.2)=11.76\,\text{N}$ . This is the greatest horizontal force that static friction can balance. When $P=10\,\text{N}<11.76\,\text{N}$ , friction adjusts to $10\,\text{N}$ ; it is not automatically equal to $\mu R$ .

(4 marks)

Tier 3 · Hard

10\,\text{kg}

particle rests on a rough plane inclined at

20^\circ

to the horizontal. The coefficient of friction is

0.30

. A force of magnitude

P\,\text{N}

acts up the plane. Using

g=9.8\,\text{m s}^{-2}

, find the complete range of values of

P

for which the particle remains in equilibrium.

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$5.89\leq P\leq61.1$	7
Notes
The normal reaction is $R=98\cos20^\circ$ , so limiting friction is $\mu R=29.4\cos20^\circ$ . The component of weight down the plane is $98\sin20^\circ$ . Equilibrium requires friction of magnitude $\|98\sin20^\circ-P\|$ , so $\|98\sin20^\circ-P\|\leq29.4\cos20^\circ$ . Hence $98\sin20^\circ-29.4\cos20^\circ\leq P\leq98\sin20^\circ+29.4\cos20^\circ$ , giving $5.89\leq P\leq61.1$ .

(7 marks)

Tier 1 · Easy

A particle is in limiting equilibrium on a rough surface. The friction has magnitude

18\,\text{N}

and the normal reaction is

60\,\text{N}

. Find the coefficient of friction.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$0.30$	2
Notes
At limiting equilibrium $F=\mu R$ , so $18=60\mu$ and $\mu=0.30$ .

(2 marks)

Tier 2 · Standard

6\,\text{kg}

block is on a rough horizontal surface. A force of

24\,\text{N}

acts at

45^\circ

above the horizontal and the block is on the point of moving in the horizontal direction of the force. Using

g=9.8\,\text{m s}^{-2}

, find the coefficient of friction, giving your answer to 3 significant figures.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$\mu=0.406$	5
Notes
Vertical equilibrium gives $R+24\sin45^\circ=6(9.8)$ , so $R=58.8-12\sqrt2\,\text{N}$ . At limiting equilibrium the friction is $\mu R$ and horizontal equilibrium gives $\mu R=24\cos45^\circ=12\sqrt2$ . Hence $\mu=12\sqrt2/(58.8-12\sqrt2)=0.406$ to 3 significant figures.

(5 marks)

Tier 3 · Hard

5\,\text{kg}

block is in equilibrium on a rough horizontal surface with coefficient of friction

0.25

. A horizontal force of

10\,\text{N}

acts to the left. A second force of magnitude

P\,\text{N}

acts upwards and to the right at an angle

\alpha

above the horizontal, where

\tan\alpha=\dfrac34

. Given

P\geq0

and using

g=9.8\,\text{m s}^{-2}

, find the complete range of values of

P

for which equilibrium is possible. Give the upper bound to 3 significant figures.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$0\leq P\leq\dfrac{445}{19}\,\text{N}$ , or $0\leq P\leq23.4\,\text{N}$ to 3 significant figures	6
Notes
Since $\cos\alpha=4/5$ and $\sin\alpha=3/5$ , vertical equilibrium gives $R=49-3P/5$ . The friction required has magnitude $\|4P/5-10\|$ . Equilibrium is possible when $\|4P/5-10\|\leq(1/4)(49-3P/5)$ . The lower inequality gives $P\geq-45/13$ , which is replaced by the stated $P\geq0$ . The upper inequality gives $4P/5-10\leq49/4-3P/20$ , so $19P/20\leq89/4$ and $P\leq445/19=23.4\,\text{N}$ to 3 significant figures. This range also keeps $R>0$ .

(6 marks)

Tier 2 · Standard

The angle

\alpha

between a rough plane and the horizontal satisfies

\tan\alpha=\dfrac34

. A

4\,\text{kg}

block slides down the plane and the coefficient of friction is

\dfrac14

. Model the block as a particle. Take

g=9.8\,\text{m s}^{-2}

. Find its acceleration down the plane.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Acceleration $=3.92\,\text{m s}^{-2}$ down the plane	4
Notes
Since $\tan\alpha=3/4$ , $\sin\alpha=3/5$ and $\cos\alpha=4/5$ . The normal reaction is $R=4g(4/5)$ , so the friction opposing the downward motion is $\mu R=(1/4)4g(4/5)=4g/5\,\text{N}$ . Resolving down the plane, $4a=4g(3/5)-4g/5=8g/5$ , hence $a=2g/5=3.92\,\text{m s}^{-2}$ .

