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AS-Level Maths: Mechanics 1 for Edexcel M1.5 Dynamics 2 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 11 of of 67 67 © Boardworks Ltd 2005 Contents Pulleys Pulleys Internal and external forces Towing Friction on a moving particle Examination style questions 22 of of 67 67 © Boardworks Ltd 2005 Connected particles We talk about connected particles when we are considering situations where two objects are joined together. There are two types of situations that involve connected particles. 1. In the first situation, two masses are connected by a string which passes over a pulley. 2. In the second situation, is one mass tows another; for example, a car towing a caravan or a train engine pulling a carriage. 3 of 67 © Boardworks Ltd 2005 Pulleys This example shows how to find the acceleration of connected masses passing over a pulley. Applying Newton’s second law: m1g – T = m1a T – m2g = m2a Adding: 4 of 67 T = tension in Newtons a = acceleration m = mass Force = mass × acceleration F = ma m1g – m2g = m1a + m2a m1 m2 a g m1 m2 © Boardworks Ltd 2005 Pulleys This example shows how to find the acceleration of connected masses passing over a pulley, when the masses are not acting vertically. Applying Newton’s second law: m2g – T = m2a T – m1gcos(90 – ) = m1a Adding: m2g – m1gsin = m1a + m2a a 5 of 67 90 – θ is the angle between the direction of the force, and the vertical. m2 m1 sin m1 m2 g © Boardworks Ltd 2005 Pulley question 1 Particles of mass 3 kg and 7 kg are attached to each end of a light inextensible string which passes over a smooth fixed pulley. The system is released from rest. a) Find the acceleration of the system. b) Find the tension in the string. 6 of 67 © Boardworks Ltd 2005 Pulley question 1 solution a) Consider the 3 kg particle: T – 3g = 3a Consider the 7 kg particle: 7g – T = 7a Adding: 4g = 10a a = 4g ÷ 10 = 3.92 (to 3 s.f.) Therefore the acceleration of the system is 3.92 m s–2. b) T – 3g = 3a T = (3 × 9.8) + (3 × 3.92) = 41.2 (to 3 s.f.) Therefore the tension in the string is 41.2 N. 7 of 67 © Boardworks Ltd 2005 Pulley question 2 A particle A of mass 3 kg is resting on a smooth, horizontal table top, which is 2 m above the floor. This particle is connected to a particle B of mass 2 kg by a light inextensible string which hangs freely over a smooth fixed pulley at the edge of the table. Particle A is held at rest at a point 1 m from the pulley. The system is then released from rest. a) Find the acceleration of the system before A reaches the pulley. b) Find the time taken for A to reach the pulley. 8 of 67 © Boardworks Ltd 2005 Pulley question 2 solution a) Consider particle A: T = 3a Consider particle B: 2g – T = 2a Adding: 2g = 5a a = 2g ÷ 5 = 3.92 (to 3 s.f.) Therefore the acceleration of the system is 3.92 m s–2. 9 of 67 © Boardworks Ltd 2005 Pulley question 2 solution cont. b) Consider particle A: distance (m) initial velocity (m s–1) acceleration (m s–2) time (sec) s=1 u=0 a = 3.92 t is unknown Constant acceleration formula Using s = ut + ½at2: 1 = 0 + 1.96t2 t = 0.714 (to 3 s.f.) It takes 0.714 seconds for particle A to reach the pulley. 10 of 67 © Boardworks Ltd 2005 Pulley question 3 A particle A of mass 4 kg is resting on a rough plane inclined at an angle of 30° to the horizontal. The particle is connected by a light inextensible string to a particle B of mass 6 kg, which hangs freely over a smooth fixed pulley at the top of the inclined plane. The coefficient of friction between the particle and the plane is 0.15. The system is released from rest. Find the acceleration of the system. 