1. Friction
Dynamics

2. Work out the vertical and horizontal components of the Force
Friction: DYNAMICS
KUS objectives
BAT solve DYNAMIC problems with moving objects
using Friction, F=ma and the suvat model
Starter: A ball of mass 0.4 kg is thrown at an angle of 26° with a force of 46 N
Work out the vertical and horizontal components of the Force
Work out the vector of the initial acceleration of the ball
Work out the magnitude of the initial acceleration of the ball

3. Limiting Equilibrium:
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
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

4. WB 1a Finding the coefficient of friction
The same 8 kg sledge is put on two different surfaces and pulled by a 100 N force at angle of 200. Given the values of Friction, find the coefficient of friction and the acceleration of the sledge for each surface
a) Icy surface, friction is 5 N
𝑖) 𝑅=78.4−100 sin 20
= 𝑁
200
Ice
100 cos sin 20
W = 8 ͯ 9.8 = 78.4 R Fr 5
100
𝑖𝑖) 𝐹𝑟=μ𝑅
μ= Fr R = =0.113
What do we mean by a direct proportion/relationship ? – as reaction force increases so does friction
𝑖𝑖𝑖) 𝐹=𝑚𝑎
100 cos 20 −5=8𝑎
𝑎=11.1 𝑚𝑠 −2

5. WB 1b. Finding the coefficient of friction
b) Wet grassy surface, friction is 15 N
Grass
200
100 cos sin 20
W = 8 ͯ 9.8 = 78.4 R Fr 15
100
𝑖) 𝑅=78.4−100 sin 20
= 𝑁
𝑖𝑖) 𝐹𝑟=μ𝑅
μ= Fr R = =0.339
What do we mean by a direct proportion/relationship ? – as reaction force increases so does friction
𝑖𝑖𝑖) 𝐹=𝑚𝑎
100 cos 20 −15=8𝑎
𝑎=9.87 𝑚𝑠 −2

6. WB 1c. Finding the coefficient of friction
c) Dry surface, friction is 25 N
Tarmac
200
100 cos sin 20
W = 8 ͯ 9.8 = 78.4 R Fr 15
100
𝑖) 𝑅=78.4−100 sin 20
= 𝑁
𝑖𝑖) 𝐹𝑟=μ𝑅
μ= Fr R = =0.566
What do we mean by a direct proportion/relationship ? – as reaction force increases so does friction
𝑖𝑖𝑖) 𝐹=𝑚𝑎
100 cos 20 −25=8𝑎
𝑎=8.62 𝑚𝑠 −2

7. WB 2a Finding the acceleration
A wooden crate of mass 25 kg is pulled in a straight line along a horizontal floor.
The Tension in the rope pulling the crate is 35 N and the rope is at an angle of ∝ where tan ∝ = The coefficient of friction between crate and floor is 0.1
a) Find the acceleration of the crate
𝑖) 𝑅=245−35 sin ∝
= 𝑁
∝ 0
35 cos ∝ 35 sin ∝
W = 25g = 245 R Fr 35
𝑖𝑖) 𝐹𝑟=μ𝑅
𝐹𝑟=0.1×224=22.4
What do we mean by a direct proportion/relationship ? – as reaction force increases so does friction
𝑖𝑖𝑖) 𝐹=𝑚𝑎
35 cos ∝ −22.4=25𝑎
sin 𝜃 = 3 5
cos 𝜃 = 4 5 3 4 5 𝜃
𝑎= 𝑚𝑠 −2

8. WB 2b Find the acceleration
The crate is now pushed along the floor by an equal force of 35 N and at the same angle ∝ .
b) Explain why the acceleration of the crate is less than in a)
∝ 0
35 cos ∝ −35 sin ∝
W = 25g = 245 R Fr 35
The vertical component of the force is now downwards
The Reaction force goes up
So Friction is greater
So the acceleration will be less since the forward horizontal force is the same as in a)
Actual working
𝑅= sin ∝ =266 𝑁
𝐹𝑟=0.1×266=26.6 𝑁
Resultant force 35 cos ∝ −26.6=25𝑎
Acceleration 𝑚𝑠 −2
What do we mean by a direct proportion/relationship ? – as reaction force increases so does friction

