M8 Forces and Newton's laws — printable coverage pack

M8.1 · Understand the concept of a force; understand and use Newton's first law.

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.

Tier 1 · Easy

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 marks]
Answer
- Normal reaction $=35\,\text{N}$ upwards
- Resultant force $=0\,\text{N}$
Method: The stationary book has zero acceleration, so Newton's first law requires zero resultant force. The upward normal reaction therefore balances the $35\,\text{N}$ weight and has magnitude $35\,\text{N}$ .

Tier 2 · Standard

1. 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.[3 marks]
Answer
- Resistance $=8\,\text{N}$ opposite to the motion
- Constant velocity means the resultant force is zero
Method: Constant velocity means zero acceleration and hence, by Newton's first law, zero resultant force. The resistance must therefore balance the motor force, so it is $8\,\text{N}$ backwards.

Tier 3 · Hard

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 marks]
Answer
- Third force $=(-4\mathbf{i}+3\mathbf{j})\,\text{N}$
- Magnitude $=5\,\text{N}$
Method: Constant velocity requires the vector resultant to be zero. The two known forces add to $4\mathbf{i}-3\mathbf{j}$ , so the third force is its negative, $-4\mathbf{i}+3\mathbf{j}$ . Its magnitude is $\sqrt{(-4)^2+3^2}=5\,\text{N}$ .

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).

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.

Tier 1 · Easy

1. A constant resultant force of $12\,\text{N}$ acts on a particle of mass $3\,\text{kg}$ . Find its acceleration.[2 marks]
Answer
- $4\,\text{m s}^{-2}$ in the direction of the force
Method: Newton's second law gives $F=ma$ , so $a=F/m=12/3=4\,\text{m s}^{-2}$ , directed with the resultant force.

Tier 2 · Standard

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

Tier 3 · Hard

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 marks]
Answer
- Normal reaction $=42.3\,\text{N}$
- Acceleration $=4.70\,\text{m s}^{-2}$ up the plane
Method: Perpendicular to the plane, $R+50\sin15^\circ-6(9.8)\cos20^\circ=0$ , so $R=6(9.8)\cos20^\circ-50\sin15^\circ=42.3\,\text{N}$ . Parallel to the plane, $50\cos15^\circ-6(9.8)\sin20^\circ=6a$ . Therefore $a=[50\cos15^\circ-58.8\sin20^\circ]/6=4.70\,\text{m s}^{-2}$ .

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.

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.

Tier 1 · Easy

1. Find the weight of a $2.5\,\text{kg}$ particle, using $g=9.8\,\text{m s}^{-2}$ .[2 marks]
Answer
- $24.5\,\text{N}$ vertically downwards
Method: $W=mg=2.5(9.8)=24.5\,\text{N}$ , and weight acts vertically downwards.

Tier 2 · Standard

1. 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.[4 marks]
Answer
- Time $=2.0\,\text{s}$
- Speed $=19.6\,\text{m s}^{-1}$
Method: Taking downwards as positive, $s=ut+\tfrac12gt^2$ gives $19.6=0+4.9t^2$ , so $t=2.0\,\text{s}$ . Then $v=u+gt=0+9.8(2)=19.6\,\text{m s}^{-1}$ downwards.

Tier 3 · Hard

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 marks]
Answer
- $1.2\,\text{m s}^{-2}$ upwards
Method: The person's weight is $70(9.8)=686\,\text{N}$ downwards. Taking upwards as positive, Newton's second law gives $770-686=70a$ . Hence $a=84/70=1.2\,\text{m s}^{-2}$ upwards.

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.

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.

Tier 1 · Easy

1. A book pushes down on a table with force $P$ . State the Newton's third-law partner to this force.[2 marks]
Answer
- The table pushes upwards on the book with force $P$
Method: The partner must be the same interaction with the bodies reversed. It is the force of the table on the book, equal in magnitude to $P$ and opposite in direction.

Tier 2 · Standard

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

Tier 3 · Hard

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 marks]
Answer
- Acceleration $=1.4\,\text{m s}^{-2}$ , with the $3\,\text{kg}$ particle moving down
- Tension $=25.2\,\text{N}$
Method: The downslope weight component of the $4\,\text{kg}$ particle is $4(9.8)\sin30^\circ=19.6\,\text{N}$ , while the hanging weight is $29.4\,\text{N}$ , so the $3\,\text{kg}$ particle moves down. For the system, $29.4-19.6=7a$ , giving $a=1.4\,\text{m s}^{-2}$ . For the hanging particle, $29.4-T=3(1.4)$ , so $T=25.2\,\text{N}$ .

M8.5 · Understand and use addition of forces; resultant forces; dynamics for motion in a plane.

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.

Tier 1 · Easy

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 marks]
Answer
- Magnitude $=10\,\text{N}$
- Angle $=53.1^\circ$ below the positive $\mathbf{i}$ direction
Method: The magnitude is $\sqrt{6^2+(-8)^2}=10\,\text{N}$ . Since the components place the force in the fourth quadrant, the angle below the positive $\mathbf{i}$ direction is $\tan^{-1}(8/6)=53.1^\circ$ .

Tier 2 · Standard

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

Tier 3 · Hard

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 marks]
Answer
- Acceleration $=(2\mathbf{i}+6\mathbf{j})\,\text{m s}^{-2}$
- Velocity $=(8\mathbf{i}+17\mathbf{j})\,\text{m s}^{-1}$
- Displacement $=(15\mathbf{i}+24\mathbf{j})\,\text{m}$
Method: The resultant force is $10\mathbf{i}+30\mathbf{j}$ , so $\mathbf{a}=\mathbf{F}/m=2\mathbf{i}+6\mathbf{j}$ . Then $\mathbf{v}=\mathbf{u}+\mathbf{a}t=(2\mathbf{i}-\mathbf{j})+3(2\mathbf{i}+6\mathbf{j})=8\mathbf{i}+17\mathbf{j}$ . Finally $\mathbf{s}=\mathbf{u}t+\tfrac12\mathbf{a}t^2=(6\mathbf{i}-3\mathbf{j})+(9\mathbf{i}+27\mathbf{j})=15\mathbf{i}+24\mathbf{j}$ .

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.

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 $\mu R$ .

Tier 1 · Easy

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 marks]
Answer
- Friction $=10\,\text{N}$ opposite to the applied force
- Greatest possible friction $=24\,\text{N}$
Method: 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.

Tier 2 · Standard

1. 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.[5 marks]
Answer
- $0.883\,\text{m s}^{-2}$ in the direction of the applied force
Method: Vertically, $R=mg=6(9.8)=58.8\,\text{N}$ . Since the block is moving, the friction model gives $F=\mu R=0.25(58.8)=14.7\,\text{N}$ . The horizontal resultant is $20-14.7=5.3\,\text{N}$ , so $a=5.3/6=0.883\,\text{m s}^{-2}$ .

Tier 3 · Hard

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 marks]
Answer
- $5.89\leq P\leq61.1$
Method: 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 $5.89\leq P\leq61.1$ .