11 of 67 © Boardworks Ltd 2005 Pulley question 3 solution Resolve perpendicular to the plane: R = 4gcos30° = 33.9 (to 3 s.f.) F = R = 0.15 × 33.9 = 5.09 (to 3 s.f.) Consider particle A: T – 5.09 – 4gcos60° = 4a Consider particle B: 6g – T = 6a Adding: 6g – 5.09 – 4gcos60° = 10a 1 a= (6g – 5.09 – 4gcos60°) = 3.41 (to 3 s.f.) 10 Therefore the acceleration of the system is 3.41 m s–2. 12 of 67 © Boardworks Ltd 2005 Pulley question 4 Example 4: A particle A of mass 3 kg is resting on a sloping surface. It is connected to another particle B of mass 5 kg, which hangs freely by a light inextensible string that passes over a smooth pulley at the bottom of the slope. The sloping surface is assumed to be smooth and is inclined at an angle of 20° to the horizontal. Find the acceleration when the system is released from rest. 13 of 67 © Boardworks Ltd 2005 Pulley question 4 solution In this example, gravity and the tension in the string are pulling particle A in the same direction. Consider particle A: Consider particle B: Adding: T + 3gcos70° = 3a 5g – T = 5a 5g + 3gcos70° = 8a 1 a = (5g + 3gcos70°) 8 Therefore the acceleration of the system is 7.38 m s–2. 14 of 67 © Boardworks Ltd 2005 Pulley question 5 Example 5: Two particles of mass 2 kg and 5 kg are resting either side of a rough inclined slope. The side that the 2 kg particle rests on is inclined at an angle of 25° to the horizontal and the other side is inclined at an angle of 30° to the horizontal. The coefficients of friction between the particles and the surface of each side of the slope are equal. The acceleration when the system is released from rest is 2 m s–2. a) Find the coefficient of friction between the particles and the slope. b) Find the tension in the string. 15 of 67 © Boardworks Ltd 2005 Pulley question 5 Solution a) Resolve perpendicular to the plane: Resolve perpendicular to the plane: Consider the 2 kg particle: T – 2gcos65° – F1 = 2 × 2 T – 2gcos65° – 2gcos25° = 4 Consider the 5 kg particle: 5gcos60° – F2 – T = 5 × 2 5gcos60° – 5gcos30° – T = 10 16 of 67 R1 = 2gcos25° R2 = 5gcos30° mass × acceleration Remember, F = R © Boardworks Ltd 2005 Pulley question 5 solution Adding: 5gcos60° – 5gcos30° – 2gcos65° – 2gcos25° = 14 5gcos30° + 2gcos25° = 5gcos60° – 2gcos65° – 14 Therefore, 5 gcos60° – 2 gcos65° – 14 0.0368 (to 3 s.f.) 5 gcos30° 2 gcos25° b) Consider the 2 kg particle: T – 2gcos65° – 2gcos25° = 4 Therefore, T = 2gcos65° + 4 + 2gcos25° × 0.0368 T = 12.9 (to 3 s.f.) The tension in the string is 12.9 N. 17 of 67 © Boardworks Ltd 2005 Pulley question 6 Example 6: Two particles A and B are connected by a light inextensible string which passes over a smooth peg. Particle A has mass 3m kg and is resting on a smooth horizontal table. Particle B of mass m kg is hanging freely over the peg which is positioned on the edge of the table. The system is released from rest. Find the tension in the string in terms of m. 18 of 67 © Boardworks Ltd 2005 Pulley question 6 solution Consider the 3m kg particle: T = 3ma Consider the m kg particle: mg – T = ma Adding: mg = 4ma g = 4a a = ¼g = 2.45 T = 3ma = 3m × 2.45 = 7.35m (to 3 s.f.) Therefore the tension in the string is 7.35m N. 19 of 67 © Boardworks Ltd 2005 Contents Pulleys Pulleys Internal and external forces Towing Friction on a moving particle Examination style questions 20 20 of of 67 67 © Boardworks Ltd 2005 Internal and external forces Whether a force is internal or external has nothing to do with the force itself, but rather the system under consideration. So, whether a force is internal or external is not an intrinsic property of the force. The next two examples illustrate the distinction between internal and external forces. 