(4 marks)

Tier 3 · Hard

The angle

\alpha

between a rough plane and the horizontal satisfies

\tan\alpha=\dfrac34

. A

5\,\text{kg}

block moves up the plane while a horizontal force of

48\,\text{N}

acts towards the upward side. The coefficient of friction is

0.20

. Model the block as a particle. Take

g=9.8\,\text{m s}^{-2}

. Find the normal reaction and the acceleration of the block. Given that its speed is initially

4\,\text{m s}^{-1}

, find its speed and direction of motion

3\,\text{s}

later.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Normal reaction $=68\,\text{N}$ Acceleration $=0.92\,\text{m s}^{-2}$ down the plane After $3\,\text{s}$ the block moves up the plane at $1.24\,\text{m s}^{-1}$	6
Notes
Here $\sin\alpha=3/5$ and $\cos\alpha=4/5$ . Perpendicular to the plane, the weight contributes $5g(4/5)=39.2\,\text{N}$ into the plane and the horizontal force contributes $48(3/5)=28.8\,\text{N}$ into the plane, so $R=68\,\text{N}$ . Since the block moves up the plane, friction $0.20R=13.6\,\text{N}$ acts down the plane. Taking up the plane as positive, the resultant is $48(4/5)-5g(3/5)-13.6=38.4-29.4-13.6=-4.6\,\text{N}$ . Thus $a=-4.6/5=-0.92\,\text{m s}^{-2}$ . After $3\,\text{s}$ , $v=4-0.92(3)=1.24\,\text{m s}^{-1}$ , which is still up the plane.

(6 marks)

Tier 3 · Hard

8\,\text{kg}

block slides on a rough horizontal plane. The only horizontal force is friction, modelled with coefficient

\mu

. The block has speed

10\,\text{m s}^{-1}

and comes to rest after travelling

25\,\text{m}

. Take

g=9.8\,\text{m s}^{-2}

and assume the friction model applies throughout. Find the exact value of

\mu

, the friction force and the time taken to stop. Give all numerical answers exactly.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\mu=\dfrac{10}{49}$ Friction force $=16\,\text{N}$ opposite to the motion Stopping time $=5\,\text{s}$	6
Notes
Taking the direction of motion as positive, $0=10^2+2a(25)$ gives $a=-2\,\text{m s}^{-2}$ . The friction magnitude is therefore $m\|a\|=8(2)=16\,\text{N}$ . Vertically, $R=8(9.8)=78.4\,\text{N}$ , so $\mu=F/R=16/78.4=10/49$ . Finally $0=10-2t$ , giving $t=5\,\text{s}$ .

(6 marks)

Q10

Tier 3 · Hard

10.

10\,\text{kg}

block moves on a rough horizontal plane with acceleration

0.8\,\text{m s}^{-2}

to the right. The coefficient of friction is

1/4

. A force of magnitude

P\,\text{N}

acts downwards and to the right at an angle

\alpha

below the horizontal, where

\tan\alpha=3/4

. Model the block as a particle, assume the friction model applies and take

g=9.8\,\text{m s}^{-2}

. Find the exact values of

P

, the normal reaction and the friction force. Give all numerical answers exactly.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$P=50\,\text{N}$ Normal reaction $=128\,\text{N}$ Friction force $=32\,\text{N}$ to the left	6
Notes
Here $\cos\alpha=4/5$ and $\sin\alpha=3/5$ . Vertical equilibrium gives $R=98+3P/5$ , so friction is $R/4$ . Resolving horizontally to the right, $4P/5-(98+3P/5)/4=10(0.8)$ . Thus $13P/20-24.5=8$ , so $13P/20=32.5$ and $P=50\,\text{N}$ . Hence $R=98+30=128\,\text{N}$ and friction is $128/4=32\,\text{N}$ to the left.

(6 marks)

Verified exam appearances

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