Page 1 Dynamics Newton's Laws...3 Newton s First Law... 3 Example 1... 3 Newton s Second Law...4 Example 2... 5 Questions A... 6 Vertical Motion...7 Example 3... 7 Example 4... 9 Example 5...10 Example 6...13 Motion of Two Connected Particles...15 Newton's Third Law... 15 Example 7...16 Pulleys... 19 Example 8...19 Example 9...21 Example 10... 23 Example 11... 26 Extension... 27 Momentum and Impulse... 28 Momentum... 28 Example 12... 28 Change in Momentum... 28 Example 13... 28 Impulse... 29 Example 14... 30 Example 15... 30 The Principle of the Conservation of Momentum... 32
Page 2 Example 16... 32 Example 17... 34 Example 18... 35 Jerk in a String... 37 Example 19... 37 Questions B... 39
Page 3 Dynamics is the study of moving objects. In previous chapters we have considered the sum total of forces on a body (resultant) and our next consideration is the movement that these forces bring about. Forces of friction, thrust, gravity and tension will be considered and the subsequent motion will be measured by the application of constant acceleration equations and the equation of motion. As in most mechanics questions a certain amount of modeling will have to be used in our working. Our first area of study of moving objects will involve the application of Newton s laws. Newton's Laws Newton s First Law A body will remain at rest, or will continue to move with constant velocity, unless external forces force it to do otherwise. A change in state of motion of a body is caused by a force. The unit of force is the Newton, (N). Example 1 A body of mass 3500Kg moves horizontally at a constant speed of 5ms -1 subject to the forces shown. Find P and S. 5ms -1 350N P S 3500g S
Page 4 There is no vertical motion therefore: 2S = 3500g S = 1750g The horizontal acceleration is zero, therefore: P =350N Newton s Second Law The force F applied to a particle is proportional to the product of mass of the particle and the acceleration produced. A force of 1N produces an acceleration of 1ms -2 in a body of mass 1kg. Newton s Second Law is summarized by the equation: F = ma this is often termed the equation of motion. It is vitally important to realise that F is the overall resultant and not Friction (F R ).
Page 5 Example 2 If the object in Example 1 is slightly modified to take account of the fact that there is a pushing force 600N, calculate the acceleration. a ms -2 350N 600N S 3500g S The resultant of the two horizontal forces is 250N pushing the object to the right. So by setting up an equation of motion: F = ma 250 = 3500 a a = 0.07ms -2
Page 6 Questions A 1 Find the resultant force that will bring about an acceleration of 4 ms -2 for a particle of mass 3.75kg 2 A particle of mass 4.5kg is acted upon by forces (6i + 3j)N and (-2i + 7j)N. Calculate the acceleration of the particle in vector form. 3 A toy train of mass 1.25kg is pulled across a horizontal floor by a horizontal string. The tension in the string is 1.2N. Calculate the acceleration and the distance traveled by the train in the first 4 seconds. 4 A car of mass 750Kg experiences a resistive force of RN while being brought to rest in 9 seconds form a speed of 18ms -1. Calculate the magnitude of the force. 5 A car of mass 900Kg is under constant resistance to motion of 500N. Find the value of the engine force required to bring about an acceleration of 1.25ms -2. 6 A train of mass 4000kg produces a driving force of 3000N. If the train experiences constant resistance to motion of 1200N calculate the acceleration. 7 A stone of mass 1.25kg is dropped into a viscous liquid and fall vertically through it with an acceleration of 4.8ms -2. Find the resistive force acting against the stone.
Page 7 Vertical Motion If a particle is falling in the earth s atmosphere then it will accelerate at 9.8ms -2. By Newton s first law the particle must be experiencing a force and the only force present is the weight, so by considering the equation of motion: F = ma Example 3 Weight = mg A man of mass 95kg is traveling up in a lift. Given that the acceleration of the lift is 0.8ms -2, find the force exerted on the man by the floor of the lift. Calculate the same force when the lift is descending with the same acceleration. R 0.8ms -2 95g
Page 8 Taking up to be positive and setting up an equation of motion: F = ma R 95g = 95a R 931 = 95 0.8 R = 1007N When the lift is traveling in the opposite direction let down be positive so the equation of motion becomes: 95g R = 95a 931 R = 95 0.8 R = 855N It is worth pointing out that the reaction from the floor is less than the weight of the man when the lift is descending (and vice versa). The next example introduces constant acceleration equations into the problem.
Page 9 Example 4 A ball of mass 0.9 kg falls from a height of 22m above horizontal ground. The ball reaches the ground after t seconds. The ball sinks into the ground a distance of 1.9cm before coming to rest. The ground is assumed to exert a constant resistive force of magnitude F newtons. Find: a) the value of t to 3 sig fig; b) the value of F to 3 sig fig. a) Using constant acceleration equations to find t: s = 22, a = 9.8, u = 0, t =? 1 s = ut + at 2 2 22 = 0 + 0.5 9.8 t 2 t = 2.12sec b) The ball is brought to rest in 1.9cm so by using constant acceleration equations we can find the deceleration and hence the force: Firstly we need to find the speed with which the ball hits the surface u = 0, v =?, t = 2.12, a =9.8 v = u + at v = 9.8 2.12 v = 20.776ms 1
Page 10 And now for the deceleration: s = 0.019, a =?, v = 0, u = 20.776 2 2 v = u + 2as 2 0 = 20.776 + 2 a 0.019 a = 11359ms 2 And finally using an equation of motion for the particle: F = ma F = 0.9 11359 F = 10.2KN Note that we were asked for the magnitude hence the positive answer. How realistic is this answer? Ask a physics teacher for some other examples. Example 5 A ball of mass 5kg falls from a height of 6m into a jar containing a viscous liquid. The upward force exerted by the liquid is of magnitude 75N. How far will the ball sink into the liquid? Calculate the total time that the ball is in motion. Firstly we need to calculate the speed with which the ball hits the liquid.
