M8 Forces and Newton's laws — 9MA0 revision notes

M8.1

Understand the concept of a force; understand and use Newton's first law.

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A force is an interaction with magnitude and direction; common modelled forces include weight, normal reaction, tension, thrust, compression and resistance.
Draw a force diagram for the chosen body, including only forces acting on that body, then resolve forces along useful perpendicular directions.
Newton's first law says that a particle remains at rest or moves with constant velocity when the resultant force on it is zero.
A common error is to infer that a moving particle must have a forward resultant force; constant non-zero velocity also means zero resultant force.

Worked example

A powered trolley moves in a straight line with constant velocity. Its motor exerts a forward force of $8\,\text{N}$ . Find the resistance force and explain your answer.

1.Constant velocity means zero acceleration and hence, by Newton's first law, zero resultant force.
2.The resistance must therefore balance the motor force, so it is $8\,\text{N}$ backwards.

Answer: Resistance $=8\,\text{N}$ opposite to the motion; Constant velocity means the resultant force is zero

Common mistakes

Don't balance only the horizontal forces and declare equilibrium while a vertical resultant remains.
Don't assume motion implies a resultant force even when the velocity is constant.

Exam tip

For constant velocity, invoke zero acceleration and balance the forces with equal magnitudes in opposite directions.

Tier 1 · Easy

ORIGINAL

1.

A book rests on a horizontal table. Its weight is

35\,\text{N}

. State the magnitude and direction of the normal reaction on the book and the resultant force on it.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A book of mass

2.5\,\text{kg}

rests on a horizontal table. Using

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

, find the weight of the book and the normal force exerted by the table, explaining why the forces have equal magnitudes.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

A particle moves with constant velocity under three coplanar forces. Two of the forces are

(6\mathbf{i}-8\mathbf{j})\,\text{N}

and

(-2\mathbf{i}+5\mathbf{j})\,\text{N}

. Find the third force, giving also its magnitude.

(4)

(Total for Question 1 is 4 marks)

M8.2

Understand and use Newton's second law for motion in a straight line (forces in two perpendicular directions or simple 2-D vectors); extend to situations where forces need to be resolved (restricted to 2 dimensions).

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Newton's second law is $\sum\mathbf{F}=m\mathbf{a}$ ; in two dimensions it gives one scalar equation in each resolved direction.
Choose axes parallel and perpendicular to the motion where possible, resolve every force onto those axes, then apply $\sum F=ma$ separately.
For motion along a fixed smooth plane, perpendicular acceleration is zero, so the perpendicular force equation determines the normal reaction while the parallel equation determines acceleration.
A common error is to write $F=ma$ for one force instead of the resultant, or to reverse signs for only some components after choosing a positive direction.

Resolve forces in perpendicular directions, then apply

\sum\mathbf F=m\mathbf a

component by component.

Worked example

A $5\,\text{kg}$ particle is on a smooth plane inclined at $30^\circ$ to the horizontal. A force of $40\,\text{N}$ pulls it up the line of greatest slope. Using $g=9.8\,\text{m s}^{-2}$ , find its acceleration.

1.Resolve up the plane.
2.The component of weight down the plane is $5(9.8)\sin30^\circ=24.5\,\text{N}$ .
3.Hence the resultant up the plane is $40-24.5=15.5\,\text{N}$ .
4.From $15.5=5a$ , $a=3.1\,\text{m s}^{-2}$ up the plane.

Answer: $3.1\,\text{m s}^{-2}$ up the plane

Common mistakes

Don't include the normal reaction in the equation parallel to an inclined plane.
Don't resolve weight using the wrong trigonometric component relative to the inclined plane.

Exam tip

Draw a force diagram, choose axes parallel and perpendicular to the plane, then resolve every force consistently.

Tier 1 · Easy

ORIGINAL

1.

A constant resultant force of

12\,\text{N}

acts on a particle of mass

3\,\text{kg}

. Find its acceleration.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A

4\,\text{kg}

particle moves on a smooth horizontal surface. A force of

20\,\text{N}

acts at

30^\circ

above the horizontal. Using

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

, find the horizontal acceleration and the normal reaction.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

A

6\,\text{kg}

particle is pulled up a smooth plane inclined at

20^\circ

to the horizontal by a force of

50\,\text{N}

acting at

15^\circ

above the plane. Using

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

, find the normal reaction and the particle's acceleration up the plane.

