1. Friction
Equilibrium

2. Friction: Equilibrium
KUS objectives
BAT Understand the directly proportional relationship between Friction force and Reaction force
BAT Solve equilibrium problems using coefficient of friction
Starter: Write a definition for each of these forces
Weight
Tension
Reaction (or Normal contact force)

3. What is Friction? Friction is best given by Fr or Frmax
‘The resistance that one surface or object encounters when moving over another’.
Friction is best given by Fr or Frmax
Do not confuse Fr with F for Resultant Force

4. Limiting Equilibrium:
This is when the magnitude of the frictional force is just sufficient to prevent relative motion
Friction is a Force on objects caused by a touching surfaces resistance to a direction . ‘Rougher’ surfaces cause greater friction
The coefficient of friction is a measure of this ‘roughness’ of different surfaces. There is a direct relationship between friction and the reaction force when an object is touching a surface
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛≤𝜇𝑅
when a particle is in equilibrium and static
𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛=𝜇𝑅
when a particle is moving
Remember:
An object moving at a constant velocity will still be in equilibrium
An accelerating object is subject to F=ma

5. 𝑎) 𝑅=50𝑔−150 sin 25 =426.6 𝑁 R 150 Fr 250 𝑏) 𝐹𝑟=150 cos 25 𝐹𝑟=135.9 𝑁
WB 1 max friction: Finding coefficient friction
A Trolley of mass 50 kg is being dragged along the ground at a constant velocity by a force of 150 N at an angle of elevation of 25. Find the coefficient of friction between the ground and the trolley
a) What is the normal Reaction force?
b) What is the Friction force?
𝑎) 𝑅=50𝑔−150 sin 25
= 𝑁
250
150
150 cos sin 25
W = 50 ͯ 9.8 = 490 R Fr
𝑏) 𝐹𝑟=150 cos 25
𝐹𝑟= 𝑁
𝐹𝑟=μ𝑅
μ= 𝐹𝑟 𝑅 = =0.319

6. 𝑎) 𝑅=50𝑔−150 sin 25 =426.6 𝑁 R 150 𝑏) 𝐹𝑟=150 cos 25 Fr 250 =135.9 𝑁
WB 2 max friction: Finding the coefficient of friction
A Trolley of mass 50 kg is being dragged along the ground at a constant velocity by a force of 150 N at an angle of elevation of 250
a) What is the Reaction Force? b) What is the Friction force?
c) Find the coefficient of friction if friction is at its maximum
𝑎) 𝑅=50𝑔−150 sin 25
= 𝑁
250
150
150 cos sin 25
W = 50 ͯ 9.8 = 490 R Fr
𝑏) 𝐹𝑟=150 cos 25
= 𝑁
𝑐) 𝐹𝑟=μ𝑅
μ= Fr R = =0.319

7. Place the 1 kg weight on the board
Measuring the coefficient of Friction
Aim: To find values for the coefficient of friction of different surfaces
Equipment: Board calculator 1 kg weight Carpet, smooth board, sandpaper, Big protractor
OUTLINE METHOD:
Place the 1 kg weight on the board
Raise the board to the point where the weight
is about to move (point of slipping)
Measure the angle of the slope using the
Protractor
Repeat with different surfaces
Find the coefficient of friction for each surface
Comment on the validity of your results

8. WB3. Object on a slope about to go up the slope
A Block with mass 3 kg is pushed by a force parallel to the direction of the slope .The block is at a tipping point where it is about to go up the slope.
The slope surface has coefficient of friction of 0.3 and is at 280 to the horizontal. Find the values of friction and the Push force
Push
Reaction i j
i, j notation
Friction
Why is the brick modelled as a particle?
280
Weight

9. WB3 Object on a slope about to go up the slope solution
Push
280
Reaction
29.4 sin cos 28
Friction
i) R = 29.4 cos 28 = 25.96
ii) Friction =R
= 0.3 x = 7.79 N
Push =friction + component of weight downslope
Push = sin 28 = N

10. WB4 Object on a slope about to go down the slope
A Block with mass 8 kg is at the point of slipping down a rough slope. The slope is at 280 to the horizontal. Find the value of the coefficient of friction of the slope
Friction will be up the slope
Friction
Reaction
280
Weight

11. WB4 Object on a slope about to go down the slope solution
Friction
280
Reaction
78.4 sin cos 28
Weight = 8 x 9.8=78.4
i) R = 78.4 cos 28 = 69.2
ii) Equilibrium
Friction= 78.4 sin 28 = 36.8 N
iii) Friction =R
 = =0.532

12. WB5. Simultaneous equations question
A Block with mass 3 kg is held at rest on a rough slope by a horizontal force (Push). The slope has coefficient of friction of 1/3 and the slope is at 300 to the horizontal. Find the value of the Push force
Reaction
Push
Friction
Why is the brick modelled as a particle?
300
Weight

13. Reaction P cos 30 −P sin 30 Fr 300 −29.4 sin 30 −29.4 cos 30 300 300 j
WB5 (cont). Simultaneous equations question solution part 1
Reaction
P cos 30 −P sin 30
300 Fr
300 i j
i, j notation
300
Weight = 3 x 9.8=29.4
Explain why moved Push force in force diagram – avoids confusion
−29.4 sin 30 −29.4 cos 30

14. (and R = 42.06 N) P = 33.2 N R = 29.4cos30 + Psin30 (2)
WB5 (cont). Simultaneous equations question solution part 2 Fr
300
Reaction
29.4 sin cos 30
P cos 30 P sin 30
R = 29.4cos30 + Psin (2)
Friction = R = R (3)
29.4sin30 + Fr – Pcos30 = 0 (1)
So sin R – Pcos30 = 0 (1)
solve simultaneously gives
So sin cos30 + Psin30 – Pcos30 = 0
(and R = N)
P = 33.2 N

15. WB6. horizontal push, object is about to go down the slope
A Block with Weight 10 Newton’s is held at rest on a rough slope by a horizontal force (Push). The slope has coefficient of friction of 0.3 and the slope is at 300 to the horizontal. Find the value of the Push, Reaction and friction forces
Friction
Reaction i j
i, j notation
Push
300
Weight

16. Fr R P cos 30 −P sin 30 300 −10 sin 30 −10 cos 30
WB6. rough slope in equilibrium, object is about to go down the slope solution
300 R Fr
P cos 30 −P sin 30
−10 sin 30 −10 cos 30

17.  0.3(8.66 + ½ 𝑃) + 0.866 𝑃 = 5  R = 10cos30 + Psin30 (1)
WB6. rough slope in equilibrium, object is about to go down the slope solution
300 R Fr
P cos 30 −P sin 30
−10 sin 30 −10 cos 30
 R = 10cos30 + Psin (1)
 Fr + Pcos30 = 10sin (2)
Fr = R = 0.3 R (3)
Solve simultaneously
 cos30 + Psin30 + Pcos30 = 10sin (2)
 0.3( ½ 𝑃) 𝑃 = 5
Helps to draw a new diagram once all I and j components worked out
Q – how many components in the i(or j) direction in this problem?
𝑃= 𝑅= 𝐹𝑟=2.95

18. One thing to improve is –
KUS objectives
BAT Understand the directly proportional relationship between Friction force and Reaction force
BAT Solve equilibrium problems with the coefficient of friction
self-assess
One thing learned is –
One thing to improve is –

19. END