M1 Notes September 2011

Transcription

1 M1 Notes RM September 011 1

2 Lesson Topics Skills Notes Book 1 Mathematical models Use of mathematical modelling understanding terminology p. 5 pp -3 in mechanics Kinematics Scalar and vector properties. pp.6-7 pp 4-10, ex. A Motion in a straight line with constant acceleration. More constant acceleration formulae pp 8-10 pp 10-17, ex. B 3 Vertical motion under gravity pp pp 17-3, Ex. C 4 Constructing and interpreting speed-time and velocity-time graphs pp pp 4-31, Ex. D 5 Consolidation mixed questions pp 3-35 Ex E 6 Introduction to forces Forces, mass, weight pp 15 pp Dynamics Applying Newton's laws of motion F = ma pp pp Ex 3A 7 Vertical motion F = ma Ex 3B 8 Force diagrams, resultant, resolving, adding components pp pp 4-45 Ex 3C 9 Using friction and the coefficient of friction p 0 pp 48-5, Ex 3D 10 Normal and frictional forces on inclined planes pp 1-4 pp 53-56, Ex 3E 11 Statics of a particle Finding the resultant of forces pp 5-6 pp 9-96, Ex 4A 1 Solving problems involving the equilibrium of coplanar forces pp 97-10, Ex 4B 13 More equilibrium pp 7-8 pp , Ex 4C 14 More normal and frictional forces p 111, Ex 4D Further dynamics Connected particle problems p. 9 pp 56-64, Ex 3F 17 Connected particles with pulleys pp 30-3 Ex 3F 18 Consolidation 19 Momentum & impulse Understanding and using momentum and impulse p 33 pp 65-68, Ex 3G 0 Conservation of momentum, impacts, jerk in string pp pp 68-7, Ex 3H 1 Consolidation p73, Ex 3I + p80 Review Ex 1 Moments Understanding the moment of a force and the sum of moments pp pp , Ex 5A + 5B 3 Moment using centre of mass of beams p. 38 pp 1-15, Ex 5C 4 Problems involving equilibrium and moments (includes non-uniform beams + pp 16-18, Ex 5D tilting) 5 Consolidation P18, Ex 5E 6 Vectors Understanding properties of vectors. Adding and subtracting vectors. pp pp , Ex 6A + 6B 7 Use of Cartesian unit vectors. Finding the magnitude of a vector. pp 4-44 pp , Ex 6D 8 Finding position, relative positions, velocity, relative velocity and acceleration pp pp , Ex 6F vectors 9 Forces as vectors p 46 pp , Ex 6F 30 Mixed questions Ex 6H, Review Ex.

3 CONTENTS MATHEMATICAL MODELS IN MECHANICS... 5 KINEMATICS... 6 Introduction - vectors in mechanics... 6 Motion in a straight line with constant acceleration Further constant acceleration formulae... 8 Vertical motion under gravity Distance-time graphs Speed-time graphs Acceleration-time graphs FORCES ON A PARTICLE Introduction to Forces Mass and Weight Newton s Laws of Motion Force diagrams Components of a force Adding force components Friction and the coefficient of friction... 0 Particle on an inclined plane.... Statics of a Particle (equilibrium of coplanar forces)... 5 DYNAMICS OF TWO CONNECTED PARTICLES... 9 Problems involving pulleys Pulleys with slope and bouncing MOMENTUM and IMPULSE Conservation of momentum Jerk in a string EXTENDED BODY PROBLEMS (MOMENTS) Resultant moment Particles and bodies in equilibrium VECTOR POSITIONS, VELOCITIES AND FORCES Displacement problems Properties of vectors Cartesian vector notation Addition of vectors given in Cartesian notation Basic calculations using vectors Types of vectors used in mechanics Position vectors Displacement (relative position) vectors Velocity vectors Acceleration vectors Equations of motion in vector form Problems where paths intersect Force vectors FORMULAE

4 Kinematics Describing the motion of a particle Particles Dynamics How things move when pushed Mechanics 1 Statics Forces that balance Rigid bodies Moments Kinematics in dimensions Vectors 3 steps to solve any mechanics problem Draw a sketch and define variables Write an equation or two Solve them. Simples! 4

