Friction

Friction

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

Friction

In the Forces section, we mentioned that friction was calculated by the equation F \leq \textcolor{red}{\mu} R, where \textcolor{red}{\mu} is known as the coefficient of friction, which has no units. As a general rule of thumb, the rougher a surface is, the higher its coefficient of friction.

We also have the term limiting friction, when friction is at its maximum. This is shown when F = \textcolor{red}{\mu} R.

Make sure you are happy with the following topics before continuing.

Fundamentals

Example: Here’s a box on a flat surface with two people pushing and pulling in the same direction.

Let’s now say that the weight of the box, W, is 100\text{ N}, and that the friction is limiting (i.e. applying any more force in T or Thrust will cause the box to accelerate). Say, also, that the friction F is 50\text{ N}.

Then we have the normal reaction force R = 100\text{ N}.

Since F = \textcolor{red}{\mu} R, the coefficient of friction, \textcolor{red}{\mu} = \dfrac{50}{100} = 0.5.

A LevelAQAEdexcelOCR

Note:

By definition, the coefficient of friction \mu is such that 0 \leq \textcolor{red}{\mu} \leq 1.

Friction on a Tilted Plane

Example: Here’s a 1500 \text{ kg} car being towed up an incline at an angle of 30\degree. Say the slope has a coefficient of friction \textcolor{red}{\mu} = 0.4. What would be the required combined force of the tension in the rope, T, and the thrust from the car? Assume the car is travelling at constant velocity up the hill.

First, adapt this system by rotating it, to visualise the perpendicular forces a little more easily.

Perpendicular to F, we have R = W\cos 30°.

W = 1500 \times 9.8 = 14700\text{ N}, so R = 12730.57\text{ N}

 

So, parallel to F, we have T + \text{Thrust} = F + W\sin 30°.

By F_{max} = \textcolor{red}{\mu} R, the limit of friction is F_{max} = 5092.23\text{ N}

Since we have W and F_{max}, we can see that the minimum thrust and tension required to move the car up the slope is 5092.23 + W\sin 30° = 12442.23\text{ N}.

A LevelAQAEdexcelOCR

Example Questions

We have F \leq \mu R.

Since no other vertical forces are applied, we have W = R.

We also have W = mg.

Therefore, we have the expression

F \leq \mu mg

Resolving perpendicular to F:

R = W\cos 20° = mg\cos 20°

Resolving parallel to F:

Resultant Force = F_{res} = ma = 0.2m

We also have

F_{res} = W\sin 20° - \mu R

= mg\sin 20° - \mu mg\cos 20°

So,

g\sin 20° - \mu g\cos 20° = 0.2

Giving

\mu = \dfrac{9.8\sin 20° - 0.2}{9.8\cos 20°} = 0.342\text{ (to }3\text{ dp)}

For v = u + at, we have a = \dfrac{-80}{20} = -4\text{ ms}^{-2} horizontally.

From Newton’s Second Law, F = ma = 0.01 \times -(-4) = 0.04\text{ N}.

Vertically, we have R = 0.01 \times 9.8 = 0.098\text{ N}

With no other horizontal forces acting on the system, we assume the friction is limiting, so

\mu = \dfrac{0.04}{0.098} = 0.408\text{ (to }3\text{ dp)}

Related Topics

MME

SUVAT Equations

A Level
MME

Forces

A Level
MME

Resolving Forces

A Level
MME

Newton’s Laws

A Level

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

You May Also Like...

A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99
View Product

A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99
View Product

Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99
View Product