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MECHANICS M1 SLBS DYNAMICS NEWTON’S LAWS 1st LAW A particle will remain at rest or will continue to move with constant velocity in a straight line unless acted upon by a resultant force. 2nd LAW The Force F applied to a particle is proportional to the mass m of the particle and the acceleration produced. F = ma 3rd LAW Every action has an equal and opposite reaction. 1 Newton is defined as the force that gives a mass of 1 kg an acceleration of 1 ms-2 Example A force of 10N acts on a particle of mass 2kg. Find the acceleration. F = ma 10 = 2a a = 5ms JMcC –2 1 MECHANICS M1 SLBS Example A car of mass 500kg is accelerating at 3ms-2. The resistance 3ms to motion due to friction is 400N. Find the driving force of the engine. Always draw a force diagram –2 3ms 500kg –2 D 400N Note: when using F = ma, F stands for Resultant Force R(→) stands for “Resolving forces horizontally & use F = ma” R( ) D – 400 = 500 3 D = 1500 + 400 = 1900N VERTICAL MOTION Newton’s 2nd Law implies that the force on an object = mass × g Weight = mg Example A string is pulling up a particle of mass 4kg with an acceleration of 1 ms-1. Find the tension in the string T 4 kg R(up) T – 4g = 4 1 T = 4 + (4 9·8) 4g = 43·2 N Ex 3A p 40 JMcC 2 MECHANICS M1 SLBS TYPES OF FORCE Weight String in Tension W = mg Spring/Rod in Thrust Normal Reaction Friction direction of motion W R F W Worksheet on Marking Forces Ex 3B p 44 JMcC 3 F MECHANICS M1 SLBS RESOLVING FORCES INTO COMPONENTS Q F O F cos sin P But OQ = OP + PQ OP = F cos and PQ = F sin Therefore F F sin is equivalent to F cos Component of a force F in any direction = F × Cosine of the angle between the force and the specified direction Note: cos(90 – ) = sin Example 8N 60 JMcC 8 sin 60 = 6.93 is equivalent to 8 cos 60 = 4 4 30 60 60 = 11133 tan sin cos 60 = cos 30 tan sin 60 30 === 22223 MECHANICS M1 SLBS TRIG RATIOS FOR 30°, 45° and 90° 1 sin 30 = 30 2 2 2 3 cos 30 = tan 45 2 45 = 11 1 60 sincos 4545 = = 2 2 1 30sin 8 cos 30 30 1 tan 30 = 1 sin 45 = 45 2 2 1 cos 60 = tan 60 = 3 2 1 2 3 3 1 2 cos 45 = 45 sin 60 = 1 2 tan 45 = 1 1 Example 8 sin 30 8N is equivalent to 30 8 cos 30 = 8 8 sin 30 = 8 3 2 1 2 8 cos 30 = 4 3 = 4 Ex 3C p 47 JMcC 5 MECHANICS M1 SLBS FRICTION F = 2NFriction always opposes motion. Consider an object being pulled along a rough horizontal surface by a pulling force P F = 3N F When P=2N, object is at rest P = 2N F = 2N F P = 3N When P=3N, object is about to move F = 3N F = 3N This is the maximum friction force F P = 4N When P=4N, object is moving F = 3N only It can be shown experimentally that: F R where = coefficient of friction max F = R JMcC 6 MECHANICS M1 SLBS Example R = 6g A particle of mass 6kg is attempted to be pulled along a rough horizontal surface. When the pulling is 35N, the particle is about to move. Calculate the value of μ. R F R = 6g 35N 6g R( ) F = 35 and this is maximum friction = F R = 35 6g = 0·60 to 2dp NOTE: When a force is applied to an object on a horizontal surface that is inclined at an angle to the horizontal, the Normal Reaction does not equal the Weight due to the effect of this force. R P F R + Psin = W F = Pcos W JMcC 7 MECHANICS M1 SLBS Example R 14 2 N 45 F = 1 7 Find the acceleration 8g a R + 14 2 sin 45 = 8g 1 R = 8g – 14 2 F = 1 2 = 8g – 14 = 64·4 N 64·4 = 9·2 N 7 R( ) 14 2 cos 45 – F = 8a 14 2 1 2 – 9·2 = 8a 14 – 9·2 = 8a 4·8 = 8a a = 0·6 ms –2 If a body is said to be in “limiting equilibrium”, this means