Page 11 Assuming that down is positive: s = 6, a = 9.8, u = 0, v =? 2 2 v = u + 2as 2 v = 0 + 2 9.8 6 v = 10.84ms 1 Secondly we can set up an equation of motion for the ball to work out the deceleration. 75N F = ma 5g 5g 75 = 5a a = -5.2ms -2 Thirdly we need to calculate the distance that the ball travels through the liquid before coming to rest. u = 10.84, v = 0, a = -5.2 s =? 2 2 v = u + 2as 0 = 117.6 2 5.2 s s = 11.31m
Page 12 Finally the total time taken by the ball in motion must be done in two parts seeing as before the ball hits the liquid it has an acceleration of 9.8ms -2 whereas, whilst falling through the liquid, it has an acceleration of -5.2ms -2. Time to meet the surface of the liquid: a = 9.8, u = 0, v = 10.84, t =? v = u + at 10.84 = 0 + 9.8t t = 1.1 sec Time to come to rest: a = -5.2, u = 10.84, v = 0, t =? v = u + at 0 = 10.84-5.2t t = 2.1 sec Therefore the total time in motion is 3.3 sec. The liquid is rather viscous, how realistic is the resistance force? What about a ball falling in to a jar of syrup?
Page 13 The next example introduces friction on a slope. The questions are beginning to get more challenging but a good diagram is always the best place to start. Examiners regularly report that the most successful candidates in mechanics M1 and M2 always draw diagrams. Example 6 A ball of mass 2kg is projected up a line of greatest slope inclined at an angle of 30º to the horizontal. The coefficient of friction between the plane and the ball is 0.4. The initial speed of the ball is 5 ms -1. Find: a) the frictional force acting whilst the ball moves up the plane. b) the distance moved up the plane by the ball before it comes to instantaneous rest. R 5ms -1 F R 30º 2g a) Since the ball is moving then Friction must be at its maximum value. Resolving perpendicular to the plane gives: R = 2g cos 30º R = 16.97N
Page 14 Using F R = μr: F R = 0.4 16.97 F R = 6.79N b) The frictional force and the weight component of the ball are trying to slow the ball down. By setting up an equation of motion for the ball we can calculate the deceleration. Assuming that uphill is positive: F = ma - F R weight comp down the plane = ma -6.79-2g cos 30º = 2a a = -8.295ms -2 At the point the ball comes to instantaneous rest it will have zero velocity and we can use constant acceleration equations to calculate the distance traveled. u = 5, v = 0, a = -8.295, s =? 2 2 v = u + 2as 0 = 25 2 8.295 s s = 1.51m
Page 15 Motion of Two Connected Particles Newton's Third Law Before we can consider the motion of two connected particles we need to discuss Newton s Third Law. This law states that action and reaction are equal and opposite. If two bodies A and B are in contact and exert forces on each other, then the force exerted by A on B is equal in magnitude and opposite in direction to the force exerted by B on A. This principle will be applied to tow truck problems and pulleys to name but two. Consider the situation below where the cartoon car is towing a racing car. T T The racing car is pulled forward by tension in the tow bar. The racing car will exert an equal but opposite force on the car. If the car is slowing down and there are no breaks on the racing car then some force must be acting in the opposite direction to the direction of motion of the two cars. In this case the tow bar will exert a thrust on both cars (arrows change directions).
Page 16 Example 7 The AA man is towing a car along a straight horizontal road. The truck has a mass of 1500kg and the car has a mass of 850kg. The truck is connected to the car by a bar which is to be modelled as a light inextensible string. The truck s engine produces a constant driving force of 2500N. The resistance to motion of the truck and the car are constant and of magnitude 750N and 400N respectively. Find: a) the acceleration of the truck and the car; b) the tension in the rope. When the truck and the car are traveling at 22ms -1 the tow bar breaks. If the magnitude of the resistance to motion of the truck remains at 750N calculate: c) the time difference in achieving a speed of 30ms -1 with and without the car in tow. 2500N 400N 750N 850kg T T 1500kg a) Setting up equations of motion for the car and truck separately gives: Car T 400 = 850a Truck 2500 750 T = 1500a Adding the two equations gives: 1350 = 2350a a = 0.574ms -2
Page 17 b) Finding the tension: Substituting the value into the car s equation of motion gives: T 400 = 850 0.574 T = 888N c) If we assume that the bar doesn t break then the time required to reach 30ms -1 is calculated by using the constant acceleration equations. u = 22, v = 30, a = 0.574, t =? v = u + at 30 = 22 + 0.574t t = 13.9 sec At the point that the tow bar breaks, the tension in the bar is no longer acting against the truck. Therefore the equation of motion of the truck becomes: 2500 750 = 1500a a = 1.167ms -2 u = 22, v = 30, a = 1.167, t =? v = u + at
Page 18 30 = 22 + 1.167t t = 6.9 sec Therefore there is a time difference of 7 seconds.
Page 19 Pulleys In all questions in M1 the pulley system will be smooth. This implies that the motion of the particles at the end of the string are unaffected by the string passing over the pulley. A further assumption is that the string is light and inextensible. These modelling assumptions make the problem simpler but we can still get pretty realistic answers. If two particles are connected by a string where the string passes over a smooth pulley, then we can assume that the particles will have equal accelerations but in opposite directions. Example 8 Two particles P and Q are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest. Find: a) the magnitude of the acceleration; b) find the tension in the string. a ms -2 P T T Q 8g 20g
Page 20 To start the problem set up an equation of motion for particle P. The particle will accelerate upwards hence: F = ma T 8g = 8a (1) When considering particle Q, its weight will cause it to accelerate down hence the equation of motion is: F = ma 20g T = 20a (2) By adding the two equations the tension will be eliminated: 12g = 28a a = 4.2ms -2 Substituting the value of the acceleration into equation (1) will give the tension. T 8g = 8 4.2 T = 112N The following example is more algebraic but this does not mean that it is more complicated.
