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

Variable definitions

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 $$F \le \mu R$$, where $$\mu$$ is the coefficient of friction and has a value between $$0$$ and $$1$$. $$\mu$$ is dependent on the amount of friction between the surfaces of contact between the body and the plane and has a value closer to $$1$$ for rougher surfaces.​

Procedure

Example 1

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

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. 

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

$$F = 10 \cos(45)$$​​​

Resolving vertically gives:

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

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Substitute into $$F = \mu R$$ to give:

​$$\begin{aligned}F &= \mu R \\10 \cos(45) &= \mu \times (10g-10\sin(45)) \\\mu &= \underline{0.0778 \ (3 \ s.f.)}\end{aligned}$$​​

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Example 2

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

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: 

​​​

Resolving perpendicular to the plane gives:

$$R = 0.4g \cos(25)$$​​

Resolving parallel to the plane gives:

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

Substitute into $$F= \mu R$$ to give:

$$\begin{aligned}F &= \mu R \\3-0.4g \sin(25) &= \mu \times 0.4g\cos(25) \\\\\mu &= \dfrac {3-0.4g \sin(25)}{0.4g\cos(25)} \\\\\mu &= \underline{0.378 \ (3 \ s.f.)}\end{aligned}$$​​