21 of 67 © Boardworks Ltd 2005 Towing example Example: A car of mass m1 pulls a trailer of mass m2 along a straight horizontal road. The resistance to the motion of the car is F1 and the resistance to the motion of the trailer is F2. The engine of the car produces a driving force of magnitude R. Find, in terms of m1, m2, F1, F2 and R, a) The acceleration of the system. b) The tension in the coupling between the car and the trailer. 22 of 67 © Boardworks Ltd 2005 Towing example solution a) When considering the motion of the car and trailer as a single entity, the tension in the coupling is an internal force and does not appear on the diagram. Applying Newton’s second law: R – F1 – F2 = (m1 + m2)a a R F1 F2 m m2 1 23 of 67 © Boardworks Ltd 2005 Towing example solution In order to calculate the tension in the coupling, the motion of the car and trailer must be considered separately. In this system, the tension is now an external force. The tension can be calculated by looking at either the motion of the car or the trailer. You could use the unused method to check the answer. Forces acting on the trailer: T is now an external force and must appear on the diagram. 24 of 67 © Boardworks Ltd 2005 Towing example solution Applying Newton’s second law: T – F2 = m2a Substituting in T = F2 + m2a answer to part a). R F1 F2 T F2 m2 m1 m2 m F m F m Rm F m F 2 2 2 2 1 2 2 T 1 2 m m 1 T 25 of 67 m F m Rm F 1 2 2 m m 1 2 2 2 2 © Boardworks Ltd 2005 Towing example solution Forces acting on the car: Applying Newton’s second law: R – T – F1 = m1a T = R – F1 – m1a R F1 F2 T R F m1 1 m1 m2 m R m R m F m F m1R m1F1 m1F2 2 1 1 2 2 T 1 m m 1 2 m F m Rm F 2 2 2 T 1 2 m m 1 26 of 67 2 © Boardworks Ltd 2005 Lift example Example: A man of mass m1 is travelling upwards in a lift of mass m2 which has constant acceleration a. The tension in the cable of the lift is T. Find in terms of m1, m2, g and T, a) The acceleration of the lift. b) The normal contact force, R, between the man and the floor of the lift. 27 of 67 © Boardworks Ltd 2005 Lift example solution When considering the motion of the man and the lift together the normal contact force is an internal force and does not appear on the diagram. a) Applying Newton’s second law: T – m1g – m2g = (m1 + m2)a a 28 of 67 T m1g m2 g m1 m2 © Boardworks Ltd 2005 Lift example solution In order to calculate the normal contact force between the man and the floor of the lift, the motion of the man and the lift must be considered separately. The normal contact force can be calculated by looking at either the man or the lift. You could use the unused method to check the answer. b) Forces acting on the man: R is now an external force and must appear on the diagram. 29 of 67 © Boardworks Ltd 2005 Lift example solution Applying Newton’s second law: R – m1g = m1a R = m1a + m1g Substituting in answer to part a). R m1 R R 30 of 67 T m1g m2 g m1 m2 m1g m1T m12 g m1m2 g m12 g m1m2 g m1 m2 m1T m1 m2 © Boardworks Ltd 2005 Lift example solution Forces acting on the lift: Newton’s third law dictates that the force that the man exerts on the floor of the lift is equal and opposite to the force on the man. Applying Newton’s second law: T – N – m2g = m2a N = T – m2g – m2a N T m g m 2 N N 31 of 67 2 So, N = R T m1g m2 g m1 m2 m1T m T m1m2 g m22 g m T m1m2 g m 2 g 2 m1T m1 m2 m1 m2 2 2 © Boardworks Ltd 2005 Contents Pulleys Pulleys Internal and external forces Towing Friction on a moving particle Examination style questions 32 32 of of 67 67 © Boardworks Ltd 2005 Towing question 1 A car is pulling a caravan along a straight horizontal road. The mass of the car is 1200 kg and the mass of the caravan is 900 kg. The car and caravan are subject to resistive forces of 600 N and 550 N respectively. There is a constant driving force of 2200 N acting on the car. a) Find the magnitude of the acceleration of the car and caravan. b) Find the tension in the towbar. 