1. FRICTION

2. When two rough surfaces are in contact, friction acts to oppose motion. Friction cannot exceed a maximum value called limiting friction and, until it reaches this value, it is just sufficient to prevent motion. When sliding occurs, or is on the point of occurring, friction takes its limiting value. The limiting value is the coefficient of friction, μ, multiplied by the normal reaction. i.e. Generally, F ≤ μR In limiting equilibrium, or in motion: F = μR *

3. Example 1: A parcel of mass 4 kg is at rest on a rough plane inclined 30 ° to the horizontal. Find the normal reaction and the friction. Given that the parcel is on the point of sliding down the plane find the coefficient of friction. 30 ° R 4g4g F Resolving R – 4g cos 30 = 0 R = 4g 2 Resolving 4g sin 30 – F = 0 Now, as particle is in limiting equilibrium, F = μR 2g = μ (2 g)μ = 1 F = 2g R = 2 g

4. Example 2:A particle of mass 2 kg is given a velocity 8 ms -1 on a rough horizontal surface. It comes to rest after moving a distance of 10 metres. Find the coefficient of friction. 2g2g R F a We can find the acceleration: v = u = a = s = t = 8 0 10 Using v 2 = u 2 + 2as 0 = 8 2 + 2a(10) a = Now apply Newton’s second law: ‘F = ma’ Now, as particle is in motion, F = μR Resolving R = 2g F = μ(2g) –μ(2g) = 2(–3.2) μ = 3.2 9.8 –3.2 = 0.327

5. Example 2:A particle of mass m is given a velocity u ms -1 on a rough horizontal surface. It comes to rest after moving a distance of d metres. Find the coefficient of friction. mg R F a We can find the acceleration: v = u = a = s = t = u 0 d Using ‘v 2 = u 2 + 2as’ 0 = u 2 + 2ad a = u 2 2d – Now apply Newton’s second law: ‘F = ma’ Now, as particle is in motion, F = μR Resolving R = mg F = μmg –μmg = m u 2 2d – ( ) μ = u 2 2gd A harder version of the previous example:

6. Example 3: A particle of mass 5 kg slides down a rough plane inclined 3434 at tan –1. The coefficient of friction is 1212. Find the acceleration of the particle down the slope. α°α° 5g5g R F a α°α° Resolving 3 4 5 α R = 5g cos α R = 5g 4545 R = 4g Now, as particle is in motion, F = μR F = 2g Now apply Newton’s second law: ‘F = ma’ 5g sin α – F = 5a 5g5g 3535 – 2g = 5a 3g – 2g = 5a a = g5g5 For the slope:

7. Summary of key points: This PowerPoint produced by R. Collins; © ZigZag Education 2008 When two rough surfaces are in contact, friction acts to oppose motion. When sliding occurs, or is on the point of occurring, friction takes its limiting value. In limiting equilibrium, or in motion: F = μR *