that it is “about to move” or is “on the point of moving.” Ex 3D p 52 JMcC 8 MECHANICS M1 W cos sin SLBS COMPONENTS OF THE WEIGHT OF AN OBJECT PLACED ON A SLOPE is equivalent to W sin W cos W Component of Weight down the slope = W sin Component of weight into the slope = W cos You must remember these! CALCULATING THE COEFFICIENT OF FRICTION R F Limiting Equilibrium W R = W cos F = W sin F = R = F R = W sin W sin = tan If tan > object will accelerate down the slope If tan object will remain at rest JMcC 9 MECHANICS M1 SLBS –2 25 Example 2ms Force P is pulling a particle of mass 8kg up an inclined smooth slope of 25° at an acceleration of 2ms-2. Find P 2ms R –2 P 8 kg 25 8g R(up slope) P – 8g sin 25 = 8 2 P = 16 + 8g sin 25 = 49·1 N Sometimes, the angle of the slope may be given in the form 3 4 x 4angle whose sine is x 4 arcsin i.e. the sin = and tan = = arccos y 55 y 3 Do not use Trigonometry and calculate the angle, rather use Pythagoras: = arccos 3 5 5 4 3 sin = 4 5 and tan = 4 3 If you find θ = 53.13…° and use 53.1, you will lose accuracy JMcC 10 MECHANICS M1 SLBS Example 30 A particle of mass 10kg rests in limiting equilibrium on a slope inclined at 30° to the horizontal. Calculate the coefficient of friction between the object and the slope. R F 10kg 30 10g There is no motion, so the forces up and down the slope, must balance each other out. R(up slope) F = 10g sin 30 R( slope) R = 10g cos 30 Max F F = R = F R = 10g sin 30 10g cos 30 = tan 30 = 1 3 Ex 3E p 55 JMcC 11 MECHANICS M1 SLBS LIFTS Example 2ms A lift if mass 500kg contains a parcel of mass 50kg. Find the tension in the lift cable and the reaction between the parcel and the lift floor if the lift is accelerating: a) upwards at 2 ms-2 b) downwards at 2 ms-2 –2 T a) Lift + Parcel 2ms R(up) T – 550g = 550 2 –2 (1) T = 1100 + 550g T = 6490 N 2ms –2 550g Lift Only 2ms T R(up) T – 500g – R = 500 2 (2) 1590 – R = 1000 –2 R = 1590 – 1000 = 590N 2 ms –2 R 500g Parcel Only R(up) R – 50g = 50 2 2 ms –2 R (3) R = 100 + 50g = 590N 50g Note: Equation (2) + Equation (3) gives Equation (1) Therefore options! JMcC 12 MECHANICS M1 2 ms –2 SLBS Always state which objects you are considering and mark up an appropriate diagram b) Lift & Parcel T R(down) 550g – T = 550 2 550g – 1100 = T 2 ms 2 ms T = 4290N –2 –2 550g Parcel Only R R(down) 50g – R = 50 2 50g – 100 = R R = 390N 2 ms –2 50g McEx 5A JMcC 13 MECHANICS M1 SLBS CONNECTED PARTICLES The tension in the tow-bar of a car towing a caravan is represented by: T T Example 0.5ms –2 A car of mass 1200 kg tows a caravan of mass 700 kg along a horizontal road. The car is accelerating at 0.5 ms-2. Ignoring any resistances, find the driving force of the engine and the tension in the tow-bar. 0.5ms 700kg T R( ) Car & Caravan T –2 1200kg D D = 1900 0·5 = 950 N R( ) Caravan only T = 700 0·5 = 350 N JMcC 14 MECHANICS M1 SLBS In practice, there will be resistance to motion. Often, it is proportional to the mass of the objects. Looking again at the previous example where there is a total resistance to motion of 380 N Resistance on car = 1200k where k = constant of proportion Resistance on caravan = 700k 0.5ms –2 Total resistance = 1900k = 380N k = 0·2 Resistance on car = 1200 0·2 = 240 N Resistance on caravan = 700 0·2 = 140 N 0.5ms 700kg T T –2 1200kg D 240 N 140 N R( ) Car & Caravan D – 140 – 240 = 1900 0·5 D = 950 + 380 = 1330 N R( ) Caravan only T – 140 = 700 0·5 T = 350 + 140 = 490 N JMcC 15 MECHANICS M1 SLBS If the car brakes, the