Coefficient of Friction (AQA A-Level Mathematics): Revision Notes

3.3.2 Coefficient of Friction

Concept:

• In A-Level Maths, friction is modelled as being proportional to the normal contact force between objects.

Mathematical Representation:

Equation:

Diagram:

• The frictional force opposes the driving force, and mg is the gravitational force acting on the object.

Nature of Friction:

• Friction "fights" and resists any driving force up to a limit $F_{\text{max}} = \mu R$.
• When an object is on the point of slipping, it is said to be in "limiting equilibrium." At this point, the frictional force is at its maximum value $F_{\text{max}}$.

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Example 1: Finding the Maximum Force without Causing Motion (Limiting Equilibrium)

Problem Statement:

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Part (a): Force P is Horizontal

3. Given:

• Coefficient of friction $\mu = 0.5$
• Weight $mg = 8g$

4. Equilibrium in the vertical direction:

• The normal reaction force $R = 8g$

5. Frictional force at the point of slipping:

$$F_{\text{max}} = \mu R = 0.5 \times 8g = 4g$$

1. Maximum force $P$:

$$P = F_{\text{max}} = 4g \, \text{N}$$

Answer (a): $P = 4g \, \text{N}$

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Part (b): Force P Acts at an Angle of $60^\circ$ above the Horizontal

6. Resolve the force $P$:

• Horizontal component: $P \cos 60^\circ$
• Vertical component: $P \sin 60^\circ$

7. Vertical equilibrium:

$$R + P \sin 60^\circ = 8g \quad \Rightarrow \quad R = 8g - \frac{\sqrt{3}}{2} P$$

2. Frictional force at the point of slipping:

$$F_{\text{max}} = \mu R = 0.5 \times \left(8g - \frac{\sqrt{3}}{2} P\right)$$

3. Horizontal equilibrium:

$$P \cos 60^\circ = F_{\text{max}} = 0.5 \times \left(8g - \frac{\sqrt{3}}{2} P\right)$$

Simplifying:

$$\frac{1}{2} P = 4g - \frac{\sqrt{3}}{4} P$$ $$\frac{1}{2} P + \frac{\sqrt{3}}{4} P = 4g$$ $$P\left(\frac{1}{2} + \frac{\sqrt{3}}{4}\right) = 4g$$

4. Solve for $P$:

$$P = \frac{4g}{\frac{1}{2} + \frac{\sqrt{3}}{4}} \approx 42.01 \, \text{N}$$

Answer (b): $P \approx 42.01 \, \text{N}$

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