Understanding Friction Dynamics in Physics
Explore the concepts of friction dynamics in physics, including solving problems with moving objects using principles such as F=ma and the suvat model. Learn about limiting equilibrium, coefficients of friction, and acceleration calculations for various surfaces. Dive into scenarios involving forces, accelerations, and equilibrium in this engaging physics topic.
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Friction Dynamics
Friction: DYNAMICS BAT solve DYNAMIC problems with moving objects using Friction, F=ma and the suvat model KUS objectives 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 ???????? ?? when a particle is in equilibrium ???????? = ?? 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 ?) ? = 78.4 100sin20 = 44.20 ? 100cos20 100sin20 100 Ice R ??) ?? = ? Fr5 200 =Fr 5 R= 44.20= 0.113 W = 8 9.8 = 78.4 ???) ? = ?? 100cos20 5 = 8? ? = 11.1 ?? 2
WB 1b. Finding the coefficient of friction b) Wet grassy surface, friction is 15 N 100cos20 100sin20 Grass 100 R ?) ? = 78.4 100sin20 = 44.20 ? Fr15 200 ??) ?? = ? =Fr 15 R= 44.208= 0.339 W = 8 9.8 = 78.4 ???) ? = ?? 100cos20 15 = 8? ? = 9.87 ?? 2
WB 1c. Finding the coefficient of friction c) Dry surface, friction is 25 N 100cos20 100sin20 Tarmac 100 R ?) ? = 78.4 100sin20 = 44.20 ? Fr15 200 ??) ?? = ? =Fr 25 R= 44.208= 0.566 W = 8 9.8 = 78.4 ???) ? = ?? 100cos20 25 = 8? ? = 8.62 ?? 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 a) Find the acceleration of the crate 4 . The coefficient of friction between crate and floor is 0.1 ?) ? = 245 35sin = 224 ? 35cos 35sin 35 R ??) ?? = ? Fr 0 ?? = 0.1 224 = 22.4 ???) ? = ?? W = 25g = 245 35cos 22.4 = 25? ? = 0.224 ?? 2 sin? =3 cos? =4 5 5 3 ? 4 5
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) 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) 35cos 35sin R 35 Fr 0 Actual working ? = 245 + 35sin = 266 ? ?? = 0.1 266 = 26.6 ? Resultant force 35cos 26.6 = 25? Acceleration 0.056 ?? 2 W = 25g = 245
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 R Fr ? = 3?c??30 Frmax= ?? = 0.35 (3? ??? 30) 30 Resultant Force ? = ?? 3?sin30 3?cos30 3????30 0.35 3? ??? 30 = 3? ? =3????30 0.35 3? ??? 30 = 1.93 ?? 2 3
WB 4 Parcel moving down a slope, find ? 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?? 2, find the coefficient of friction ? between the particle and the slope R Fr ? = 2?c??30 Frmax= ?? = ?(2? ??? 30) 30 Resultant Force ? = ?? 2?sin30 2?cos30 2????30 ? 2? ??? 30 = 2 1 ? =2????30 2 2???? 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 R Fr ? = 3?cos30 = 25.46 30 ?? = 3?sin30 = 14.7 3?sin30 3?cos30 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 R Step 2 Bigger slope Fr ? = 3?cos40 = 22.52 40 Coefficient of Friction is the same 3?sin40 3?cos40 Friction = 0.577 22.52 = 12.99 Resultant Force ? = ?? 3?sin40 12.99 = 3? ? = 1.97 ?? 2
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 ? = 0.3 find the acceleration of the box R ? = 2?c??10 + 25sin10 25cos10 25sin10 ? = 23.64 25 Frmax= 0.3 ? = 7.093 Fr 30 Resultant Force ? = ?? 2?sin10 2?cos10 25???10 2???? 10 7.093 = 2? ? = 7.06 ?? 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 = 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 ? = 5?cos? = 39.2 R Fr Friction = ?? = 39.2 ? use suvat to find the acceleration ? = ? + ?? ? =4.9 3 arcsin3 5 = 1.63 ?? 2 5?sin? 5?cos? 29.4 39.2 = Now use F = ma 5? ??? ? 39.2? = 5 1.63 sin? =3 cos? =4 5 5 ? =5? ??? ? 5 1.63 3 = 0.542 ? 39.2 4 5
WB 8 Parcel find acceleration and distance, suvat A particle is held at rest on a rough plane inclined at an angle of inclination where tan =3 is 0.5 The particle is released an slides down the plane a) Find the acceleration of the particle 4 . The coefficient of friction between the parcel and the plane ? = ??cos =4 5?? R Fr Friction = ?? =2 5?? use F=ma to find the acceleration ??sin 2 5?? = ?? ??sin? ??cos? 3 5? 2 5? = ? sin? =3 cos? =4 ? =1 5? =49 25= 1.96 ?? 2 5 5 3 ? 4 5
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 where tan =3 is 0.5 The particle is released an slides down the plane b) Find the distance it slides in the first 2 seconds 4 . The coefficient of friction between the parcel and the plane Down slope ? = ?? +1 ? = ? = 0 ? = ? =1 2??2 ? = 0 +1 2 1 5? 4 = 3.92 ? 5? ? = 2
WB 9Parcel: 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 R ? = 3?cos? Fr ?? = ?? = 0.6 3?cos? ? Resultant force ? = ?? 3?sin? 3?cos? 3?sin? 0.6 3?cos? = 3 4 29.4sin? 17.64cos? = 12 This is solvable using the addition formula
WB 9Parcel: 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 29.4sin? 17.64 cos? = 12 addition formula ? ??? ? ? = ?sin?cos ?sin cos? = 12 = arctan17.64 ? sin = 17.64 29.4 = 31.0 ? cos = 29.4 17.642+ 29.42= 34.29 ? = So 29.4sin? 17.64 cos? = 34.29 ??? ? 31.0 = 12 12 34.29 ??? ? 31.0 = ? 31.0 = 20.5 angle of inclination ? = 51.5
BAT solve DYNAMIC problems with moving objects using Friction, F=ma and the suvat model KUS objectives self-assess One thing learned is One thing to improve is
WB 7 find coefficient friction solution A parcel of mass 5 Kg is released from rest on a rough ramp of inclination = 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 R Fr ? = ? = 0 ? = 4.9 ? = ? = 3 ? = ? + ?? arcsin3 Suvat to find a 5 4.9 = 0 + ? 3 5?sin? 5?cos? 29.4 39.2 = ? = 1.63 ?? 2 resultant force in the i direction Equilibrium in the j direction ? = ?? ? = 39.2 ? 29.4 ?? = 5 1.63 ?? = 21.23 21.23 = ? 39.2 ?? = ?? Coefficient of friction ? = 0.542