Page 21 Example 9 Two particles P and Q of masses have masses 8m and Km, where K > 8. They are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut and the hanging parts of the string vertical, as shown below. Initially P has an acceleration of magnitude of 3 g 4. a) Find, in terms of m and g, the tension, T, in the string. b) Find the value of K. a) Adding forces to the system: P Q 3 g 4 T T Q P Kmg 8mg
Page 22 Setting up an equation of motion for particle P: F = ma T 8mg = 3 8 m 4 g T = 14mg (1) b) Considering particle Q: Kmg T = 3 km 4 g Using (1): 1 Kmg 4 = 14mg K = 56 The next problem involves a pulley system where one of the particles is being dragged across a horizontal table as the other particle is falling. The problem will be made more complex when the horizontal table is considered to be rough. The questions are increasing in complexity but the same basic principles apply.
Page 23 Example 10 A particle A, of mass 0.9kg, rests on smooth horizontal table and is attached to one end of a light inextensible string. The string passes over a smooth pulley P fixed at the edge of the table. The other end of the string is attached to a particle B of mass 1.8kg which hangs freely below the pulley. The system is released from rest with the string taut and B at a height of 3.2m above the ground. In the subsequent motion A does not reach the pulley before B reaches the ground. Find: a) the tension in the string before B reaches the ground. b) the time taken by B to reach the ground. Then, to make the model more realistic, assume that the coefficient of friction between the particle and the table is 0.3. Using this modification find the time taken by B to reach the ground. A R T P 0.9g T B 3.2m 1.8g a) Setting up equations of motion for the two particles gives: A F = ma B F = ma T = 0.9a 1.8g T = 1.8a
Page 24 Adding the two equations gives: 1.8g = 2.7a Therefore: a = 6.53ms -2 T = 0.9 6.53 T = 5.88N b) The particle B is falling with acceleration 6.53ms -2. So by using constant acceleration equation we can find the time it takes to reach the floor. u = 0, a = 6.53, s = 3.2, t =? 1 s = ut + at 2 2 1 3.2 = 6.53 t 2 2 t = 1.01sec Seeing as the particle is moving friction must be at its maximum value and hence F R = μr. Setting up equations of motion for the two particles with friction included gives: A F = ma B F = ma T - F R = 0.9a (1) 1.8g T = 1.8a (2)
Page 25 Resolving vertically for A: R = 0.9g Using F R = μr F R = 0.27g Adding equations (1) and (2) gives: 1.8g - 0.27g = 2.7a a = 5.553ms -2 The new value of acceleration can now be used to calculate the new time: u = 0, a = 5.553, s = 3.2, t =? 1 s = ut + at 2 2 1 3.2 = 5.553 t 2 2 t = 1.07 sec In the next problem a particle is being pulled up a rough inclined plane by the motion of another particle falling towards a floor. This is very similar to an exam question and would be worth in excess of 10 marks on an M1 paper.
Page 26 Example 11 A particle, A of mass 6kg, rests on a rough plane inclined at an angle of 35º to the horizontal. The particle is attached to one end of a light inextensible string which lies in a line of greatest slope of the plane and passes over a light smooth pulley P fixed at the top of the plane. The other end of the string is attached to a particle B of mass 16kg. The particles are released from rest with the string taut. The particle B moves down with an acceleration of 2 g 5. R F A T P T 35º 6g B 16g x Find: a) the tension, T, in the string. b) the coefficient of friction between the plane and A. a) Setting up equations of motion for A and B gives: A F = ma B F = ma T - 6g sin 35º - F R = 6 2 g 5 (1) 16g T = 16 2 g 5 (2) Using (2) to find the tension: T = 48 g = 94.08N 5
Page 27 Resolving vertically for A: R = 6gcos35º = 48.166 Using F R = μr F R = 48.166μ Therefore equation (1) becomes: 12 94.08 33.726-48.166μ = g 5 Rearranging for μ: μ = 12 60.35 - g 5 48.166 μ = 0.765 Extension Assume that in the example above the particle is 5m above a level surface and that after 0.5sec the string breaks. Calculate the total time that the particle B is in flight and the distance that A moves up the plane before it comes to instantaneous rest (assume that it does not reach the pulley). This is a rather large value for μ and this obviously accounts for the low value for the acceleration. What is the resultant force on the pulley?
Page 28 Momentum and Impulse Momentum The momentum of a body of mass m, having a velocity v is mv. The units of momentum are Newton seconds (Ns). Momentum = mv The momentum of a body is dependent upon its velocity therefore momentum is a vector quantity. This implies that direction is very important and great care must be taken with signs. Example 12 Find the momentum of a hockey ball of mass 0.9kg hit at 18ms -1. Momentum = mass velocity Momentum = 0.9 18 = 16.2Ns Change in Momentum If a particle experiences a change in velocity then, by definition, its momentum must change. Let the initial velocity be u and the final velocity be v then the change in momentum is given by: Example 13 Change in Momentum = mv mu = m( v - u ) If the hockey ball from example 12 hits a wall directly and returns with a velocity of 12ms -1. Find the change in momentum.
Page 29 Always draw a diagram. Before 18ms -1 After 12ms -1 0.9kg 0.9kg Taking left to right as positive, therefore u = 18, v = -12 Change in Momentum = m( v - u ) = 0.9 (-12 18 ) = -27Ns The wall in the above example has experienced a force as the hockey ball hits it. This action takes place in a very short time period and is called the impulse. Impulse When a force F, is applied to a particle for a period of time t, then this quantity is defined as the impulse of the force. Obviously an impulse will bring about a change in velocity and therefore momentum will change. Therefore: Impulse = F t = m(v- u )
Page 30 The derivation of the formula comes from the equation of motion and constant acceleration equations: F = ma v = u + at F = m v t u a = v u t Therefore Ft = m(v u) Example 14 A particle of mass 7.5kg is acted on by a force for 6 seconds and in the process its velocity increases from 6ms -1 to 15ms -1. Find the magnitude of the force. Impulse = change in momentum F t = m(v- u ) F 6 = 7.5 (15 6) Example 15 F = 11.25N A ball of mass 1.2kg is moving vertically with a speed of 14 ms -1 when it hits a smooth horizontal floor. It rebounds with a speed 8 ms -1. Find the magnitude of the impulse exerted by the floor on the ball.