(7)

(Total for Question 1 is 7 marks)

M8.3

Understand and use weight and motion in a straight line under gravity; gravitational acceleration, g, and its value in S.I. units to varying degrees of accuracy.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Weight is the gravitational force $W=mg$ , directed vertically downwards; mass is measured in kilograms and does not depend on location.
The value of $g$ is not universal and depends on location; use the stated value, or the Edexcel default $9.8\,\text{m s}^{-2}$ when none is supplied.
For free motion near Earth's surface in the constant- $g$ model, acceleration is $g$ downwards, but a support or tension changes the resultant acceleration.
A common error is to call mass a force or to assume that the normal reaction always equals weight when the body has vertical acceleration.

Worked example

A particle is released from rest and falls freely through $19.6\,\text{m}$ . Using $g=9.8\,\text{m s}^{-2}$ , find the time taken and its speed after falling this distance.

1.Taking downwards as positive, $s=ut+\tfrac12gt^2$ gives $19.6=0+4.9t^2$ , so $t=2.0\,\text{s}$ .
2.Then $v=u+gt=0+9.8(2)=19.6\,\text{m s}^{-1}$ downwards.

Answer: Time $=2.0\,\text{s}$ ; Speed $=19.6\,\text{m s}^{-1}$

Common mistakes

Don't use $g=9.8$ with distances in centimetres and times in seconds without converting units.
Don't use positive gravitational acceleration with an upward-positive displacement equation, producing inconsistent signs.

Exam tip

Declare the positive direction before substituting gravitational acceleration into a constant-acceleration equation.

Tier 1 · Easy

ORIGINAL

1.

Find the weight of a

2.5\,\text{kg}

particle, using

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

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A particle is projected vertically upwards with speed

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

. Using

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

, find its greatest height above the point of projection and the time taken to return to that point.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

A person of mass

70\,\text{kg}

stands on a scale in a lift. The scale exerts an upward force of

770\,\text{N}

on the person. Using

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

, find the magnitude and direction of the lift's acceleration.

(4)

(Total for Question 1 is 4 marks)

M8.4

Understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; apply to smooth pulleys and connected particles; resolve forces in 2 dimensions; equilibrium of a particle under coplanar forces.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Newton's third-law forces are equal and opposite, act on different bodies and arise from the same interaction; they therefore do not cancel on one body's force diagram.
For a particle in equilibrium under coplanar forces, resolve in two independent directions and set both component resultants equal to zero.
For connected particles, draw separate force diagrams, use a common acceleration magnitude while the string is taut, and use one tension for a light string over a smooth pulley before solving simultaneous $F=ma$ equations.
A common error is to treat weight and normal reaction as a third-law pair; both act on the same body, whereas a third-law pair acts on different bodies.

Two particles connected by a light string over a smooth pulley share one tension magnitude while the string is taut.

Worked example

A particle is in equilibrium under three coplanar forces. One force is $14\,\text{N}$ east and another is $10\,\text{N}$ at $120^\circ$ anticlockwise from east. Find the third force as a vector in east-north components and find its magnitude.

1.The first two forces have resultant $(14+10\cos120^\circ)\mathbf{i}+(10\sin120^\circ)\mathbf{j}=9\mathbf{i}+5\sqrt3\mathbf{j}$ .
2.Equilibrium requires zero resultant, so the third force is $-9\mathbf{i}-5\sqrt3\mathbf{j}$ .
3.Its magnitude is $\sqrt{9^2+(5\sqrt3)^2}=\sqrt{156}=2\sqrt{39}\,\text{N}$ .

Answer: Third force $=(-9\mathbf{i}-5\sqrt3\mathbf{j})\,\text{N}$ ; Magnitude $=2\sqrt{39}\,\text{N}$

Common mistakes

Don't assign different tension magnitudes to the two ends of one light inextensible string over a smooth pulley.
Don't treat a Newton third-law pair as two forces on the same particle or assume connected-particle tensions without a force diagram.

