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Friction with static particles

Friction with static particles

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Explainer Video

Tutor: Meera

Summary

Friction with static particles

​​In a nutshell

Usually, you have to take into account the effect of any frictional force when solving static equilibrium problems. It is critical to know whether a body is on the point of moving or not, in order have a complete study of the forces. 


Equations

DESCRIPTION

EQUATION

Maximum value of the frictional force reached whenever the body is on the point of moving.
​Fmax=μRF_{max} = \mu RFmax​=μR​​
General definition of the force of friction.
​F≤μRF \leq \mu RF≤μR​​


Variable definitions

QUANTITY NAME

SYMBOL

UNIT NAME

UNIT

​Maximum value of the frictional force\textit{Maximum value of the frictional force}Maximum value of the frictional force​​
​FmaxF_{max}Fmax​​​
​NewtonsNewtonsNewtons​​
​NNN​​
​Force of friction\textit{Force of friction}Force of friction​​
​FFF​​
​NewtonsNewtonsNewtons​​
​NNN​​
​Coefficient of friction\textit{Coefficient of friction}Coefficient of friction​​
​μ\muμ​​
​no unitsno \ unitsno units​​

​Reaction force\textit{Reaction force}Reaction force​​
​RRR​​
​NewtonsNewtonsNewtons​​
​NNN​​
​Weight\textit{Weight}Weight​​
​WWW​​
​NewtonsNewtonsNewtons​​
​NNN​​



Friction with static particles

Problems can be solved in a similar way to other static equilibrium problems you have already studied. The friction can be calculated by using F≤μRF \le \mu RF≤μR, where μ\muμ is the coefficient of friction and has a value between 000 and 111. μ\muμ is dependent on the amount of friction between the surfaces of contact between the body and the plane and has a value closer to 111 for rougher surfaces.​


Procedure

1.
Draw a force diagram.
2.
Resolve the forces into different components. 
3.
Use ∑F=0\sum F = 0∑F=0 for every direction of movement to form equations.
4.
Solve the equations. 


Example 1

A book of mass m=10 kgm=10 \ kgm=10 kg rests on a rough horizontal table. If a force of magnitude P=10 NP= 10 \ NP=10 N acts on it, at an angle of θ=45∘\theta= 45^{\circ}θ=45∘ to the horizontal in the upwards direction, then the book is in equilibrium. Find out the value of the coefficient of friction μ\muμ.​


First, you should draw a force diagram, in order to visualize the forces involved. Resolve the forces into different components, parallel and perpendicular to the plane. You should add labels to your diagram as follows. 

Maths; Application of forces; KS5 Year 13; Friction with static particles


Set vertically upwards and horizontally to the right each as positive. Resolving horizontally gives:

F=10cos⁡(45)F = 10 \cos(45)F=10cos(45)​​​


Resolving vertically gives:

R+10sin⁡(45)=10gR=10g−10sin⁡(45)\begin{aligned}R+10 \sin(45) &= 10g\\R &= 10g - 10\sin(45)\end{aligned}R+10sin(45)R​=10g=10g−10sin(45)​​​

​

Substitute into F=μRF = \mu RF=μR to give:

​F=μR10cos⁡(45)=μ×(10g−10sin⁡(45))μ=0.0778 (3 s.f.)‾\begin{aligned}F &= \mu R \\10 \cos(45) &= \mu \times (10g-10\sin(45)) \\\mu &= \underline{0.0778 \ (3 \ s.f.)}\end{aligned}F10cos(45)μ​=μR=μ×(10g−10sin(45))=0.0778 (3 s.f.)​​​​

​​

Example 2

A particle of mass m=0.4 kgm=0.4 \ kgm=0.4 kg is on an inclined plane at an angle θ=25°\theta = 25 \degreeθ=25° to the horizontal. The particle is at rest on the plane due to a force P=3 NP=3 \ NP=3 N, upwards and to the right. Friction acts down the plane. Calculate the coefficient of friction μ\muμ.


First, draw a force diagram, in order to visualize the forces involved. Resolve the forces into different components, parallel and perpendicular to the inclined plane. You should add labels to your diagram as follows: 

Maths; Application of forces; KS5 Year 13; Friction with static particles

​​​

Resolving perpendicular to the plane gives:

R=0.4gcos⁡(25)R = 0.4g \cos(25)R=0.4gcos(25)​​


Resolving parallel to the plane gives:

3=F+0.4gsin⁡(25)F=3−0.4gsin⁡(25)\begin{aligned}3 &= F + 0.4g \sin (25) \\F &= 3-0.4g \sin(25)\end{aligned}3F​=F+0.4gsin(25)=3−0.4gsin(25)​​​


Substitute into F=μRF= \mu RF=μR to give:

F=μR3−0.4gsin⁡(25)=μ×0.4gcos⁡(25)μ=3−0.4gsin⁡(25)0.4gcos⁡(25)μ=0.378 (3 s.f.)‾\begin{aligned}F &= \mu R \\3-0.4g \sin(25) &= \mu \times 0.4g\cos(25) \\\\\mu &= \dfrac {3-0.4g \sin(25)}{0.4g\cos(25)} \\\\\mu &= \underline{0.378 \ (3 \ s.f.)}\end{aligned}F3−0.4gsin(25)μμ​=μR=μ×0.4gcos(25)=0.4gcos(25)3−0.4gsin(25)​=0.378 (3 s.f.)​​​​



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Friction and terminal velocity

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Friction and terminal velocity

The coefficient of friction

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The coefficient of friction

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Friction with static particles

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Friction with static particles

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FAQs - Frequently Asked Questions

What is the best way to start solving a problem that involves the force of friction?

The best way to solve this kind of exercises is by drawing a force diagram.

What is the general definition of the force of friction?

The force of friction F is less than or equal to the product of the coefficient of friction by the reaction force.

What is the value of the frictional force when a body is on the point of moving?

The product of the coefficient of friction and the reaction force.

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