Page 31 Take care with signs Assuming down to be positive Impulse = Change in Momentum = m( v u ) = 1.2( -8 14 ) = 1.2( 22 ) = -26.4 Ns
Page 32 The Principle of the Conservation of Momentum When a collision occurs between two bodies, A and B, then the force exerted on A by B will be equal and opposite to the force exerted on B by A (by application of Newton s third law). If no other forces are present then the change in momentum in one particle will equate to the loss of momentum in the other particle. Momentum is conserved and therefore the sum of the momentum of the particles before collision must equal the sum of momentum after the collision. This is referred to as the Principle of Conservation of Momentum. If two particles of masses, m 1 and m 2, with initial velocities u 1 and u 2, collide then, given that their final velocities are v 1 and v 2 we can say that: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 The following examples illustrate the principle and once again it is always best to draw a diagram as this will help to avoid mistakes with signs and direction. Example 16 Two particles A and B have masses of 1.2kg and 1kg respectively. Particle A is moving towards a stationary particle B with a velocity of 3 ms -1. Immediately after the collision the speed of A is 2.1ms -1 and its direction is unchanged. Find: a) the speed of B after the collision; b) the magnitude of the impulse exerted on A in the collision.
Page 33 a) Before After 3ms -1 0ms -1 2.1ms -1 v ms -1 1.2kg 1kg 1.2kg 1kg A B A B By conservation of momentum: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 3 1.2 = 2.1 1.2 + v 3.6 2.52 = v v = 1.08 ms -1. b) Impulse = Change in Momentum = m( v u ) = 1.2( 2.1 3 ) = -1.08 Ns
Page 34 Example 17 Two small balls A and B have masses 1.2kg and 2.5kg respectively. They are moving in opposite directions on a smooth horizontal surface when they collide directly. Immediately before the collision, the speed of A is 4.5ms -1 and the speed of B is 0.9ms -1. The speed of A immediately after the collision is 1.3ms -1. The direction of A remains unchanged after the collision. Find: a) the speed of B immediately after the collision; b) the magnitude of the impulse exerted on B in the collision. Before 4.5ms -1 0.9ms -1 After 1.3ms -1 v ms -1 1.2kg 2.5kg 1.2kg 2.5kg A B A B a) By conservation of momentum: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 1.2 4.5 2.5 0.9 = 1.2 1.3 + 2.5v v = 0.636ms -1 b) Impulse = Change in Momentum = m( v u ) = 2.5 ( 0.636 - - 0.9 ) = 3.84Ns
Page 35 Example 18 A locomotive A, of mass 1800kg is moving along a straight horizontal track with a speed of 9ms -1. It collides directly with a stationary coal truck, B, of mass 1000kg. In the collision, A and B are coupled and move off together. a) Find the speed of the combined train. b) After collision a constant breaking force of magnitude R Newtons is applied. The train comes to rest after 15 seconds. Find the value of R. 9ms -1 0ms -1 1800kg 1000kg a) By conservation of momentum: m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 9 1800 = 2800 v v = 5.79ms -1 b) The combined train comes to rest in 15 seconds therefore we need to calculate the acceleration for use in an equation of motion. v = 0, u = 5.79, a =?, t = 15 Using: v = u + at 0 = 5.79 + 15a a = -0.386ms -2
Page 36 Assuming that the train acts as one body: Equation of motion: F = ma R = -2800 0.386 R = 1080.8N = 1.10KN
Page 37 Jerk in a String If two particles are connected by a light inextensible string and one of the particles is projected away from the other then at some point there will be a jerk in the string. At the instant before the jerk one of the particles will have momentum. As soon as the string becomes taut the particles will move onwards with the same velocity. The overall momentum must be conserved so therefore the velocity after the jerk must be lower than the initial velocity. This idea is best explained through an example. Example 19 Two particles P and Q of masses 4kg and 7.5kg respectively are connected by a light inextensible string which is initially slack. Q is projected away from P with velocity 5ms -1. When the string becomes taught the two particles move on together with a common speed. Find the common speed and the impulse exerted on P by the string. Using the conservation of momentum where v 1 = v 2 : m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 1 4 0 + 7.5 5 = 4 v 1 + 7.5 v 1 37.5 = 11.5v 1 v 1 = 3.26ms -1 So the common speed is 3.26ms -1
Page 38 Impulse is the change in momentum so considering particle P: Impulse = m(v u) = 4(3.26 0) = 13.0Ns
Page 39 Questions B 1 A hockey ball of mass 0.125kg, is moving horizontally at 22ms -1 when it hits a vertical kick board at right angles. The ball rebounds horizontally at 12ms -1. a) Find, in Ns, the impulse of the force exerted by the ball on the kick board. Given that the ball is in contact with the kick board for 0.15s: b) Find, in N, the force, assumed constant exerted by the ball on the kick board. 2 Two particles of mass 5kg and 10kg are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut. Find the acceleration of the system and the tension in the string. 3 A particle of mass 0.8kg rests on rough horizontal table and is attached to one end of a light inextensible string. The string passes over a smooth pulley P fixed at the edge of the table. The other end of the string is attached to a particle P of mass 1.4kg which hangs freely below the pulley. The coefficient of friction between the particle and the table is 0.45. The system is released from rest with the string taut and B at a height of 1.2m above the ground. At the point of release A is 1.8m from P. Find: a) the acceleration of the particles; b) the time taken by B to reach the ground. c) the speed with which A hits P.
Page 40 A R T P 0.8g T B 1.2m 1.4g 4 Thomas the tank engine has a mass of 12000Kg and is moving along horizontal rails at 1.2ms -1, strikes buffers and is brought to rest in 0.45s. a) Calculate the impulse, in Ns, of the force exerted by the buffers on Thomas in bringing him to rest. b) Calculate the magnitude of this force assuming it to be constant. 5 Two particles A and B, of masses 4m and 2m respectively are moving towards each other on a smooth horizontal surface with speeds 9v and 3v respectively. The particles collide directly and after the collision A continues to move in the same direction but its speed is halved. Find: a) the speed of B after the impact. b) the magnitude of the impulse exerted by A on B. 6 A bullet is fired horizontally with a speed of 450ms -1 into a block of wood of mass 0.15kg that is placed on a smooth horizontal surface. Given that the block begins to move with a velocity of 9ms - 1, find, in kg the mass of the bullet.
