What is M8.6: Friction?

Friction acts to oppose actual or impending relative motion; in equilibrium its magnitude adjusts within 0≤ F≤ 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.

Explanation

Friction acts to oppose actual or impending relative motion; in equilibrium its magnitude adjusts within $0\leq F\leq\mu R$ .
Find the normal reaction first, decide the likely direction of motion, and use $F=\mu R$ only when friction is limiting or the particle is moving in this model.
For a range of equilibrium values, write the required friction in terms of the applied force and impose $|F|\leq\mu R$ before solving the resulting inequality.
A common error is to set $F=\mu R$ in every static problem; away from limiting equilibrium, friction may be strictly smaller than $\mu R$ .

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

F\leq\mu R

Worked example

A $6\,\text{kg}$ block is moving on a rough horizontal surface with coefficient of friction $0.25$ . A horizontal force of $20\,\text{N}$ acts in the direction of motion. Using $g=9.8\,\text{m s}^{-2}$ , find its acceleration.

1.Vertically, $R=mg=6(9.8)=58.8\,\text{N}$ .
2.Since the block is moving, the friction model gives $F=\mu R=0.25(58.8)=14.7\,\text{N}$ .
3.The horizontal resultant is $20-14.7=5.3\,\text{N}$ , so $a=5.3/6=0.883\,\text{m s}^{-2}$ .

Answer: $0.883\,\text{m s}^{-2}$ in the direction of the applied force

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.

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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Other points in M8 Forces and Newton's laws

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