Friction Dynamics

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

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

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 =44.20 𝑁 200 Ice 100 cos 20 100 sin 20 W = 8 ͯ 9.8 = 78.4 R Fr 5 100 𝑖𝑖) 𝐹𝑟=μ𝑅 μ= Fr R = 5 44.20 =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

WB 1b. Finding the coefficient of friction b) Wet grassy surface, friction is 15 N Grass 200 100 cos 20 100 sin 20 W = 8 ͯ 9.8 = 78.4 R Fr 15 100 𝑖) 𝑅=78.4−100 sin 20 =44.20 𝑁 𝑖𝑖) 𝐹𝑟=μ𝑅 μ= Fr R = 15 44.208 =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

WB 1c. Finding the coefficient of friction c) Dry surface, friction is 25 N Tarmac 200 100 cos 20 100 sin 20 W = 8 ͯ 9.8 = 78.4 R Fr 15 100 𝑖) 𝑅=78.4−100 sin 20 =44.20 𝑁 𝑖𝑖) 𝐹𝑟=μ𝑅 μ= Fr R = 25 44.208 =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

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 ∝ = 3 4 . The coefficient of friction between crate and floor is 0.1 a) Find the acceleration of the crate 𝑖) 𝑅=245−35 sin ∝ =224 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=0.224 𝑚𝑠 −2

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=245+35 sin ∝ =266 𝑁 𝐹𝑟=0.1×266=26.6 𝑁 Resultant force 35 cos ∝ −26.6=25𝑎 Acceleration 0.056 𝑚𝑠 −2 What do we mean by a direct proportion/relationship ? – as reaction force increases so does friction

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 −0.35 3𝑔 𝑐𝑜𝑠 30 =3𝑎 𝑎= 3𝑔 𝑠𝑖𝑛 30 −0.35 3𝑔 𝑐𝑜𝑠 30 3 =1.93 𝑚𝑠 −2

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

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= 14.7 25.46 =0.577

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

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 10 25 sin 10 𝑅=2𝑔 c𝑜𝑠 10 +25 sin 10 𝑅=23.64 Frmax =0.3 𝑅 =7.093 Resultant Force 𝐹=𝑚𝑎 25 𝑐𝑜𝑠 10 −2𝑔𝑠𝑖𝑛 10 − 7.093 =2𝑎 𝑎=7.06 𝑚𝑠 −2

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= 4.9 3 =1.63 𝑚 𝑠 −2 = 29.4 39.2 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−5×1.63 39.2 =0.542

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 ∝ = 3 4 . The coefficient of friction between the parcel and the plane is 0.5 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= 49 25 =1.96 𝑚𝑠 −2

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 ∝ = 3 4 . The coefficient of friction between the parcel and the plane is 0.5 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=0+ 1 2 × 1 5 𝑔×4 =3.92 𝑚

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

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 −17.64 cos 𝜃 =12 addition formula 𝑅 𝑠𝑖𝑛 𝜃−𝛼 =𝑅 sin 𝜃 cos ∝ −𝑅 sin ∝ cos 𝜃 =12 ∝= arctan 17.64 29.4 =31.0° 𝑅 sin ∝ =17.64 𝑅 cos ∝ = 29.4 𝑅= 17.64 2 + 29.4 2 =34.29 So 29.4 sin 𝜃 −17.64 cos 𝜃 = 34.29 𝑠𝑖𝑛 𝜃−31.0 =12 𝑠𝑖𝑛 𝜃−31.0 = 12 34.29 𝜃−31.0 =20.5 angle of inclination 𝜃=51.5° 18

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 –

END

WB 7 find coefficient friction solution 5𝑔 sin 𝜃 5𝑔 cos 𝜃 = 29.4 39.2 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=39.2 𝑁 29.4−𝐹𝑟=5×1.63 𝐹𝑟=21.23 𝐹𝑟=𝜇𝑅 21.23=𝜇×39.2 Coefficient of friction 𝜇=0.542 21