9. WB 3a moving down a slope, find acceleration
A parcel of mass 3 kg is sliding down a rough slope of inclination 300
The coefficient of friction between the parcel and the slope is 0.35.
Find the acceleration of the particle
3𝑔 sin 30 3𝑔 cos 30 R
30° Fr
𝑅=3𝑔 c𝑜𝑠 3 0
Frmax =𝜇𝑅=0.35 (3𝑔 𝑐𝑜𝑠 30)
Resultant Force 𝐹=𝑚𝑎
3𝑔 𝑠𝑖𝑛 30 − 𝑔 𝑐𝑜𝑠 30 =3𝑎
𝑎= 3𝑔 𝑠𝑖𝑛 30 − 𝑔 𝑐𝑜𝑠 =1.93 𝑚𝑠 −2

10. R Fr 𝑅=2𝑔 c𝑜𝑠 3 0 Frmax =𝜇𝑅=𝜇(2𝑔 𝑐𝑜𝑠 30) 2𝑔 sin 30 2𝑔 cos 30
WB 4 Parcel moving down a slope, find 𝜇
A particle of mass 2 kg is sliding down a rough slope that is inclined at 30 to the horizontal. Given that the acceleration of the particle is 1 𝑚𝑠 −2 , find the coefficient of friction 𝜇 between the particle and the slope
2𝑔 sin 30 2𝑔 cos 30 R
30° Fr
𝑅=2𝑔 c𝑜𝑠 3 0
Frmax =𝜇𝑅=𝜇(2𝑔 𝑐𝑜𝑠 30)
Resultant Force 𝐹=𝑚𝑎
2𝑔 𝑠𝑖𝑛 30 −𝜇 2𝑔 𝑐𝑜𝑠 30 =2×1
𝜇= 2𝑔 𝑠𝑖𝑛 30 −2 2𝑔𝑐𝑜𝑠 30 =0.460

11. WB5 two-step problem, change of angle
A 3 kg particle rests in limiting equilibrium on a plane inclined at 300 to the horizontal. Determine the acceleration with which the particle will slide down the plane when the angle of inclination is increased to 400
Step 1 Equilibrium
3𝑔 sin 30 3𝑔 cos 30 R
30° Fr
𝑅=3𝑔 cos 30 =25.46
𝐹𝑟=3𝑔 sin 30 =14.7
𝜇= =0.577

12. WB5 Parcel two-step problem, change of angle (cont)
A 3 kg particle rests in limiting equilibrium on a plane inclined at 300 to the horizontal. Determine the acceleration with which the particle will slide down the plane when the angle of inclination is increased to 400
3𝑔 sin 40 3𝑔 cos 40 R
40° Fr
Step 2 Bigger slope
𝑅=3𝑔 cos 40 =22.52
Coefficient of Friction is the same
Friction =0.577×22.52=12.99
Resultant Force 𝐹=𝑚𝑎
3𝑔 sin 40 −12.99=3𝑎
𝑎=1.97 𝑚 𝑠 −2

13. R 25 Fr 𝑅=2𝑔 c𝑜𝑠 10 +25 sin 10 𝑅=23.64 Frmax =0.3 𝑅 =7.093
WB 6 moving up a slope, find acceleration
A box of mass 2 kg is pushed up a rough slope by a horizontal force of 25 N. the slope is inclined at 10 to the horizontal. Given that 𝜇=0.3 find the acceleration of the box
2𝑔 sin 10 2𝑔 cos 10 R
30° Fr 25
25 cos sin 10
𝑅=2𝑔 c𝑜𝑠 sin 10
𝑅=23.64
Frmax =0.3 𝑅 =7.093
Resultant Force 𝐹=𝑚𝑎
25 𝑐𝑜𝑠 10 −2𝑔𝑠𝑖𝑛 10 − =2𝑎
𝑎= 𝑚𝑠 −2

14. WB 7 down slope, find coefficient friction, suvat
A parcel of mass 5 Kg is released from rest on a rough ramp of inclination
θ = arcsin 3/5 and slides down the ramp. After 3 secs it has a speed of 4.9 ms-1
Treating the parcel as a particle, find the coefficient of friction between the parcel and the ramp
5𝑔 sin 𝜃 5𝑔 cos 𝜃 R
arcsin 3 5 Fr
𝑅=5𝑔 cos 𝜃 =39.2
Friction =𝜇𝑅=39.2 𝜇
use suvat to find the acceleration 𝑣=𝑢+𝑎𝑡
𝑎= =1.63 𝑚 𝑠 −2 =
Now use F=ma
5𝑔 𝑠𝑖𝑛 𝜃−39.2𝜇=5×1.63
sin 𝜃 = 3 5
cos 𝜃 = 4 5 3 4 5 𝜃
𝜇= 5𝑔 𝑠𝑖𝑛 𝜃−5× =0.542