tow-bar will go into THRUST T T Example In the previous example, the car reduces the driving force to zero and applies the brakes. Given that the resistances to motion are unchanged and the thrust in the tow-bar is 300N, find the value of the braking force a 700kg 140 N 300 300 B 1200kg 240 N Caravan only 300 + 140 = 700a 440 = 700a a = 0·62857... Car & Caravan B + 380 = 1900 0·62857... B = 1194·285... – 380 = 814N McEx 5B JMcC 16 MECHANICS M1 SLBS PULLEYS Example Particles of mass 3kg and 5kg are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. The system is released from rest. Find the acceleration of the system and the tension in the string. Find T and a T T a T T 3 kg 5 kg a 3g R(↓) 5 kg R(↑) 3 kg 5g 5g T − − (1) + (2) Subst in (2) T − T 3g = = 5a 3a (1) (2) 2g a = = 8a g/4 m/s2 (= 2.45 m/s2) 3g T T = = = 3g/4 3g/4 + 3g 15g/4 N (= 36.75 N) Note: Smooth pulley =>Tension in both portions of the string is the same. Inextensible string =>Acceleration of both particles is the same (and they have the same speeds at any given instant) Exam questions often ask where in your solutions you have used these assumptions. McEx 5C JMcC 17 MECHANICS M1 SLBS Example (Horizontal Table) Two particles A and B of masses 4kg and 10 kg respectively are connected by a light inextensible string. Particle A rests on a rough horizontal table where the coefficient of friction between A and the table is ½. The string passes over a smooth pulley fixed at the edge of the table and B is hanging freely. Find a) the acceleration of the system b) the tension in the string R a A T T Find T and a T F 4g T a 10g R(↑) R R(→) A R(↓) B = 4g T − 10g − (1) + (2) Subst. in(1) T JMcC − => F = ½ × 4g 2g = T = 4a (1) 10a (2) 8g = a = 14a 8g/14 2g = T = T = 16g/7 16g/7 + 2g 30g/7 = 42 N = = = 2g 4g/7 ms−2 5.6 ms−2 18 MECHANICS M1 SLBS Note: Force on the pulley T sin 45 T is equivalent to T cos 45 T T sin 45 T cos 45 T sin 45 components cancel each other out! Force on Pulley = = = = 2T cos 45 2 × 30g/7 × 1/√2 30g√2/7 59.4 N McEx 5D JMcC 19 MECHANICS M1 SLBS Example (Pulley on a Slope) A package A of mass 26 kg rests on a smooth inclined plane of 5 angle arcsin . A light inextensible rope is attached to A 13 and passes over a smooth fixed pulley at the top of the plane. Another package B of mass 30 kg hangs freely from the other end of the rope. The system is released from rest. Find the acceleration of the system and the tension in the rope. a T R T 26g a 30g R(down) B R(up slope) A 30g – T = 30a (1) T – 26gsin = 26a T – 26g 5 13 = 26a T – 10g = 26a (1) + (2) 20g = 56a a = Subst. in (2) 20g 56 = T – 10g = 26 T = JMcC (2) 270g 14 = 135g 7 5g 14 5g 14 = 3·5 ms = –2 130g 14 = 189 N 20 MECHANICS M1 SLBS PROBLEMS WHERE FORCES CHANGE Example Two particles A and B of masses 4kg and 5kg respectively, are connected by a light inextensible string which passes over a smooth fixed pulley. They are originally held so that both particles are 2m above the ground. The system is released from rest. B hits the ground and does not rebound. Find A’s greatest height above the ground. a T 5kg 4g (1) + (2) 5g − T − B A 4kg R(↓) B R(↑) A T T = 4g = a 5a (1) 4a (2) g = 9a a = JMcC 5g 2m g 9 21 MECHANICS M1 SLBS We need to find the speed that A and B have when B hits the ground Use constant acceleration equations 2 u = 0 a = 2 v – u = 2as g 2 g 2 v – 0 = 2 9 s = 2 v = 9 2 4g g 0 h v A Now A acts under gravity only, so its acceleration = -g 4m B u = 4g 9 a = – g 2 2 v – u = 2as 0 – v = 0 