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DYNAMIC FORCES Equations of motion 1, Acceleration (rate of change of Velocity) velocity is the vector version of speed, (speed in a straight line) v–u t =a Can be rearranged to v = u + at u = initial velocity (in m/s) v= final velocity (in m/s) t = time (in seconds (s)) a = m/s2 Sample question • A car increases it’s velocity from 10 m/s to 20m/s in 5 seconds. What is it’s acceleration? Sample question (answer) v–u t =a v = 20m/s u =10m/s t = 5 sec 20 – 10 5 = 2m/s2 Sample question A car moving at 10 m/s accelerates at 2m/s2 for 4 seconds. What is it’s final velocity? Sample question (answer) Can be re-arranged to v = u + at u = 10m/s a = 2m/s2 t = 4 sec v = 10 + 2 x 4 = 18m/s Sample question •A stone block with a mass of 800Kg is lifted from rest with a uniform acceleration by a crane such that it reached a velocity of 14m/s. after 10 seconds. Calculate the tension in the lifting cable Sample question • To do this calculation we need to use another equation • Force = mass x acceleration • F = ma • Weight (force acting downwards) = mass x acceleration due to gravity • Wt = mg Wt = m x 9.81 (acceleration due to gravity on earth) Sample question (answer) First find the acceleration Tension acceleration U = 0 from standing start v–u =a t v t =a 14 = 1.4m/s2 10 Weight mg Force due to acceleration F = ma Sample question (answer) F = ma F = 800 x 1.4 = 1120N Tension Wt of box mg = 800 x 9.81 = 7848N (7.8kN) acceleration Tension in cable = 1120 + 7848 = 8938N (8.9kN) Weight mg Force due to acceleration F = ma Sample question • If the stone is lowered at the same rate of acceleration the tension is less than the weight of the box Force due to Tension acceleration F = ma Tension in cable = 7848 - 1120 = 6728N (6.7kN) acceleration Weight mg Lift You get the same effect when you travel in a lift. You feel heavier when the lift begins and accelerates upwards and lighter when the lift accelerates downwards A man standing on weighing scales in a lift has a mass of 60 Kg. The lift accelerates uniformly at a rate of 2m/s2 Calculate the reading on the scales during the period of acceleration . acceleration Weight mg Force due to acceleration F = ma F = ma F = 60 x 2 = 120N Wt of man mg = 60 x 9.81 = 588.6N Force acting on scale = 120 + 588.6 = 708.6N The reading on the scale is 708.6 ÷ 9.81 = 72.23Kg 2, Displacement (s) is the vector version of distance (distance in a straight line) Displacement = average velocity x time s = (v + u) x t 2 Sample question •A car moving at 10 m/s increases it’s velocity to 20m/s in 4 seconds. How far will it have travelled during this time? Sample question (answer) s = (v + u) x t 2 s = (20 + 10) x 4 2 v =20m/s u = 10m/s t = 4 sec S = 60m Sample question • How long will it take for an athlete to accelerate from rest to 4 m/s over 8m ? Sample question (answer) s = (v + u) x t 2 t = 2s v u =0 so t = s= vxt 2 16 4 = 4 seconds 3, Displacement related to acceleration s = ut + at2 2 With zero acceleration (constant velocity) a=0 s = ut From a standing start ut = 0 s = at2 2 Example •A car accelerates uniformly from rest and after 12 seconds has covered 40m. What are its acceleration and its final velocity ? Example 2s = at2 80 = a 122 80 = a 144 2s = a t2 a = 0.56m/s2 Finding v From a standing start u = 0 a=v–u t v= axt a=v t v= axt v = 0.56 x 12 = 6.7 m/s Or alternatively • • U = 0 from standing start s = (v + u) x t 2 s= v = 2s t vxt 2 v = 80 = 6.7m/s 12 4, final velocity, initial velocity acceleration and displacement 2 •v = 2 u + 2 as Example If a car travelling at 10m/s accelerates at a constant rate of 2m/s2, what is it’s final velocity after it has travelled 10m? Sample Question (answer) 2 v = 2 u + 2 as V2 = 102 + (2 x 2 x 10) = 100 + 40 = 2 140m/s D’Alembert’s Principle Acceleration (a) Applied Force (F) mass F + Fi = 0 so F = -Fi F = ma so ma = - Fi or Fi = - ma Inertia Force (Fi) When an object is accelerating the applied force making it accelerate has to overcome the inertia. This is the force which resists the acceleration (or deceleration) and is equal and opposite to the applied force. This means that the total force acting on the body is zero D’Alembert’s Principle Applied Force 200N Mass 20Kg Spring Force 150N Acceleration Friction Force 20N Free body diagram F (applied) 200N W (weight) Fs (spring) 150N Mg = 20 x 9.81 =196.2N Fr (Friction) 20N D’Alembert’s Principle • Downward forces are minus and upward forces are positive. • From the diagram • -200N – 196.2N +20N +150N • = -226.2N (which is downward as to be expected) Conservation of energy (Gravitational potential energy) GPE = mass x gravity x height (mgh) = 5 x 9.81x10 10m Work done (in lifting the mass) = force x distance 49.05 x 10 = 490.5joules 490.5joules (F = mg) 5Kg Work done = energy gained Kinetic energy 5Kg 10m GPE = mgh = 490.5 joules Neglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottom KE (at the bottom) = 490.5J = mv2 2 (v = velocity) Kinetic energy 5Kg 10m GPE = mgh = 490.5 joules Neglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottom The velocity at the bottom v =√(2Ke÷m) = √ (2 x 490.5 ÷ 5) √196.2 = 14m/s Work and energy Work done (in joules) = Force x distance moved (in direction of force) Work done = energy used Force distance Example • A car of mass 800Kg is stood on a uniform 1 in 10 slope when it’s handbrake is suddenly released and it runs 30 metres to the bottom of the slope against a uniform frictional force of 50N. What is the car’s velocity at the bottom of the slope? Example • First find the angle of the slope Tan of the angle = opposite/adjacent = 1/10= 0.1 Tan-1 0.1 = 5.7o 1 10 Example • Then find the height of the slope • sin 5.7o = opposite/hypotenuse • Opposite = hypotenuse x sine5.7o 30m ht • 30 x sine 5.7o 5.7o • =2.98m Example 1 (conservation of energy method) • Find the gravitational potential energy of the car at the top of the slope 30m ht 5.7o • 800 x 9.81 x 2.98 • = 23.4kN Example 1 (conservation of energy method) • Work done against friction = force x distance = 50N x 30m = 1.5kj 30m ht 5.7o Example 1 (conservation of energy method) • Kinetic energy of the car at the bottom of the slope 30m ht 5.7o mv2/2 = 23.4 -1.5 = 21.9kj v= √(2 x Ke/m) v= √(2 x 21900/800) v= √54.75 v = 7.4 m/s • Example 2 (Resolving forces method) • Work out the forces involved 30m ht 5.7o Weight of car (force acting vertically downward) = mass x gravity = 800x 9.81 7848N Example 2 (Resolving forces method) Wt = weight of car (7848N) Fn is the normal reaction force of the slope on the car Fs is the force on the car down the slope Fs wt Fn 5.7o 5.7o Example 2 (Resolving forces method) • Work out the forces involved Sine 5.7o = Fs ÷ wt Fs = Wt x sin 5.7o Fs = 7848 sin 5.7o Fs Fs = 779.5N (Force down the slope wt Fn 5.7o 5.7o Example 2 (Resolving forces method) • Work out the forces involved Resultant force down the slope = Fs – friction force 779.5 – 50 =749.5N Fs wt Fn 5.7o 5.7o Example 2 (Resolving forces method) Acceleration down the slope a= F/m 749.5/800 = 0.94m/s2 Fs wt Fn 5.7o 5.7o Example 2 (Resolving forces method) Velocity at the bottom of the slope v2 = u2 + 2as (u2 = 0) so v2 = 2as v2 = 2 x 0.94 x 30 v =√56.4 v = 7.5m/s Fs wt Fn 5.7o 5.7o Linear Momentum and Collisions • Conservation of Energy • • Momentum = mass x velocity • The total momentum remains the same before and after a collision • Momentum is a vector quantity Linear Momentum and Collisions • A railway coach of mass 25t is moving along a level track 36km/hr when it collides with and couples up to another coach of mass 20t moving in the same direction at 6km/hr. Both of the coaches continue in the same direction after coupling. What is the combined velocity of the two coaches? Linear Momentum and Collisions •Let the mass of the first coach be M1 and the mass of the second coach be M2 and the velocity of the first coach be V1 and the velocity of the second be V2 Linear Momentum and Collisions • Before coupling the momentum of the first coach is 25 x 36 = 900tkm/hr and the momentum of the second is 20 x 6 =120 tkm/hr • Which is a total of 1020tkm/hr Linear Momentum and Collisions • After the coupling the momentum of both is the same as before the coupling which is 1020 tkm/hr • And the combined mass is 45t Linear Momentum and Collisions • Velocity after coupling is momentum divided by mass • 1020÷45 • 22.6 km/hr Example A hammer of mass 200Kg falls 5m on to a pile of mass 300Kg and drives it 100mm into the ground a) Calculate the loss of energy on the impact. b) Calculate the work done by the resistance of the ground. c) calculate the average resistance to penetration. Example • Before falling the GPE of the hammer is 200kg x 9.81 x 5m = 9810j • Kinetic energy of the hammer just before impact = 9810j (0.5 x m x v2) v2 = (9810)/0.5 x m (9810)/0.5 X 200 =98.1 v = √98.1 = 9.9m/s • Velocity of the hammer just before impact • • Example • Momentum of hammer just before impact = mass x velocity • = 200kg x 9.9m/s = 1980 kgm/s • Momentum before collision = momentum after collision • mass of hammer plus pile after collision = 200kg + 300kg = 500kg • thus 500kg x velocity after collision = 1980 kgm/s • Velocity after collision = 1980 ÷ 500 = 3.96m/s Example • Kinetic energy after collision (0.5 x mass x v2) = 0.5 x 500 x 3.962 • = 3920j • Loss of energy = 9810 – 3920 = 5890j • To find deceleration of pile hammer • v2 = u2 +2as (v2 =0) • u2 = - 2as • Example • a = - u2/2s • 3.962/0.2 • - 78.4m/s2 • The minus sign means deceleration, Resistive force of the ground = mass x deceleration • 500 x78.4 • 39.2kN Example • Work done by ground = Force x distance • 39.2kN x 0.1 m • 3.92kj (this agrees with the fact that fact that the kinetic energy of the hammer and pile after impact was 3.92kj and zero when the pile stopped moving in the ground