5 MATHEMATICAL MODELS IN MECHANICS Mechanics is the science concerned with the action of forces on bodies, in particular: If an object is in equilibrium (not accelerating), the forces on it must balance If there is a net force on it, how does it move? Mathematical Modelling. We have to make simplifying assumptions so that we can approximate a complex physical situation with a mathematical model that is simple enough to allow quick and easy calculations. The choices between effects that matter and those that probably do not are known as modelling assumptions. It is often useful to make some very considerable assumptions (such as ignoring wind resistance) to obtain a simple formula that can be easily used to a reasonable accuracy. Examples: In real life, an engineer might model a piston in a cylinder as: (a) A point mass moving up and down (centre of mass vibration problems) (b) A cylindrical lump of metal sliding on an oil film (lubrication problems) (c) Aluminium alloy with heat transfer in and out (metal temperature studies). These are all modelling assumptions. Similarly in M1 when modelling the motion of a projectile one might (a) Neglect wind resistance (b) Assume the earth is flat and that gravity acts straight down Basic terminology in modelling in Mechanics (for help in interpreting questions). A body is any object to which a force can be applied. A particle is a body whose dimensions, except for mass, are negligible. The mass of the body can be considered to be concentrated at a single point. o A bead is a particle with a hole in it for threading on a wire or string. o A rigid body is one whose shape is unaltered by any force applied to it. Unlike a particle, it might rotate. A wire is like a metal string. A rod is an object all of whose mass is concentrated along a line. We pretend it is infinitely thin even though it is usually stiff. o A uniform rod is one in which equal lengths have equal masses, so the centre of mass is half-way along (a good model for a plank of constant thickness). A lamina is a flat body having area but negligible thickness. o A uniform lamina is one in which equal areas of lamina have equal masses (a good model for a metal sheet of constant thickness) A pulley is a wheel with a rope passing over it. The wheel is free to turn. o A smooth pulley is one with no friction in its bearings. A peg is a pin from which a body can be hung or on which it can rest. o A smooth peg (with a rope over it) is equivalent to a smooth pulley A plane surface is a completely flat surface. A smooth surface is frictionless, an object can slide with no frictional force. A rough surface is one where friction has to be taken into account. A light object (body, string, lamina, etc.) is one whose mass is negligible. Even if it is accelerating, the forces on it balance. A rigid rod or an inextensible or inelastic string is one whose length never changes (it is not stretchy, all points along it have the same speed and acceleration). In M1 questions about modelling assumptions, you are simply expected to spot simplifications and describe them using terms such as no wind resistance, particle, light, inextensible etc. 5

6 KINEMATICS Introduction - vectors in mechanics Physical quantities are of two kinds: 1. Scalar quantities or scalars which have magnitude but no associated direction, for instance: Mass (10 kg) Temperature (30 C) Pressure (1 bar) Distance (0 miles), Speed (5 m/s). Vector quantities or vectors which have both magnitude and an associated direction, for instance Displacement from one point to another Velocity (16 m/s in the y-direction, 16j m/s ) Acceleration Force Notes. Speed is the magnitude of velocity and distance is the magnitude of displacement. We use many vector formulae e.g. force = mass acceleration, F = ma vector scalar vector Gravitational force F Acceleration a } same direction Books use normal text for scalar quantities (m) and bold for vectors (F, a). When writing one tends to underline vector letters (since one cannot write in bold) F = ma. Motion in a straight line with constant acceleration. For the simple case where an object is moving in a straight line and has a constant acceleration in the line direction we can use scalar values u, v instead of velocity vectors u, v. These are components of the velocity vector in the straight line direction (so they can be negative, whereas speed is always positive). We tend to still call them velocity; really we mean velocity in the line direction. Let: s = displacement (metres) a = acceleration ( ms ) t = time (seconds) 1 u = initial velocity ( ms ) 1 v = final velocity ( ms ) With constant acceleration, change in velocity acceleration time 6