Page 41 7 Two particles P and Q of masses have masses 8kg and m, they are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut and the hanging parts of the string vertical. After 1.5 seconds P has fallen 2.5m. Assuming that Q does not reach the pulley, calculate the tension in the string and the value of m. 8 Two particles A and B of masses 2kg and 3kg are connected by a light inextensible string which passes over a smooth fixed pulley, as outlined in the diagram below. The system is released from rest with the string taut. Find the acceleration of the system and the tension in the string if: a) both planes are smooth; b) both planes are rough and the coefficient of friction between the particles and the plane is 0.1. R F A T P T B S 35º 2g 3g 55º
Page 42 9 A particle, A of mass 4kg, rests on a rough plane inclined at an angle of 30º to the horizontal. The particle is attached to one end of a light inextensible string which lies in a line of greatest slope of the plane and passes over a light smooth pulley P fixed at the top of the plane. The other end of the string is attached to a particle B of mass 13kg. The particles are released from rest with the string taut. Given that the coefficient of friction between the particle A and the inclined plane is 0.35 calculate: a) the tension in the string and the acceleration of the system. b) the angle of inclination required to increase the acceleration by 50%. R F A T P T 30º 4g B 13g 10 A body of mass 6kg is moving with velocity (4i + 7j)ms -1 when an impulse is applied. The impulse causes its velocity to change to (-3i - 5j)ms -1. Find the impulse. 11 A body of mass 7.5kg is initially at rest on a smooth horizontal surface, experiences a force (8i - 13j)N for 3 seconds. Find the final velocity of the body and its speed.
Page 43 12 A pile driver of mass 250 Kg strikes a pile of mass 450kg and drives it into the ground. The pile driver strikes the pile directly with a velocity of 8.5ms -1. The driver does not rebound and in the subsequent motion the pile and the drive move as one. a) Calculate the common speed of the pile and driver immediately after impact. b) Calculate the impulse exerted by the driver on the pile. c) The pile and the driver penetrate 0.55m into the ground before coming to rest. Assuming that the ground exerts a constant resistive force on the pile and driver, calculate the magnitude of the force.
Page 1 Kinematics Introduction and examples... 1 Definitions... 1 Example 1... 3 Example 2... 3 Example 3... 4 Vertical Motion Under Gravity... 5 Example 4... 5 Example 5... 6 Example 6... 7 Questions A... 8 Speed Time Graphs... 11 Definitions... 11 Example 1... 11 Example 2...12 Example 3...13 Problems involving two vehicles... 14 Example 4...14 Questions B... 15 Introduction and examples Kinematics is the study of the motion of particles. In M1 all motion will have constant acceleration. This leads to the development of the constant acceleration equations. Definitions a = acceleration (ms -2 ) u = initial velocity (ms -1 ) v = final velocity (ms -1 ) t = time (second) s = displacement (metres) If we know any of the three we can find the other two.
Page 2 From GCSE you should remember that: change in velocity acceleration = time a = v-u t so v = u + at (1) As mentioned above the acceleration is constant hence the average velocity is simply the average of u and v. u + v average velocity = 2 Another definition is that: displacement s average velocity = = time t Therefore: u + v s = 2 t u + v s = t (2) 2 By using equation (1) we can eliminate v. u + u + at s = t 2 1 2 s = ut + at (3) 2 If equation (1) is rearranged to make t the subject: v-u t = a Then by substituting into equation (2): u + v v-u s = 2 a 2as = v - u 2 2 2 2 v = u + 2as (4)
Page 3 You must learn all four equations above and remember that they only apply to constant acceleration problems. Since velocity is a vector quantity, getting the direction right in these problems is vital. Example 1 A particle is moving in a straight line from O to P with a constant acceleration of 4ms -2. Its velocity at P is 48ms -1 and it takes 12 seconds to travel from O to P. Find (a) the particle s velocity at O and (b) the distance OP. With all problems of this nature, write down what you are given and the one required. It should then be obvious as to which equation you need to use. a) a = 4, u =?, v = 48, t = 12 Obviously we need to use equation (1): v = u + at 48 = u + 4 12 Therefore u = 0ms -1 b) a = 4, u = o, v = 48, t = 12, s =? Using equation (3): 1 2 s = ut + at 2 1 s = 4 12 2 2 Example 2 s = 288m Percy starts from rest at a station S and moves with constant acceleration. He passes a signal box B 14 seconds later with a speed of 84kmh -1. Modelling the train as a particle, find the acceleration of the train in ms -2 and the distance in metres between the station and the signal box.
