A Level Maths: Resolving Forces
Introduction
When it comes to A Level Mechanics one of the main areas that you will be looking at is forces. This is the case whether you are in year 12 or in year 13. Questions involving forces can be horizontal, on an inclined slope, they can also be a part of a SUVAT question and within 2 or 3 dimensions.
Within this article we are going to consider:
What is meant by a resultant force
How to correctly draw a force diagram and to determine the resultant force
How to be able to resolve forces
Resolving Forces - Some Terminology
A change in the state of motion of a body is caused by a force. The unit of force is Newton and this is abbreviated to N.
One force which everything is subject to because of the earth’s gravitational pull is weight. Weight = mass x g. And g=9.8 ms-2
If you have a block that is resting on a table and the weight of the book is M, then an equal force is acting upwards from the table onto the book to ensure that it does not fall through the table. This is called the normal reaction and is generally represented by the letter R.
Resultant forces
A resultant force is a single force that can be used to replace two or more forces. For instance if we consider the diagram below where we have a block that is being pulled by the two forces as shown:
Then we can replace the above system with diagram shown below:
Resolving Forces - An Example
A body is a rest when subjected to the forces shown in the diagram. Find X and Y.
Solution
If a body is at rest we can say that it is in a state of equilibrium and so opposite forces must be equal.
Horizontal forces must balance so: X = 50 + 10 = 60 N
Vertical forces must balance so: Y + 10 = 50 🡪 Y = 40 N
Resolving Forces – Components of a force
From the above diagram, consider a force F acting at an angle above the horizontal. If we were to find the horizontal and vertical components then we have resolved the forces.
The components of the force F can be found by using trigonometry and in particular SOHCAHTOA
\text { Horizontal component: } \quad \cos \theta=\frac{x}{F} \rightarrow x=F \cos \theta\text { Vertical component: } \quad \sin \sin \theta=\frac{y}{F} \rightarrow y=F \sin \thetaExample
Find the components of the force 10N acting at 30° above the horizontal.
Solution
Finding the components means we need to find both the horizontal and vertical components.
\begin{gathered} x=10 \cos 30^{\circ}=8.66 \mathrm{~N} \\ y=10 \sin 30^{\circ}=5 \mathrm{~N} \end{gathered}When it comes to determining what the resultant force is for two forces this is generally done using Pythagoras.
Given the components above we have:
The resultant force is the single force that can replace these two and this is calculated by \sqrt{8.66^2+5^2}=\sqrt{100}=10
Finding the resultant of more than two forces
Consider the following system of forces. We want to determine the magnitude of the resultant force:
In order to find the resultant force we need to resolve all the forces horizontally and vertically. It is important to also first determine a positive direction. Everything to the right will be taken to be positive and everything vertically upwards will also be taken to be positive as shown in the diagram.
Resolving horizontally we have:
4 \cos 30+3 \cos 10-5 \sin 20=4.708 \mathrm{~N}Resolving vertically we have:
4 \sin 30-3 \sin 10-5 \cos 20=-3.219 \mathrm{~N}The resultant force can then be found by Pythagoras i.e. F=\sqrt{4.708^2+3.219^2}=5.703 \mathrm{~N}
Having found the resultant force we can also find its direction and it is always best to draw a diagram of the horizontal and vertical components as shown below:
To find the direction we simply perform the following calculation:
\alpha=\tan ^{-1}\left(\frac{3.219}{4.708}\right)=34.36^{\circ}So we can say that the direction of the resultant force is 34.36below the horizontal.
For many students, resolving forces can be challenging but you need to remember that the skills required involve the use of trigonometry which you learnt at GCSE. If you think you need additional support in this area then our online maths tutors for A Level are available to give clear and detailed explanations and to help you prepare for your summer examinations which exam board you may be doing.
resolving Forces and vector notation
Forces can also written in the form ai+bj where i represents the horizontal and j represents the vertical.
Example
Write in the form ai+bj the force 4N which is acting at 30° above the horizontal.
Solution
This question is basically saying to resolve the force horizontally and vertically.
Horizontally: 4 \cos 30^{\circ}=3.46 \mathrm{~N}
Vertically: 4 \sin 30^{\circ}=2 \mathrm{~N}
We can now write the force as 3.46 i+2 j
Example
Given the vectors F_1=2 i+3 j \mathrm{~N}, F_2=i-2 j \mathrm{~N} \text { and } F_3=2 i-6 j \mathrm{~N}. What is the resultant of the three forces and its magnitude?
Solution
Remember that a resultant force is a single force that can replace two or more forces. So in order to find the resultant force here we need to find the sum of the three forces. Let F be the resultant force then:
\begin{gathered} F-F_1+F_2+F_3 \\ =2 i+3 j+i-2 j+2 i-6 j \\ =5 i-5 j \end{gathered}We can then determine the magnitude using Pythagoras: |F|=\sqrt{5^2+5^2}=\sqrt{50} N
Equilibrium
Equilibrium was mentioned briefly earlier. But if a set if the resultant of a set of forces acting on a particle is zero then it is said to be in equilibrium. If a particle is in a state of equilibrium then it is either in a state of rest or moving with a constant speed.
Example
The system of forces shown is in equilibrium. Find the force P and the angle \theta .
Solution
Before we start to resolve, we need to determine a positive direction. As mentioned before, everything to the right and everything vertically upwards will be taken to be positive.
Horizontally: 6 \cos 30+P \cos \theta-9=0 \rightarrow P \cos \theta=3.8038
Vertically: 6 \sin 30-P \sin \theta=0 \rightarrow P \sin \theta=3
Dividing we have:
\frac{P \sin \theta}{P \cos \theta}=\frac{3}{3.8038} \rightarrow \tan \theta=0.78865 \therefore \theta=38.26^{\circ}And P=\frac{3}{\sin 38.26}=4.84 \mathrm{~N}
Depending on the exam board that you are going to do you will meet horizontal forces in year 12 and it is more than likely that you will deal with resolving forces in year 13 of your A Level Maths Course. You will have seen that the only maths that has been done is SOH CAH TOA which you learnt when doing GCSE maths, so this should not be difficult to grasp.
There are a number of terms that have been introduced such as equilibrium and as mentioned there will be other types of questions that will involve forces such as questions involving friction.
Questions involving mechanics can be quite big to read in terms of the actual word count, so always make sure that you have plenty of space to work and draw a diagram which is always going to assist you.
Students looking for additional support attend our A Level Maths Revision Course which takes place during the half term breaks and are classroom based rather than online simply because the interaction between students and the tutor is always better.
We will see in another article how to deal with forces on a slope.