33 of 67 © Boardworks Ltd 2005 Towing question solution a) Applying Newton’s second law to the whole system: 2200 – 600 – 550 = (900 + 1200) × a 2100a = 1050 a = 0.5 Therefore the acceleration of the car and caravan is 0.5 m s–2. 34 of 67 © Boardworks Ltd 2005 Towing question solution b) Applying Newton’s second Law to the car only, 2200 – 600 – T = 1200 × 0.5 T = 2200 – 600 – 600 T = 1000 Therefore the tension in the towbar is 1000 N. 35 of 67 © Boardworks Ltd 2005 Towing question 2 A train comprises an engine of mass 40,000 kg coupled to two trucks, each of mass 8000 kg. The train moves along a horizontal road. The resistances to the motion of the engine, middle truck and last truck are 15,000 N, 5000 N and 2000 N respectively. The tension in the coupling between the two trucks is 0. a) Show that the train is decelerating and find the magnitude of this deceleration. b) Find the tension in the coupling between the engine and the middle truck. c) Determine whether the engine is exerting a braking or driving force, and find the magnitude of this force. 36 of 67 © Boardworks Ltd 2005 Towing question solution a) Applying Newton’s second law to the last truck: 0 – 2000 = 8000a –2000 a= = –0.25 8000 Therefore the train is decelerating at a rate of 0.25 m s–2. 37 of 67 © Boardworks Ltd 2005 Towing question solution b) Applying Newton’s second law to the middle truck: T – 5000 = 8000 × –0.25 T = 5000 – 2000 T = 3000 Therefore the tension in the towbar is 3000 N. 38 of 67 © Boardworks Ltd 2005 Towing question solution Use F = ma on the whole train: E – 15,000 – 5000 – 2000 = 56,000 × –0.25 E = 15,000 + 5000 + 2000 – 14,000 E = 8000 Therefore the engine is exerting a driving force of 8000 N. 39 of 67 © Boardworks Ltd 2005 Contents Pulleys Pulleys Internal and external forces Towing Friction on a moving particle Examination style questions 40 40 of of 67 67 © Boardworks Ltd 2005 The coefficient of friction Friction is a very common force that acts on objects moving relative to each other (for example a block sliding along a table) to eventually slow them down. Many of the examples involving moving objects have involved a resistive force. This is often due to friction. Friction depends on the roughness of the bodies touching and on the normal contact force. The roughness is characterised by the coefficient of friction, μ, and the frictional force is then F = μR. As μ gets closer to 1, the rougher the contact between the surfaces. As μ gets closer to 0, the smoother the contact. 41 of 67 © Boardworks Ltd 2005 Question 5 A children’s slide is inclined at an angle of 25° to the horizontal. A child of mass 30 kg goes down the slide with a constant speed. a) Draw a diagram showing all the forces. b) Calculate R, the normal contact force. c) Calculate the coefficient of friction between the child and the slide. a) R a=0 F 25° 42 of 67 30g © Boardworks Ltd 2005 Question 5 Solution b) Resolving perpendicular to the plane, R = 30g cos25° R = 266 (to 3 s.f.) Therefore R is 266 N. c) Applying Newton’s Second Law down the plane, 30g cos65° – F = 0 F = 124 (to 3 s.f.) but, F = R = 124 ÷ 266 = 0.466 (to 3 s.f.) 43 of 67 © Boardworks Ltd 2005 Question 6 A children’s slide is inclined at an angle to the horizontal. A child of mass 25 kg goes down the slide with a constant speed. Calculate given that = 0.5. a=0 R = 0.5 F θ 44 of 67 