7 Hence v u a and v u at t Aside: This is a special case of the more general vector form handles changes in magnitude and direction of the acceleration. a du dt that properly distance travelled average speed time (this is what we mean by an average speed ) Since acceleration is constant (a straight line on a speed-time graph), speed u v s t u t v u v time Example. A body moves along a straight line from A to B with a uniform acceleration of 1s and the velocity at B is 5 ms 1. Find: 3 ms. The time taken is (a) the velocity at A (b) the distance AB First, define the question using letters as per the formulae: a = 3 ms, t = 1s, v = 5 ms 1 (a) Using v u at 5 u 1 3 u 17ms 1 (b) Using u v s t 17 5 s 1 5 m Edexcel Page 9 Ex. A 7

8 Further constant acceleration formulae The above formulae link the 5 variables u, v, a, s and t We only need 3 of these to define any problem, so we can re-arrange to eliminate u, v or t: Eliminating v from equations 1 and gives: u v s t, u v s t Hence u u at u at s t t t 1 s ut at Eliminating u from equations 1 and gives: u v at, 1 s vt at v at v v at s t t t Eliminating t from equations 1 and gives: a v u t so v u t a u v u v s t v u v u a a v u as Notes. (1) Since displacement, velocity and acceleration are really all vector values, it is necessary to decide which direction is positive and to define values in the opposite direction as negative () v u as is sometimes called the energy equation. Multiplying by 1 m gives 1 mv 1 m which we can interpret as u ma s final kinetic energy = initial kinetic energy + work done on object This particular equation is always valid with scalar speeds u, v provided a and s are both in the same direction. 8

9 Example 1. A train sets off with constant acceleration from a station S. 10 seconds later it passes a signal P at a speed of 63 km/hr. Modelling the train as a particle, find its acceleration in m/s and the distance between S and P. As a particle means that we do not need to worry about the length of the train (is it the front or back that leaves the station at t=0) and we can assume the whole train travels at the same speed there is no slack in the couplings. First get a consistent set of units: km 1000 m 1 hr 63 km/hr 63 hr 1 km 3600 seconds Note the unit multipliers which = 1 (since equivalent top and bottom). The pairs of km and hr cancel out to leave us with m/s units. km 1000 m 1 hr km/hr 63 m/s 17.5 m/s hr 1 km 3600 seconds 3600 u 0 vu m/s a t 10 Then either: or 1 1 s ut at m uv s t m Example A car travels with a constant speed of 4 ms -1 along a straight road. It passes a stationary police car which sets off seconds later in pursuit of the speeding vehicle. The police car accelerates at 6 ms -. How long after being passed does the police car take to draw level with the other car? Leave your answer in an exact form. Car. u = 4 ms -1 Police car u = 0 ms -1 a = 0 ms - a = 6 ms - v = 4 ms -1 v =? Passes police at t=0 secs Time since setting off t t s =? s =? secs Car s ut 1 at s 4t 0 s 4 t Police car 1 s ut at 1 s 0 6t 3 t 9

10 When the cars draw level, their displacements will be equal. Therefore: 4t 3 t 14t t t 18t t t 9 77 t 4t 4 Example 3. A stone slides in a straight line across a horizontal sheet of ice. It passes a point A with velocity 14 1 ms and point B.5s later. Assuming the deceleration is uniform and that AB = 30m, find: (a) the acceleration (b) the velocity at B (c) how long after passing A the stone comes to rest. Define variables: (a) using: (b) using: u 14 m/s, t.5 s, s 30 m 1 s ut at a a 8 a 1.6ms (negative so a retardation, it slows down) v u as v v 10ms (at B) (c) when the stone comes to rest, the velocity v = 0 using: v u at t t 8.75s Note. It would not make sense to continue the calculation after t = 8.75 s. When the motion stops, the frictional force becomes zero. There is no force to make it start moving again. Edexcel Page 15 Ex. B 10

