System Of Forces - A Level Maths
System Of Forces – Introduction
After reading this article you will be able to:
- To be able to resolve a system of forces involving more than three forces and on a slope
- To be able to understand friction
- To be able to calculate friction on an inclined plane
System of Forces - More than three forces
When dealing with a system of forces quite often there can be any number of forces present and you may sometimes be asked to calculate a missing force and find a missing angle when a system is in a state of equilibrium.
The process of resolving forces horizontally and vertically is still however used.
Consider the following example:
The above system is in equilibrium. Find any unknown forces and missing angles.
Solution
Taking directions to the right and vertically upwards to be positive.
Resolving horizontally:
\begin{gathered} 6 \cos 80+P \sin \theta-4 \cos 70=0 \\ P \sin \theta=0.326 \end{gathered}Resolving vertically:
\begin{gathered} 6 \sin 80+4 \sin 70-5-P \cos \theta=0 \\ P \cos \theta=4.668 \end{gathered}Dividing we have:
\begin{gathered} \frac{P \sin \theta}{P \cos \theta}=\frac{0.326}{4.668} \\ \tan \theta=\frac{0.326}{4.668} \\ \theta=4.0^{\circ} \\ \therefore P=4.68 \mathrm{~N} \end{gathered}System of Forces - Resolving forces on a plane
When it comes to dealing with problems that are related to a plane, simple trigonometry is used to allow us to resolve the components of the weight of the mass parallel and perpendicular to the plane.
Sometimes ‘plane’ and ‘slope’ are used to mean the same thing.
Example
A mass of 10kg is on a plane inclined at 20° to the horizontal. Find the components of the mass parallel and perpendicular to the plane.
Solution
First we will need to draw a diagram and we will need to determine the weight of the object by multiplying by 9.8 \mathrm{~ms}^{-2}
When it comes to resolving a force, probably one of the easiest methods is to remember that it is the “force x cosine of the angle” i.e. F ×cosθ. This is provided an angle is given otherwise it is sine. An alternative is to give yourself the angle and to use cosine.
To quickly demonstrate this, consider the following:
To resolve horizontally, the 10N need to move through an angle of 30° so we can say it is 10cos30°. The same result could have been found using trigonometry.
To resolve vertically i.e. so the 10N is on the y-axis. The same result could have been found using trigonometry and there are two options.
Either: 10sin30° or 10cos60°
We can have two results because cos 90- α = sinα
Going back to the question involving the slope we can determine further angles.
System of Forces – Friction
Friction is a force that opposes the direction of motion of the body and will therefore act in a direction that is opposite to its direction of motion.
If the above block is to be placed on a rough surface then the block will actually move if P is greater than the frictional force, F.
Frictional force is determined by: F_{\text {max }}=\mu R where is the coefficient of friction. The frictional force is only as large as necessary to prevent motion.
The frictional force, F, for a block and a surface is not constant but will actually increase as the applied force P increases until the frictional force reaches a value after which it cannot increase. At this stage the block is on the point of moving and we have what is known as a state of limiting equilibrium.
Example
Find the coefficient of friction from the following diagram:
Solution
Vertically: R=10g
Horizontally: F=10N
Using F=\mu R: 10=\mu 98 \therefore \mu=\frac{10}{98}=\frac{5}{49}
Example
What is the maximum frictional force which can act when a block of mass 2kg rests on a rough horizontal surface when the μ=0.7
Solution
Vertically: R = 2g
Horizontally: F = 0.7 x 2g = 13.72N
System of Forces – Friction on a plane
Example
A mass of 6kg rests in limiting equilibrium on a rough place inclined at 30° to the horizontal. What is the coefficient of friction between the mass and the plane?
Solution
Because the mass is in limiting equilibrium this means that it is about to slide down the slope.
The components of the weight are shown parallel and perpendicular to the slope.
Perpendicular to the plane: R=6gcos30
Parallel to the plane: F=6gcos60= μR
\therefore \mu=\frac{6 g \cos 60}{6 g \cos 30}=\frac{1}{\sqrt{3}}To help you get some additional question practice with system of forces questions here are some questions for you to try. The answers are at the very bottom. Goodluck!
System of Forces – Multiple Choice Questions
Find the unknown forces.
a) F1=7.01N, F2=17.7N
b) F1=4.01N, F2=7.7N
c) F1=4.01N, F2=17.7N
d) F1=8.01N, F2=27.7N
Find the unknown force and the missing angle.
Find the unknown forces.
Find the unknown forces.
Questions 5 – 6 relate to the following.
A mass of 50kg is supported in equilibrium as shown in the diagram below.
Questions 5 – 6 relate to the following.
Find T_1
\begin{aligned} & 566 \mathrm{~N} \\ & 733 \mathrm{~N} \\ & 82 \mathrm{~N} \\ & 555 \mathrm{~N} \end{aligned}Find T_2
\begin{aligned} & 721 N \\ & 135 N \\ & 283 N \\ & 358 N \end{aligned}A force acts parallel to and up a line of greatest slops which holds a particle of mass 10kg in equilibrium on a smooth which is inclined at 30° to the horizontal. What is the magnitude of this force and of the normal reaction between the particle and the plane?
\begin{aligned} & 49 \mathrm{~N}, 84.87 \mathrm{~N} \\ & 19 \mathrm{~N}, 14.87 \mathrm{~N} \\ & 89 \mathrm{~N}, 88.87 \mathrm{~N} \\ & 41 \mathrm{~N}, 81.87 \mathrm{~N} \end{aligned}A horizontal force P holds a body of mass 10kg in equilibrium on a smooth plane which is inclined at 30° to the horizontal. What is the magnitude of this force and of the normal reaction between the particle and the plane?
\begin{aligned} & 16.58 \mathrm{~N}, 113.16 \mathrm{~N} \\ & 56.58 \mathrm{~N}, 113.16 \mathrm{~N} \\ & 56.58 \mathrm{~N}, 11.16 \mathrm{~N} \\ & 56.58 \mathrm{~N}, 13.16 \mathrm{~N} \end{aligned}Questions 9 – 10 both have a coefficient of friction of \frac{1}{3}. Find the force X if the body is on the point of moving up the plane.
Answers
1 – C, 2 – D, 3 – A, 4 – B, 5 – A, 6 – C, 7 – A, 8 – B, 9 – C, 10 – D