Page 4 u = 0, t = 14, v = 84kmh -1 a =?, s =? Using equation (1) : v = u + at 84000/60/60 = 0 + 14a a = 1.67ms -2 Once again using equation (3) 1 2 s = ut + at 2 1 5 s = 14 2 3 2 s = 163.33m = 163m Example 3 ET is travelling up a hill on his BMX. He experiences a constant retardation of magnitude 3ms -2. Given that his speed at the bottom of the hill was 15ms -1 determine how far he will travel before he comes to rest. Retardation is simply deceleration. a = -3, u = 15, v = 0, s =? There is no mention of time therefore we use equation (4): 2 2 v = u + 2as 0 = 225-2 3 s s = 37.5m
Page 5 Vertical Motion Under Gravity A few assumptions need to be stated before continuing. 1. Objects will be treated as particles. 2. Motion will only be in a straight line. 3. No evidence of spinning or turning of objects. 4. Particles will have constant acceleration of g (9.8ms -2 ). If an object is projected vertically upwards and it falls 3m below the point of release then the time taken can be calculated by setting s = -3m and a = -9.8 and substitute the values into equation (3). A lot of students work out the time the object takes to reach the top, then the time to return to point of release and finally the -3m part, but this much more complicated than what you need to do! It is also worth noting at this point that time taken to reach maximum height and fall back down to the point of release are the same. This type of question regularly appears on exam papers and it is far quicker to use the first method. Example 4 A Kinder Surprise falls off a shelf which is 0.9 m above the floor. Find: (a) the time it takes to reach the floor (b) the speed with which it will reach the floor. a) s = 0.9, a = 9.8, u = 0, t =? Using equation (3): 1 2 s = ut + at 2 1 0.9 = 9.8 t 2 2 s = 0.429sec
Page 6 b) s = 0.9, a = 9.8, u = 0, t = 0.429, v =? Using equation (1): v = u + at v = 0 + 9.8 0.429 v = 4.20ms -1 Example 5 A bouncy ball B is projected vertically upwards from a point O with a speed of 42ms -1. Find: (a) the greatest height h above O reached by B (b) the total time before B returns to O (c) the total distance travelled by the particle. With questions of this nature one has to be careful as to which direction is positive. Max Height Acceln = -9.8 42ms -1 Start/End a) At the greatest height, v = 0, u = 42, a = -9.8, s =? Using equation (4): 2 2 v = u + 2as 2 0 = 42-2 9.8 s 2 42 s = 19.6 s = 90m
Page 7 b) When the ball returns to B its displacement will be zero! s = 0, a = -9.8, u = 42, t =? Using equation (3): 1 2 s = ut + at 2 2 0 = 42t-4.9t Factorising gives: 7t( 0.7t - 6 ) = 0 Therefore t = 0,t = 8.57sec c) In part (a) we found the distance to the top therefore we only need to double the answer. Total distance = 180m Example 6 A cricket ball is thrown vertically upwards with a velocity of 15ms -1. Modelling the ball as a particle moving under gravity alone, find for how long its height exceeds 10 m. We need to find the times at which the ball has a displacement of 10m. a = -9.8, u = 15, s = 10, t =? Using equation (3): 1 2 s = ut + at 2 10 = 15t - 4.9t 2 2 4.9t - 15t + 10 = 0 By using the quadratic formula we find the two values of t. 15 ± 225 4 10 4.9 t = 9.8 t = 2.08 or 0.98. Therefore the ball is above 10m for 1.1 second.
Page 8 Questions A 1 A car moves with constant acceleration along a straight horizontal road. The car passes the point A with speed 7ms -1 and 5 seconds later it passes the point B, where AB = 53m. a) Find the acceleration of the car. When the car passes the point C, it has a speed of 26ms -1. b) Find the distance AC. 2 A competition diver makes a dive from a highboard into a pool. She leaves the board vertically with a speed of 4ms - 1 upwards. When she leaves the board, she is 6m above the surface of the pool. The diver is modelled as a particle moving vertically under gravity alone and it is assumed that she does not hit the springboard as she descends. a) Find her speed as she reaches the surface of the pool. b) Find the time taken to reach the surface of the pool. c) State two physical characteristics that have been ignored in the model. 3 A car moves from rest at a point O and moves in a straight line. The car moves with constant acceleration 5ms -2 until it passes the point A when it is moving with speed 14 ms -1. It then moves with constant acceleration 2 ms -2 for 8 seconds until it reaches the point B. a) Find the speed of the car at B. b) Find the distance OB.
Page 9 4 An aircraft moves along a straight horizontal runway with constant acceleration. It passes a point A on the runway with speed 15ms -1. It then passes the point B on the runway with speed 39ms -1. The distance AB is 190m. a) Find the acceleration of the aircraft. b) Find the time taken by the aircraft in moving from A to B. c) Find the speed of the aircraft when it passes the mid point of A and B. 5 A racing car moves along a straight horizontal road with constant acceleration. It passes the point O with speed 11ms -1. It passes the point A, 4 seconds later with speed 55ms -1. a) Find the acceleration of the car. b) Find the distance OA. c) Find the speed of the car as it passes the midpoint of OA. 6 A body is projected vertically upwards from ground level at a speed of 49ms -1. Find the length of time for which the body is at least 78.4m above the ground. 7 A body is projected vertically upwards from ground level at a speed of 14ms -1. Find the height of the body above the level of projection after: a) 1 second of motion b) 2 seconds of motion Find the distance travelled by the body in the 2 nd second of motion.
Page 10 8 A balloon is moving vertically upwards with a steady speed 3ms -1. When it reaches a height of 36m above the ground an object is released from the balloon. The balloon then accelerates upwards at a rate of 2ms -2. Find a) the greatest height of the object above the ground. b) the speed of the object as it hits the ground. c) the time taken by the object from leaving the balloon to striking the ground. d) the speed of the balloon as the object hits the ground. 9 A cyclist travels on a straight road with a constant acceleration of 0.6 ms -2. P and Q are 120m apart. Given that the cyclist increases his speed by 6ms -1 as he travels from P to Q. Find: a) the speed of the cyclist at P b) the time taken to travel from P to Q 10 Oblivion at Alton Towers has a vertical drop of 60m. Assuming that the ride starts from rest, calculate the speed of the ride at the bottom of the pit.