11 Vertical motion under gravity The uniform acceleration formulae can be used when considering the vertical motion of a body falling under gravity. It is really important to define which direction is positive for u, v, a and s (upwards or downwards?). On Earth, the acceleration due to gravity is nominally m/s. This is a measure of the strength of the gravitational field and it defines the weight of stationary objects as well as the acceleration of those in free-fall. We always use the letter g for this in formulae. We assume that wind resistance is negligible. For interest: In real life, every object falling through the atmosphere asymptotes towards a terminal velocity at which the wind resistance is equal to its weight. Heavy, dense, streamlined objects have a very high terminal velocity (Tallboy WWII bomb 1100 m/s) Small, lightweight, unstreamlined objects have lower terminal velocities (Parachutist 5m/s) Tiny particles (pollen, bacteria etc) have terribly low terminal velocities, less than 1 mm/second v No resistance, v gt Terminal velocity Man Cat? t Further very minor assumptions (extension, for Oxbridge applicants!) 1. g is constant. In fact, g varies very slightly 1 r it reduces as one goes away from the centre of the Earth (so < 9.81m/s for a satellite). It is effectively less at the equator due to the rotation of the Earth ( centripetal force ) and because the Earth s radius is greater at the equator (it is slightly flat at the poles). It varies slightly depending on the density of the local rocks So there is no point in using g to 6 significant figures. People usually use g = 9.81 or g=9.8 m/s.. If the object was falling hundreds of miles we would have to model it as following part of an orbit, not going straight down. 11

12 Example 1. A brick is thrown vertically downwards from the top of a building and has an initial velocity of 1.5ms -1. If the height of the building is 19 7 m, find: (a) the velocity with which the brick hits the ground (b) the time taken for the brick to fall Defining positive quantities downwards: u 1.5 m/s (downwards, so positive) s 19 m 7 (downwards from top of building, so positive) a 9.8 m/s (gravity pulling downwards, so positive) (a) Using v u as v b15. g b9. 8g v 19. 5ms (b) Using v = u + at 19.5 = t t 184. s. Example. A particle is projected vertically upwards with a velocity of 34.3ms -1. (a) Find how long after projection the particle is at a height of 49m above the point of projection for: (i) the first time (ii) the second time (b) Find its maximum height and the time it takes to reach it. Define upwards as positive. Then (a) Using: u 34.3 m/s a 9.8 m/s s 49 m 1 s ut at 1 Note the - (acceleration is negative): Gravity makes the upwards velocity fall to zero and eventually go negative t 9.8 t clearly a quadratic equation, 4.9t 34.3t 49 0 perhaps divide by 4.9? t 7t10 0 t5 t 0, t = 5 s or s So the particle is 49m above the point of projection: (i) after s (on its way up), (ii) after 5 s (on its way down). (b) At the maximum height, v = 0. as v u v u s m, also v a 9.8 u at, t, t = 3.5 s 1

13 Example 3. 1 A ball is thrown vertically upwards with a velocity of 14.7 ms from a platform 19.6m above ground level. Find: (a) the time taken for the ball to reach the ground (b) the velocity of the ball when it hits the ground Define upwards as positive u 14.7 m/s a 9.8 m/s s 19.6 m at ground level (a) s ut 1 at t 9.8 t 4.9t 14.7t divide by 4.9? t 3t 4 0 t 4 t 1 t 4 or t 1 The ball reaches the ground after 4s. [The t = -1 means that the ball could alternatively have been thrown upwards from ground level at time t = -1 and would have then been level with the platform, moving up at 14.7 m/s, at time t = 0] (b) v u at v m/s The ball hits the ground with a downward velocity of 4.5 m/s. Alternatively we could use v m/s v u as 14.7 ( 9.8)( 19.6) Edexcel Page 3 Ex. C Distance-time graphs Distance-time graphs (as GCSE) show constant velocity (zero acceleration) journeys as straight lines. 1 ds s ut at but a = 0 so s ut and the speed is the gradient of the line, u dt Distance s ds Gradient u dt Time t They are less useful for questions with acceleration since it is harder to sketch and interpret a curve. 13