25g © Boardworks Ltd 2005 Question 6 Solution Resolving perpendicular to the plane, R = 25g cos Applying Newton’s Second Law down the slope, 25g cos(90 – )° – F = 0 F = 25g cos(90 – )° We know that cos(90 – )° = sin, so F = 25g sin. =F R 0.5= 25g sin 25g cos We know that sin ÷ cos = tan, so cancelling gives us: tan = 0.5 45 of 67 = 26.6° (to 3 s.f.) © Boardworks Ltd 2005 Question 7 A small wooden block is projected across a horizontal floor. Initially it has a speed of 3 ms–1, and it comes to rest after travelling 2 m. a) Find the acceleration of the block. b) Find the coefficient of friction between the block and the floor. R m a F mg 46 of 67 © Boardworks Ltd 2005 Question 7 Solution a) Using v2 = u2 + 2as with s = 2, u = 3 and v = 0: R m a = –2.25 F 0 = 9 + 4a a = –2.25 The block is therefore decelerating at 2.25 ms–2. mg b) Resolving perpendicularly, R = mg Applying Newton’s Second Law: –F = m × –2.25 F = 2.25m m = 0.230 (to 3 s.f.) μ = F = 2.25 mg R 47 of 67 © Boardworks Ltd 2005 Question 8 A lift is accelerating upwards at 2 ms–2. A woman of mass 60 kg is standing in the lift. Find the normal contact force between the woman and the floor of the lift. Applying Newton’s Second Law: R – 60g = 60 × 2 R 60g 48 of 67 a = 2 ms–2 R = 60g + 120 R = 708 Therefore the normal contact force is 708 N. © Boardworks Ltd 2005 Question 9 A particle of mass m kg is sliding down a rough plane inclined at 20° to the horizontal. The coefficient of friction between the particle and the surface of the plane is 0.2. Find the acceleration of the particle down the slope. a=0 μ = 0.2 R F 20° 49 of 67 mg © Boardworks Ltd 2005 Question 9 Solution Resolving perpendicular to the plane, R = mg cos20° F = R = 0.2mg cos20° Applying Newton’s Second Law down the plane: mg cos70° – F = mga F = mg cos70° – mga = 0.2mg cos20° mg cos70 0.2mg cos 20 a= = cos 70° – 0.2 cos 20° mg = 0.154 (to 3 s.f.) the particle is accelerating at 0.154 ms–2. 50 of 67 © Boardworks Ltd 2005 Contents Examination-style questions Pulleys Internal and external forces Towing Friction on a moving particle Examination-style questions 51 51 of of 67 67 © Boardworks Ltd 2005 Examination-style question 1 Two particles, A and B, of masses 1 kg and 2 kg respectively, are connected by a light, inextensible string which passes over a smooth, fixed pulley. This pulley is at the top of a rough plane inclined at an angle of 30° to the horizontal. A is resting on the plane 2.5 m from the pulley. B is hanging freely over the pulley 1 m above the ground. The coefficient of friction between particle A and the slope is 0.25. a) Find the acceleration of the system when it is released from rest. b) Find the time taken for B to reach the ground. c) Find the distance travelled by A up the slope after B has reached the ground. 52 of 67 © Boardworks Ltd 2005 Solution 1 a) Resolving perpendicular to the plane, R = gcos30° Using F = R: F = ¼gcos30° = 2.12 (to 3 s.f.) Consider A: T – 2.12 – gcos60° = a Consider B: 2g – T = 2a Adding: 2g – 2.12 – gcos60° = 3a 1 a = (2g – 2.12 – gcos60°) = 4.19 (to 3 s.f.) 3 Therefore the acceleration of the system is 4.19 m s–2. 53 of 67 © Boardworks Ltd 2005 Solution 1 For B , u=0 a = 4.19 s=1 t=? Using s = ut + ½at2 1 = 0 + 0.5 × 4.19 × t2 t2 = 0.477… t = 0.691 (to 3 s.f.) Therefore B takes 0.691 seconds to reach the ground. 54 of 67 © Boardworks Ltd 2005 Solution 1 55 of 67 © Boardworks Ltd 2005 Solution 1 Consider A until B reaches the ground: u=0 a = 4.19 v = velocity t = 0.691 v=? Using v = u + at: v = 0 + 4.19 × 0.691 = 2.90 (to 3 s.f.) So, the velocity of particle A at the moment that B reaches the ground, is 2.90 m s–1. 