Page 11 Speed Time Graphs Definitions For constant acceleration problems the speed time graph will be a straight line. The gradient of the graph is the acceleration. The area under the graph represents the distance traveled. Example 1 A body starts from rest, accelerates uniformly to a velocity of 8ms -1 in 2 seconds, maintains that velocity for a further 5 seconds, and then retards uniformly to rest. The entire journey takes 11 seconds. Find: a) the acceleration of the body during the initial part of the motion. b) The retardation of the body in the final part of the motion. c) The total distance traveled by the body In these type of questions it is vital that you make a sketch of the motion. 9 8 7 6 velocity (m/s) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 time (s)
Page 12 a) As the definitions suggested earlier the acceleration is simply the gradient of the velocity time graph. This is: grad = 8/2 = 4ms -2. b) Similarly for the retardation: grad = 8/4 = 2ms -2. c) The total distance travelled by the body is equal to the area of under the graph. Area = 0.5 8 ( 11 + 5 ) = 64m Example 2 A car accelerates uniformly from rest to a speed of 10ms -1 in T seconds. The car then travels for 4T seconds and finally decelerates uniformly to rest in a further 70s. The total distance traveled by the car is 1250m. Find: a) the value of T. b) the initial acceleration of the car. Once again, sketch the journey. 10ms -1 T 5T 5T+70 a) the area under the graph must equal 1250. 1250 = 10T + 40T + 700/2 900 = 50T T = 18sec. b) The car accelerates to 10ms -1 in 18 seconds, therefore the acceleration is: Acceln = 10/18 = 0.556ms -2
Page 13 Example 3 A train is traveling along a straight track between the points P and Q. It starts from rest at P and accelerates at 3ms -2 until it reaches a speed of 54ms -1. It continues at a constant speed of 54ms -1 for a further 180seconds and then decelerates at a constant deceleration of 6ms -2. a) sketch a speed time graph for the train s journey. b) calculate the total time for the journey from P to Q. c) calculate the distance between P and Q. a) The train will take 54/3 seconds to reach the constant speed (18seconds). It will take 54/6 seconds to decelerate to rest (9 seconds). Hence the graph will have the following shape. 54ms -1 18 198 207 b) the total time for the journey is 207 seconds. c) the total distance is once again equal to the area under the graph. Area = (54 18)/2 + 54 180 + (54 9)/2 Distance = 10449m = 10.4Km
Page 14 Problems involving two vehicles Example 4 Two particles A and B are traveling along a straight path PQ of length 20m. A leaves P, heading for Q, from rest with acceleration of 2ms -2 and at the same time B leaves Q, in the direction of P, from rest with a constant acceleration of 5ms -2. Find how far from A the two particles collide? Let the point of collision be x metres from P and hence (20 x)m from Q. So for particle A: u = 0 a = 2 t = T s = x Using: 1 2 s = ut + at 2 x = t 2 (i) So for particle B: u = 0 a = 5 t = T s = 20 - x Using 1 2 s = ut + at 2 2 20 - x = 2.5t Substituting (i) gives: 20 - t = 2.5t 2 2 Using (i) again: t = 2.39sec x = 2.39 2 x = 5.71m
Page 15 Questions B 1 The diagram shows the sketch of a velocity time graph for a particle moving in a straight line. Find the value of T given that: 10ms -1 a) the distance covered is 100m. b) the constant speed attained by the particle is 10ms -1 c) the rate of acceleration is twice the rate of retardation. d) The time spent at constant speed is equal to the total time spent accelerating and decelerating. T 2 Two trains, P and Q, run on parallel straight tracks. Initially both trains are at rest. At time t = 0, P moves off with constant acceleration for 15 seconds until it reaches a speed of 30ms -1. P then continues at a constant speed. At time t = 35seconds Q moves off with the same acceleration until it reaches a speed of 55ms -1. Q then continues at this constant speed. Train Q overtakes P whilst both trains are travelling at constant speeds after time T. Sketch a speed time graph for the journeys and find the value of T. 3 A particle Q starts from rest at a point O and accelerates at 3ms -2 until it reaches a speed V ms -2. Q then continues at the constant speed for 90 seconds before decelerating at 1.5 ms -2 to come to rest at the point R. If the entire journey takes 150 seconds find the distance QR. Make sure that you draw a speed time graph.
Page 16 4 A racing car moves in a straight line. The car accelerates at 8ms -2 for 5 seconds, maintains a steady speed for 15 seconds and then decelerates to rest in 9 seconds. Sketch a velocity time graph for the car and calculate the total distance travelled.
Page 1 Moments Introduction and examples... 1 Example 1... 1 Example 2... 2 Example 3... 3 Example 4... 4 Equilibrium of a lamina under parallel forces... 5 Example 5... 5 Uniform Beams... 6 Example 6... 6 Non uniform beams...8 Example 7... 8 Questions... 9 Extension... 11 Introduction and examples The moment of a force is the measure of its capacity to turn the body on which it is acting. Example 1 Moment = Force Perpendicular Distance A door of width 1.2m is being pushed by a force of 25N. Find the moment about the hinge. Moment = Force Perpendicular Distance = 25 1.2 = 30Nm
Page 2 Example 2 Forces of magnitude 6N, 8N and 5N are applied to a light rod AB, of length 8m, as outlined in the diagram below. A 2m 8m 6N 8N 5N Calculate the sum of the moments about the midpoint of the rod. Moment of 6N force is given by: 6 4 = 24Nm clockwise Moment of 8N force is given by: 8 2 = 16Nm clockwise Moment of 5N force is given by: 5 4 = 20Nm anti-clockwise B Therefore the sum of the moments is 20Nm clockwise. It is more likely however that the forces are not being applied at right angles to the object. F θ d x P The diagram above shows a force F, acting on an object P at a given angle θ and given distance d. The force brings about a turning effect at P. The dotted line is the perpendicular distance. x = d Sinθ.
Page 3 Therefore the moment about P is given by: Moment = Force Perpendicular Distance from line of action of force to pivot point Example 3 Moment = FdSinθ Two forces are applied to a light rod as outlined in the diagram below. Find the sum of the moment about P. 30º 3m P 8N 5m 6N N The 6N force gives a 30Nm clockwise moment. 30º 3m. P 8N 5m 6N N The dotted line has been added to show the perpendicular distance. The 8N force gives an anticlockwise moment of magnitude: 8 3 Sin30 = 12Nm.
Page 4 Therefore the system is experiencing a clockwise moment of magnitude 18Nm. Example 4 A point Q(4,6) is acted upon by a force (2i + 5j)N. Calculate the sum of the moment about the origin. 5j 6 Q 2i 3 R 4 7 The 2i force gives a clockwise moment of: 2 6 = 12Nm The 5j force gives a anticlockwise moment of: 5 4 = 20Nm Therefore the sum of the moments is 8Nm anticlockwise. Find the sum of the moments about the point R(7,3). The 2i force gives a clockwise moment of: 2 3 = 6Nm The 5j force gives a clockwise moment of: 5 3 = 15Nm Therefore the sum of the moments is 21Nm clockwise.