4g 9 = – 2gh h = 2 9 m s=h Greatest Height reached by A = 4 + h = 4 2 9 m Old M1 Book Ex 5D Q 13-16 Ex 3F p64 JMcC 22 MECHANICS M1 SLBS MOMENTUM and IMPULSE v = u + at a = F = ma = v – u t m(v – u) t Ft = mv – mu (*) MOMENTUM = mass × velocity = mv IMPULSE = Force × time Ft = Both quantities are measured in Newton-seconds (Ns) (*) => Impulse = Change in Momentum Example A particle of mass 3 kg increases its speed from 2 ms-1 to 5 ms-1 Find its gain in momentum Initial momentum Final momentum Gain in momentum = = = 3×2 3×5 15 − 6 = = = 6 NS 15 Ns 9 Ns Example A ball of mass 100g hits a wall at 10m/s and rebounds at 8 m/s. Find the impulse exerted by the wall on the ball. 10 m/s 0.1 kg 8 m/s Ex 3G p 68 JMcC Initial Momentum = 0.1 10 = 1 Ns Final Momentum = 0.1 ( – 0.8) = – 0.8 Ns Impulse = Change in Momentum = 1 – ( – 0.8) = 1 + 0.8 = 1.8 Ns Note: Momentum may be positive or negative 23 MECHANICS M1 SLBS The Principle of Conservation of Momentum When two particles A and B collide, they exert equal and opposite forces, and hence impulses, on each other. The impulse that A exerts on B (equal to B’s change in momentum) is therefore equal and opposite to the impulse that B exerts on A (equal to A’s change in momentum). If these changes are equal and opposite, there is no change in the total amount of momentum of both particles taken together. Hence, we have the Principle of Conservation of Momentum: Unless acted upon by external forces, the total momentum of a system remains constant. So, the total momentum before impact equals the total momentum after impact. Example A particle A of mass 3kg travelling in a straight line at 5 m/s catches up with another particle B of mass 2kg moving in the same direction at 2 m/s. After the collision, A has a speed of 1 m/s in its original direction. Find the speed of particle B Before 3kg After C of M 5 m/s 2 m/s 2kg 1 m/s v 2v + (3 1) = (2 2) + (3 5) 2v + 3 = 4 + 15 2v = 16 v = 8 m/s JMcC 24 MECHANICS M1 SLBS If two objects stick together, they are said to COALESCE Example A particle of mass 4kg travelling at 3 m/s catches up with a particle of mass 2kg travelling in the same direction at 2 m/s. On impact, the particles coalesce. Find the speed of the combined particle. Before 3 m/s 4kg 2kg After 6kg C of M 2 m/s v 6v = (4 3) + (2 2) 6v = 12 + 4 6v = 16 v = JMcC 8 3 m/s 25 MECHANICS M1 SLBS Remember that momentum can be negative Example Two particles of masses 8kg and 5kg are approaching each other with speeds of 7 m/s and 3 m/s respectively. After the collision, the lighter particle reverses its direction and has a speed of 6 m/s. Find the speed of the heavier particle after the collision. Before 8kg After 3 m/s 7 m/s 5kg v 6 m/s Decide on which direction to take as positive. In this case, we shall take from left to right (→) as positive. C of M 8v + (5 6) = (8 7) – (5 3) 8v + 30 = 56 – 15 8v = 11 v = 11 8 m/s Ex 3H p72 JMcC 26 MECHANICS M1 SLBS JERKS The principle of conservation of momentum also applies when a string is jerked taut. Example Two particles of masses 4kg and 5kg are joined together by a light inextensible string. Initially they are at rest with the string slack. The 5kg particle is projected away from the other particle with a speed of 2 m/s. Find their common speed when the string jerks taut. Before 5kg 4kg After 2 m/s v v After the jerk, both particles move with the same speed. C of M 4v + 5v = 5 2 9v = 10 v = 10 9 m/s Old M1 Book Ex 5F p 123 Q 13-18 Mixed Ex 3I p73 JMcC 27