14 Speed-time graphs. The gradient of a speed-time graph gives the acceleration, du a dt The area under the speed-time graph gives the distance covered. b s udt. It is usually much easier to find the area from our knowledge of geometry than by defining line equations and integrating. a Example. On the London Underground, Oxford Circus and Piccadilly Circus are 0.8km apart. A train accelerates uniformly to a maximum speed when leaving Oxford Circus and maintains this speed for 90s before decelerating uniformly to stop at Piccadilly Circus. The whole journey takes minutes. Draw a sketch of the motion and find the maximum speed. Speed (m/s) 90 s Using the rule for the area of a trapezium, v v 7.6 m/s (to 3 sig. fig.) 10 Acceleration-time graphs. On an acceleration-time graph we can identify regions of zero acceleration (constant velocity), positive acceleration (velocity increases) and negative acceleration (velocity reduces). Area under curve = change in velocity Example. A car accelerates uniformly from rest at 1. m/s for 10 seconds. It then travels at a uniform speed for 5 seconds. Finally, it decelerates uniformly at 0.8m/s for 15 seconds and comes to rest. Sketch an acceleration-time graph of the motion of the car. accel.( ms ) 1. Time (s) time(s) -0.8 Edexcel p.30 Ex. D + p. 3 Ex. E, Summary p

15 FORCES ON A PARTICLE In the book these topics are split between chapters 3 and 4. Both chapters use the same equations for force components, resultant force and friction. These notes contain: Explanation of the theory (with brief examples) Further examples, grouped by geometry Introduction to Forces Forces push or pull objects. In terms of physics, all the forces we experience stem from 3 fundamental effects. These are non-contact forces: Electrostatic forces ( like charges repel, opposite charges attract ) Magnetism (commonly thought of as field lines) Gravity (masses attract each other) Everyday contact forces include: Fluid pressure (e.g. air on an aeroplane wing) Fluid shear (drag on an aeroplane) Deformation of a solid: Tension: Compression: Tension T, string gets longer (T mg if in equilibrium) mass of weight mg Compressive force T making rod shorter Weight mg Friction acting along a surface (e.g. friction between car tyres and road, making the car go around corners without skidding straight on) (All these result from the electrostatic forces between atoms). The unit of force is the Newton, defined as the force needed to make 1 kg accelerate at 1 m/s. All forces are vector quantities that have direction as well as magnitude. We often also need to know where the force is applied (if I push down on a see-saw, it matters where I push, or ) Mass and Weight 1 kg was originally defined as the mass of 1000 cm 3 of water. Nowadays it is defined by a standard kilogram made of platinum alloy and kept in Paris. Mass has two properties: All masses attract each other (gravity) Massive objects have inertia (a force is needed to make them accelerate) The weight of an object is the force you need to prevent it falling under gravity. You weigh less on the Moon because the Moon has a weaker gravitational field than the Earth. weight W mg E.g. My mass is 75 kg so my weight is = 735 N. Weight always acts straight down (towards the centre of the Earth). If I am in orbit, I am weightless but I still have mass 15

16 Newton s Laws of Motion Newton's First Law of Motion. A body will remain at rest, or will continue to move with constant velocity, unless external forces cause it to do otherwise. All that matters is the net force i.e. the resultant sum of all the forces in each direction. If a body has an acceleration, there must be a resultant force acting on it. If a body has no acceleration, there is no resultant force acting on it. Note: although there is no such thing as an absolute velocity, there is such a thing as an absolute acceleration (you feel it - an acceleration of over 1000 m/s is likely to be fatal!) and this results from the forces acting on the body which are independent of any frame of reference. Example The Voyager spacecraft is now 8000 million miles from Earth and moving at mph. The Sun s gravity is very weak at this distance and it will continue moving at constant velocity for tens of thousands of years. Newton's Second Law of Motion. The acceleration of a body is proportional to the external force acting on it and has the direction of that force. F ma Example An artillery shell has mass 0 kg and an acceleration of m/s. We can calculate the force causing the acceleration: F ma N 3.5 MN (Remember prefixes: M = mega = 10 6 ). Newton's Third Law of Motion. Action and reaction are equal in magnitude and opposite in direction. This means that if two bodies, A and B, are in contact and exert forces on each other, then the force exerted on B by A is equal in magnitude and opposite in direction to the force exerted on A by B. Example. A box of mass 5kg is lowered vertically by a rope with an acceleration of 4ms -. Find the force in the rope. T 5 kg a 4 m/s Note double arrow showing direction of positive acceleration Weight = = 4.9 N but nicer to use exact values in the diagram W 5g Use F ma where F W T is the resultant force 5gT 54 0 kgm/s 0 N T N We will write weights as 5g and not worry too much about showing units. It is not 5g N since g itself has units. Writing 5g kg, though dimensionally correct, would confuse everyone. Edexcel Page 40 Ex. 3A. 16