Page 5 Equilibrium of a lamina under parallel forces If a system is in equilibrium then the following must hold: 1. The component of the resultant force in any direction must be zero. 2. The algebraic sum of the moments about any point must be zero. For uniform non light systems the weight acts through the centre of mass. A rod is said to be uniform if it has even weight distribution and therefore the centre of mass acts at the centre of the rod. Example 5 A uniform rod PQ of length 7m and mass 12kg is pivoted at the point R where PR is 2.5m. Calculate the mass of the particle that must be attached at P to maintain the rod in a horizontal position. P R 7m G Q 2.5m xgn 12gN The moment at P about R must equate to the moment at G about R. The mass of the rod gives a clockwise moment of: 12g 1 = 12gNm The mass added at P gives an anticlockwise moment of: xg 2.5 = 2.5xNm. Therefore: 12gNm = 2.5xNm
Page 6 x = 4.8Kg Uniform Beams Example 6 A uniform beam AB of mass 40kg and length 12m is supported in a horizontal position at C and D, where AC = 1.5m and DB = 1m. A man of mass 80Kg stands on the beam at the point E where EB = 2.5m. Find the reactions at C and D. Examiners always suggest that a diagram is VITAL. A X Y 12m 1.5m E 1m B C D 40g 80g Resolving vertically gives: X + Y = 120g (1) Taking Moments about C gives: The mass of the beam gives a clockwise moment of: 40g 4.5 = 180gNm The mass of the man gives a clockwise moment of: 80g 8 = 640gNm The reaction at D gives an anticlockwise moment of: 9.5YNm Therefore: 9.5Y = 820g
Page 7 Y = 1640 g = 845.9N 19 Using equation (1) from above X + Y = 120g X + 845.9 = 120g X= 330N
Page 8 Non uniform beams Example 7 A non uniform beam AB of mass 20kg and length 12m has an object of mass 40kg placed at a point 8m from A. The beam is in equilibrium in a horizontal position resting on a support C at the midpoint of AB. Find the position of the centre of mass. 12m X A y E B C 20g 40g Taking moments about C gives: 40g 2 = 20g y Y = 4m
Page 9 Questions (some diagrams have forces missing, you need to figure out which ones!) 1 A uniform rod AB of weight 70N and length 5m. It rests in a horizontal position supported at point C and D, where AC = 0.4m. the reaction on the rod at C has magnitude 9N. Find a) the magnitude of the reaction on the rod at D b) the distance AD. A 9 Y B C D 2 A uniform rod AB of length 6m and mass 40Kg. It is supported by two smooth pivots in a horizontal position at A and C where AC = 3m. A woman of mass 75Kg stands on the rod which remains in equilibrium. The magnitudes of the reactions at the two pivots are equal to R Newtons. A B C Find a) the value of R. b) the distance of the woman from A.
Page 10 3 A non uniform plank of wood AB of length 10m and mass 100Kg is smoothly supported in a horizontal position at A and B. An object of mass 90Kg is put on the plank at C, where AC = 6m. The plank is in equilibrium and the magnitudes of the reactions at A and B are equal. Find: a) the magnitude of the reaction R, on the plank at B. b) the distance, x, of the centre of mass of the plank from A. A 6m C B 4 A uniform plank AB has weight 80N and length xm. The plank rests in equilibrium on two supports at A and C, where AC = 3m. A rock of weight 20N is placed at B and the plank remains in equilibrium. The reaction on the plank at C has magnitude 70N. a) find the value of x A xm B C The support at A is now moved to a point D on the plank and the plank remains in equilibrium with the rock at B. The reaction on the plank at C is now two times the reaction at D. b) find the distance AD. 5 A non uniform rod, AB, of length 7m and mass 10Kg is suspended in equilibrium in a horizontal position by ropes attached to the points E and F of the rod, where AE = 2m and AF = 6m. The tensions in the ropes are equal. Find the distance of the centre of mass from A.
Page 11 Extension 1 A large log AB is 10m long. It rests in a horizontal on two supports C and D, where AC = 1m and BD = 1m. An estimate of the weight of the log is required, but the log is too heavy to lift off the supports. When a force of magnitude 1100N is applied vertically to the log at A, the log is about to tilt about D. a) state the value of the reaction on the log at C for this case. b) by modeling the log as a uniform rod, estimate the weight of the log. A 1m X 10m Y 1m B C D??g The force at A is removed and a force vertically upwards is applied at B. The log is about to tilt about C when the force has a magnitude of 1600N. By modeling the log as a non uniform rod, with the distance of the centre of mass of the log being x metres from A, find: c) a new estimate for the weight of the log. d) the value of x.
Page 1 Statics Resolving Forces... 2 Example 1... 3 Example 2... 5 Resolving Forces into Components... 6 Resolving Several Forces into Components... 6 Example 3... 7 Equilibrium of Coplanar Forces...8 Example 4... 8 Questions A... 9 Friction... 13 Rough and Smooth surfaces... 13 Limiting Equilibrium... 13 Coefficient of Friction (μ)... 14 Example 5...15 Non Horizontal Forces...16 Example 6...17 Objects on Inclined Planes... 18 Example 7...18 Varying Values of Friction... 19 Other external forces on inclined planes...20 Example 8... 20 Example 9... 22 Questions B... 24 Solutions to Exam type questions... 27 Statics is the study of stationary objects. We will consider a variety of situations where bodies are acted upon by a number of forces. A few of the concepts introduced in our work on vectors will be built upon in this unit.
Page 2 Resolving Forces In the vectors unit we were made aware of the fact that the resultant of two vectors is the diagonal of a parallelogram, as highlighted in the diagram below. This idea can be applied to forces: F 2 R F 1 In a real world sense the path R is the direction that a particle would take if it were to be acted upon by the forces F 1 and F 2. This principle can be applied to more than two forces. Suppose that a particle is acted upon by the forces F 1, F 2 and F 3. F 3 F 2 The diagrams below should explain the path that the particle will follow. F 1