56 of 67 © Boardworks Ltd 2005 Solution 1 When B reaches the ground the string becomes slack and there is no longer a force pulling A up the slope. Using F = ma up the slope: –2.12 – gcos60° = a a = –7.02 (to 3 s.f.) For A: u = 2.90 a = –7.02 v=0 s=? Using v2 = u2 + 2as: (v 2 u 2 ) 2.902 s= 2a 14.04 s = 0.599 (to 3 s.f.) Therefore, A travels a further 0.599 m up the slope after B stops. 57 of 67 © Boardworks Ltd 2005 Examination-style question 2 Two particles, A and B, have masses 3m kg and km kg respectively, where k>3. The particles are connected by a light, inextensible string which passes over a smooth, fixed pulley. g The system has an acceleration of m s–2 when it is 5 released from rest. a) Find the tension in the string in terms of m and g. b) Find the value of k. 58 of 67 © Boardworks Ltd 2005 Solution 2 3mg T – 3mg = 18mg 5 T= 5 b) Consider particle B: kmg – 18mg kmg 5 5 4kmg 18mg so, 4kmg = 18mg 5 5 4k = 18 Therefore, k = 4.5 a) Consider particle A: 59 of 67 © Boardworks Ltd 2005 Examination-style question 3 A car of mass 850 kg is being towed by a tow-truck of mass 1500 kg. The car and the tow-truck are joined by a light towbar, which is at an angle of 20° to the ground. The car and the tow-truck experience resistances to motion of 250 N and 600 N respectively. The driving force exerted by the tow-truck is 3200 N. a) Find the acceleration of the car and tow-truck. b) Find the tension in the towbar. 60 of 67 © Boardworks Ltd 2005 Solution 3 Applying Newton’s second law to the whole system: 3200 – 600 – 250 = 2350 × a 2350a = 2350 a=1 Therefore the acceleration of the system is 1 m s–2. 61 of 67 © Boardworks Ltd 2005 Solution 3 Applying Newton’s second law to the car: Tcos20° – 250 = 850 × 1 850 250 T= cos20 T = 1170 (to 3 s.f.) Therefore the tension in the towbar is 1170 N. 62 of 67 © Boardworks Ltd 2005 Examination-style question 4 A particle of mass m kg slides down a rough plane inclined at 25° to the horizontal. The particle passes through a point A with speed 3 ms–1, and 2 seconds later passes a point B with speed 9 ms–1. a) Find the acceleration of the particle. b) Find F in terms of m. c) Find the coefficient of friction between the particle and the surface of the slope. 63 of 67 © Boardworks Ltd 2005 Solution 4 93 a) Acceleration = = 3 ms–2 2 a = 3 ms–2 b) R F 25° mg Applying Newton’s Second Law down the plane: mg cos65° – F = m × 3 F = mg cos65° – 3m c) Resolving perpendicular to the plane: R = mg cos25° mg cos65 3m F = = = 0.129 (to 3 s.f.) R mg cos25 64 of 67 © Boardworks Ltd 2005 Examination-style Question 5 A sledge of mass 70 kg is pulled up a slope inclined at an 7. angle to the horizontal where tan = 40 The slope is modelled as a rough inclined plane and the rope as a light, inextensible string acting parallel to the line of greatest slope. The coefficient of friction between the sledge and the slope is 0.3 and the sledge is accelerating up the slope with an acceleration of 0.4 ms–2. Find the tension in the rope. 65 of 67 © Boardworks Ltd 2005 Solution 5 a = 0.4 ms–2 R T = 0.3 F θ 70g Before we resolve the forces, look at tan = 7 40 . From the right-angled triangle with smaller sides of length 7 and 40, which has hypotenuse length 402 + 72 ≈ 41, we can deduce the values of sin and cos : sin ≈ 7 ÷ 41 66 of 67 and cos ≈ 40 ÷ 41 © Boardworks Ltd 2005 Solution 5 Resolving perpendicular to the plane, R = 70g cos R = 669.3 F = R = 0.3 × 669.3 = 200.8 Applying Newton’s Second Law up the plane, T – 200.8 – 70g cos(90 – )° = 70 × 0.4 T – 200.8 – 70g sin = 70 × 0.4 T = 200.8 + 117.1 + 28 T = 346 (to 3 s.f.) Therefore the tension in the rope is 346 N. 67 of 67 © Boardworks Ltd 2005