15. WB 8 Parcel find acceleration and distance, suvat
A particle is held at rest on a rough plane inclined at an angle of inclination
θ ∝ where tan ∝ = The coefficient of friction between the parcel and the plane is The particle is released an slides down the plane
a) Find the acceleration of the particle
𝑅=𝑚𝑔 cos ∝ = 4 5 𝑚𝑔
𝑚𝑔 sin 𝜃 𝑚𝑔 cos 𝜃 R ∝ Fr
Friction =𝜇𝑅= 2 5 𝑚𝑔
use F=ma to find the acceleration
𝑚𝑔 sin ∝ − 2 5 𝑚𝑔=𝑚𝑎
3 5 𝑔 − 2 5 𝑔=𝑎
sin 𝜃 = 3 5
cos 𝜃 = 4 5 3 4 5 𝜃
𝑎= 1 5 𝑔= = 𝑚𝑠 −2

16. WB 8 Parcel find acceleration and distance, suvat (cont)
A particle is held at rest on a rough plane inclined at an angle of inclination
θ ∝ where tan ∝ = The coefficient of friction between the parcel and the plane is The particle is released an slides down the plane
b) Find the distance it slides in the first 2 seconds 𝑠=
𝑢=0 𝑣=
𝑎= 1 5 𝑔
𝑡=2
Down slope 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2
𝑠= × 1 5 𝑔×4 = 𝑚

17. WB 9 Parcel: finding the angle of inclination
A parcel of mass 3 kg is sliding down a rough inclined plane with an acceleration of 4 ms-2 . Find the angle of inclination of the plane if the coefficient of friction between the parcel and plane is 0.6
3𝑔 sin 𝜃 3𝑔 cos 𝜃 R 𝜃° Fr
𝑅=3𝑔 cos 𝜃
𝐹𝑟=𝜇𝑅=0.6 3𝑔 cos 𝜃
Resultant force 𝐹=𝑚𝑎
3𝑔 sin 𝜃 −0.6 3𝑔 cos 𝜃 =3×4
29.4 sin 𝜃 −17.64 cos 𝜃 =12
This is solvable using the addition formula 17

18. WB 9 Parcel: finding the angle of inclination (solution cont)
A parcel of mass 3 kg is sliding down a rough inclined plane with an acceleration of 4 ms-2 . Find the angle of inclination of the plane if the coefficient of friction between the parcel and plane is 0.6
29.4 sin 𝜃 − cos 𝜃 =12
addition formula
𝑅 𝑠𝑖𝑛 𝜃−𝛼 =𝑅 sin 𝜃 cos ∝ −𝑅 sin ∝ cos 𝜃 =12
∝= arctan =31.0°
𝑅 sin ∝ =17.64
𝑅 cos ∝ = 29.4
𝑅= =34.29
So sin 𝜃 − cos 𝜃 =
34.29 𝑠𝑖𝑛 𝜃−31.0
=12
𝑠𝑖𝑛 𝜃− =
𝜃−31.0 =20.5
angle of inclination 𝜃=51.5° 18

19. One thing to improve is –
KUS objectives
BAT solve DYNAMIC problems with moving objects
using Friction, F=ma and the suvat model
self-assess
One thing learned is –
One thing to improve is –

20. END

21. WB 7 find coefficient friction solution
5𝑔 sin 𝜃 5𝑔 cos 𝜃 = R
arcsin 3 5 Fr
A parcel of mass 5 Kg is released from rest on a rough ramp of inclination
θ = arcsin 3/5 and slides down the ramp.
After 3 secs it has a speed of 4.9 ms-1
Treating the parcel as a particle, find the coefficient of friction between the parcel and the ramp
𝑠= 𝑢=0 𝑣=4.9 𝑎= 𝑡=3
𝑣=𝑢+𝑎𝑡
Suvat to find a
4.9=0+𝑎×3
𝑎=1.63 𝑚 𝑠 −2
resultant force in the i direction
𝐹=𝑚𝑎
Equilibrium in the j direction
𝑅= 𝑁
29.4−𝐹𝑟=5×1.63
𝐹𝑟=21.23
𝐹𝑟=𝜇𝑅
21.23=𝜇×39.2
Coefficient of friction
𝜇=0.542 21