17 Example. A body of mass kg, subject to the forces shown, accelerates uniformly in the direction indicated. Find the acceleration and the value of force P. P a kg 4 N 10 N Generally we can write separate F ma equations for each possible direction of motion. In the vertical direction there is zero acceleration so our equation is P g 0 0 P g The resultant force in the direction of motion (horizontally) is: (10-4) N g F ma hence 10 4 a a 3 m/s Example A car travels 8 m along a road while accelerating uniformly from rest to 1 m/s. (i) Find the acceleration, (ii) Given that the car has mass 1050 kg and the engine provides a propulsive force of 3000 N, find the resistance D to its motion. (i) (ii) v u as, 1 0 a8 D a.57 m/s, 3000 N 144 a.57 m/s 56 Net force = 3000 D ma N D N Example A fireman of mass 80 kg slides down a pole in a fire station. The resistance to him sliding is 600 N. After seconds he lands on the floor below. If g 10 m/s, how far has he descended? Resultant force = 80g N downwards. F 00 Using F ma, a.5 m/s m s ut at m Edexcel Page 44 Ex. 3B 17

18 Force diagrams We usually need a sketch, both to help us understand a written question and as a way of communicating with examiners and defining what our letters F, R, P etc mean. Use: Capital letters for forces and positions, lower case for distances, Greek letters for angles. Arrows show the direction we are defining as positive for each force. 1. Make the diagram large enough to show clearly all of the forces acting on the body and to enable any necessary geometry and trigonometry to be done.. Show only forces which are acting on the body being considered. 3. Weight always acts on a body unless the body is described as 'light'. 4. Contact with another object or surface gives rise to a normal reaction and sometimes friction. 5. Check that no forces have been forgotten or included more than once. Examples F R T W is the weight of the object A tension T in a string is trying to make it slide over the surface The surface exerts a normal reaction force R upwards on the object and a friction force F opposing the direction of motion. W A R W B S x A uniform bar of weight W is supported at points A and B by reaction forces R and S. B is a distance x from one end. Components of a force We will deal properly with the mathematics of vectors in chapter 6 but we need before then to understand how forces in different directions add up. When two or more forces act on a particle, we speak of a single resultant force that has the same effect as all the separate forces. One possible approach for finding the resultant would be to make an accurate scale drawing, with arrows in the direction of each force and length proportional to the force, and measure the overall length and direction. In practice we do a rough drawing to help us think about the formulae and then calculate the resultant exactly using trigonometry. Example 5 N 4 N 3 N Forces of 4 N and 3 N acting at 90 to each other have a resultant of 5 N acting at an angle where 3 3 tan, sin and 4 5 (so 36.9). 4 cos. 5 Note how the arrows join nose to tail when we add the forces. 18

19 Reversing this process, we could describe a single force of 5 N at an angle of 36.9 as being the sum of two or more imaginary components. We can choose any pair of directions (n.b. different directions, not parallel). The case where our chosen directions are perpendicular to each other is especially easy to calculate. o If we choose to have just two components with one horizontal and one vertical, they obviously need to be 4 N and 3 N. We think of the components as simply being the lengths on a scale drawing. Example A force of 18 N acts at 5 above the horizontal. Find the horizontal and vertical components. 18 N 5 18cos 5 N 18sin 5 N I am using solid lines for real forces and dashed lines for their equivalent in terms of components and resultants. Adding force components To find the resultant of two or more forces that may not be perpendicular, we split each force into components, add the horizontal components and add the vertical components. Example. Find the sum of the components in the x-direction and the y-direction of the following system of forces. y 6 N 5 48 x 1 N 13 N Resolving Net horizontal force = 13cos48 1cos5.18 N (to 3 s.figs.) Resolving Net vertical force 61sin 513sin N (to 3 s.figs.) We could then go on to find a resultant:.18 N 8.73 N N (to 3 s.figs.) 19