Exam tip

Draw a separate force diagram for each particle and distinguish interaction pairs from forces that can balance.

Tier 1 · Easy

ORIGINAL

1.

A book pushes down on a table with force

P

. State the Newton's third-law partner to this force.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Particles of masses

3\,\text{kg}

and

5\,\text{kg}

hang at the ends of a light inextensible string passing over a smooth pulley. Using

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

, find the acceleration of the system and the tension in the string.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

A

4\,\text{kg}

particle on a smooth plane inclined at

30^\circ

is connected by a light inextensible string over a smooth pulley to a freely hanging

3\,\text{kg}

particle. Using

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

, find the acceleration of the system and the tension in the string.

(7)

(Total for Question 1 is 7 marks)

M8.5

Understand and use addition of forces; resultant forces; dynamics for motion in a plane.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Forces add as vectors, so a resultant may be written in component form and then converted to magnitude-direction form.
Resolve each force along two perpendicular axes, add corresponding components, and apply $\sum\mathbf{F}=m\mathbf{a}$ to obtain the acceleration vector.
If the resultant is $X\mathbf{i}+Y\mathbf{j}$ , its magnitude is $\sqrt{X^2+Y^2}$ and its direction must be placed in the correct quadrant from the component signs.
A common error is to add force magnitudes without accounting for their directions, or to quote an inverse-tangent angle in the wrong quadrant.

The resultant force is the vector sum of the component forces.

Worked example

Two forces of magnitudes $12\,\text{N}$ and $10\,\text{N}$ act at an angle of $120^\circ$ to each other. Find the magnitude of their resultant and the angle the resultant makes with the $12\,\text{N}$ force.

1.Take the $12\,\text{N}$ force along the positive horizontal direction.
2.The other force has components $10\cos120^\circ=-5$ and $10\sin120^\circ=5\sqrt3$ .
3.The resultant is $7\mathbf{i}+5\sqrt3\mathbf{j}$ , with magnitude $\sqrt{49+75}=2\sqrt{31}\,\text{N}$ .
4.Its angle is $\tan^{-1}(5\sqrt3/7)=51.1^\circ$ .

Answer: Magnitude $=2\sqrt{31}\,\text{N}$ ; Angle $=51.1^\circ$ towards the $10\,\text{N}$ force

Common mistakes

Don't find a resultant magnitude using Pythagoras even though the force components are not perpendicular.
Don't add force magnitudes arithmetically despite a non-zero angle between them.

Exam tip

Resolve forces into perpendicular components or use the cosine rule, then find the resultant direction from its components.

Tier 1 · Easy

ORIGINAL

1.

A force is

(6\mathbf{i}-8\mathbf{j})\,\text{N}

. Find its magnitude and its angle below the positive

\mathbf{i}

direction.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

A particle of mass

2\,\text{kg}

is initially at rest. Constant forces

(8\mathbf{i}-2\mathbf{j})\,\text{N}

and

(-2\mathbf{i}+10\mathbf{j})\,\text{N}

act on it. Find the resultant force, its acceleration, and its speed after

2\,\text{s}

.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

A particle of mass

5\,\text{kg}

has initial velocity

(2\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}

. Constant forces

(15\mathbf{i}+20\mathbf{j})\,\text{N}

and

(-5\mathbf{i}+10\mathbf{j})\,\text{N}

act on it. Find its acceleration, velocity after

3\,\text{s}

and displacement during those

3\,\text{s}

.

(7)

(Total for Question 1 is 7 marks)

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.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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.

Tier 1 · Easy

ORIGINAL

1.

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)

Tier 2 · Standard

ORIGINAL

1.

A

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 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

A

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 1 is 7 marks)

Understand the concept of a force; understand and use Newton's first law.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Understand and use Newton's second law for motion in a straight line (forces in two perpendicular directions or simple 2-D vectors); extend to situations where forces need to be resolved (restricted to 2 dimensions).

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Understand and use weight and motion in a straight line under gravity; gravitational acceleration, g, and its value in S.I. units to varying degrees of accuracy.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; apply to smooth pulleys and connected particles; resolve forces in 2 dimensions; equilibrium of a particle under coplanar forces.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Understand and use addition of forces; resultant forces; dynamics for motion in a plane.

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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