A Level Maths: Newton's Laws
Introduction
When you do mechanics in both year 1 and year 2 of your maths course you are going to meet A Level Maths Newton’s laws and this area of maths can also include SUVAT, gravity, questions on a slope as well as horizontally as well as questions that involve fraction.
A Level Maths Newton’s Laws: First
A change in the state of motion of a body is caused by a force. The unit of force is the Newton and is shortened to N.
Newton’s first law of motion states:
A body will remain at rest, or will continue to move with constant velocity unless external forces cause it to do otherwise.
Example
If a block of mass 3kg slides down a rough plane (μ=0.5) inclined at an angle to the horizontal, with constant speed, what is the value of \alpha ?
Solution
Here we need to resolve parallel and perpendicular to the plane.
Perpendicular: R=3 g \cos \alpha
Parallel: 3 g \sin \alpha=F
Now F=\mu R \rightarrow \mu R=3 g \sin \alpha \rightarrow R=\frac{3 g \sin \alpha}{0.5}
\begin{aligned} & 3 g \cos \alpha=\frac{3 g \sin \alpha}{0.5} \\ & 0.5 \cos \alpha=\sin \alpha \\ & 0.5=\tan \alpha \rightarrow \alpha=26.6^{\circ} \end{aligned}A Level Maths Newton’s Laws: Second
When a resultant force acts on a body then it will cause an acceleration which is proportional to the force.
The acceleration that is produced depends very much on the mass of the body.
In general we can say that a force of F Newton’s acting on a body of mass m kg has an acceleration of a m s^{-2} where F = ma and this is also known as the equation of motion.
F = ma is a vector equation which means that the acceleration produced is in the same direction as the force.
Example
A body of mass 8kg is acted upon by a force of 10N. What is the acceleration?
Solution
Using F = ma we have:
10=8 a \rightarrow a=1.25 \mathrm{~ms}^{-2}Example
A body of mass 2kg is subject to the forces shown in the diagram below. Find the acceleration and the value of P.
Solution
Here we need to resolve vertically and horizontally:
Vertically: P = 20
Horizontally: Using F = ma
10-4=2 a \rightarrow \frac{6}{2}=a \therefore a=3 \mathrm{~ms}^{-2}Example
Find in vector form the acceleration produced by a body of mass 5kg which is subject to the following forces (4i +j)N and (-i + j)N. State also the magnitude of the acceleration and its direction with the horizontal.
Solution
Resultant force = (4i +j) + (-i + j) = 3i + 2j
Using F = ma:
\begin{aligned} & 3 \mathbf{i}+2 \mathbf{j}=5 \mathrm{a} \rightarrow \mathrm{a}=\frac{1}{5}(3 i+2 j) \mathrm{ms}^{-2} \\ & |\mathrm{a}|=\sqrt{\frac{3}{5}^2+\frac{2}{5}^2}=0.72 \mathrm{~ms}^{-2} \end{aligned}When you are doing A Level Maths Newton’s Laws questions in vectors remember not to remove the vector notation.
Example
A particle of 20kg is pulled across a rough horizontal surface by a string which is inextensible and it is inclined at 30° to the horizontal. The tension in the string is 50N and the acceleration is 0.5 m s^{-2}. What is the frictional force and the coefficient of friction?
Solution
Horizontally:
\begin{aligned} & 50 \cos 30-F=20(0.5) \\ & F=50 \cos 30-10=33.03 N \\ & \text { Vertically: } \\ & R+50 \sin 30=20 g \\ & R=20 g-55 \sin 30=168.5 \\ & F=\mu R \\ & 33.03=\mu R \\ & \therefore \mu=\frac{33.03}{168.50}=0.198 \end{aligned}xample
A sledge of mass 30kg is accelerating down a hill whilst a boy is trying to prevent it from sliding by pulling on a rope which is attached to the sledge with a force of 40N. The hill is inclined at 28° to the horizontal and the rope is inclined to the hill at 10°. The coefficient of friction between the sledge and the hill is 0.35.
Find:
a) The normal reaction
b) The resultant force acting down the slope
c) The acceleration of the sledge
Solution
a) Perpendicular to the plane:
\begin{aligned} & R+40 \sin 10=30 g \cos 28 \\ & R=30 g \cos 28-40 \sin 10 \\ & R=252.64 N \end{aligned}
b) Parallel to the plane down the slope:
\begin{aligned} &\begin{aligned} & 30 g \sin 28-40 \cos 10-F \\ & =30 g \sin 28-40 \cos 10-0.35(252.64) \end{aligned}\\ &=10.208 \mathrm{~N} \end{aligned}
c) Using F = ma
\frac{10.208}{30}=\mathrm{a} \rightarrow \mathrm{a}=0.34 \mathrm{~ms}^{-2}
A Level Maths Newton’s Laws: Third
This law states that: Action and Reaction are equal and opposite.
What this means is that if two bodies A and B are in contact and exert forces on each other then the force exerted on B by A is equal in magnitude and opposite in direction to the force exerted on A by B.
To illustrate consider the diagram below:
The diagram above shows a 5kg block sitting on a horizontal floor. The box exerts a force on the floor and for the floor to prevent the block from falling through it reacts with an equal and opposite force. Because the box is at rest the reaction force must be equal to the weight of the box.
The reaction force is known as the normal reaction as it acts at right angles to the surface it contacts.
In all the examples that have been done within the article, a diagram has been drawn. It is important when doing A Level Maths Newton’s Laws questions that you are drawing diagrams and that they are clear and well labelled. If a diagram is given to you in a question then you are more than likely going to need to include all the relevant forces.
If your question involves a slope then you need to make sure that you are resolving all forces parallel and perpendicular to the plane.