20 Edexcel Page 47 Ex. 3C. Friction and the coefficient of friction. Two solid bodies exert equal and opposite forces on one another when they are in contact. There is always a normal reaction, R, which is perpendicular to the common tangent of the two surfaces. If one then applies a force parallel to the surface in an attempt to make one body slide over the other, the applied force is resisted by an equal and opposite frictional force. If we increase the applied force, the frictional force will rise until it reaches some limiting value F max. Beyond this point, the applied force is bigger than the frictional force, there is a net force and the body starts to accelerate. push P reaction R weight mg friction force F (resisting motion) Experiments show, for a given pair of materials, that the limiting force reaction R. The limiting equilibrium case follows the line F max Fmax is proportional to the normal R where is the coefficient of friction for the two surfaces in contact. After motion has started the frictional force is equal to R, no matter how much the applied force is increased, F F max hence 0 F R F starts slipping Fmax R R pushing harder Some typical coefficients of friction (for interest) Aluminium on aluminium 1. Rubber on concrete (dry) 1 Wood on concrete 0.6 Rubber on concrete (wet) 0.3 Diamond on diamond 0.1 Teflon on steel 0.04 In real life the coefficient of friction depends more on the choice of materials than the smoothness of the surfaces the roughness determines the rate of wear (coarse sandpaper removes paint quicker than fine sandpaper). In M1 however when we speak of a surface being smooth or rough we simply mean 0 - it tells us whether we need to calculate a friction force. or 0 0

21 Example. A car of mass 1000 kg is parked on a dry road ( 0.9 ). The police want to drag it away. What force is needed? R Fmax R 1000 kg T ResolvingA R g = 0, R = 1000g = 9800 N W 1000g Now F max = R = 0.9x9800 = 880 N (assuming all wheels locked, e.g. a 4 4). Example. A 100 kg Diesel generator lies on a horizontal concrete floor. The coefficient of friction between the 3 generator and floor is. Calculate the magnitude of the force, P, which is necessary to overcome 4 friction if P is applied at 30 o to the horizontal. F R R 30 P Resolving A, F ma, R Psin30 100g 0 (zero vertical acceleration) Taking 100g g 9.8 m/s, R P (sin ) Resolving Pcos30 R 0 (zero horizontal acceleration, the generator only just starts to move) Now we have a pair of simultaneous equations in P and R. We want P so substitute to get rid of R: so with Pcos P 0 1 equation in 1 unknown good! 3 cos30 and P 3 P P P 39 N 4 5 Edexcel Page 5 Ex. 3D, 4 Forces handout 1

22 Particle on an inclined plane. As in the previous example we have two unknowns so we need two equations. Need to write F ma equations in two directions, usually (but not necessarily) perpendicular. Could try (parallel and perpendicular to slope ) or (horizontal, vertical ). If accelerating, must use parallel & perpendicular. Limiting equilibrium example. A car is parked on a 0 hill. The road is slippery and it is about to slide down ( limiting equilibrium ). Find the coefficient of friction between the tyres and the road. State any assumptions made. R F R 0 mg 0 Resolving perpendicular to the plane: R mg cos 0 0, R mg cos0 Resolving parallel to the plane: mg sin 0 R 0, R mg sin 0 mg sin 0 tan mg cos 0 This is a general result in limiting equilibrium tan We have assumed that all the wheels are locked (perhaps it was left in gear). More realistically, if the front wheels are free to roll and only half the weight is on the back wheels (held by hand brake) the coefficient of friction would have to be twice this. Edexcel Page 109 Ex. 4C. Mixed exercise Page 111. Acceleration when sliding uphill or downhill If you are sliding uphill, both friction and gravity are slowing you down If you are sliding downhill, gravity makes you go faster, friction makes you go slower. Which wins? tan, steep Good for skiing tan (constant velocity slope) tan accelerate Gentle slope or high friction slow down

23 Accelerating example Find the force needed to accelerate a kg block at 3 ms - up a rough plane, with coefficient of friction and inclined at angle sin to the horizontal, if the force is parallel to the slope. 13 F R R g a 3 m/s P sin sin, cos, tan Resolving perpendicular to slope: Rgcos 0, R g 13 4 Now since sliding, F R 0. g N 13 Resolving F ma along slope: P F g sin ma 3 6 N 5 P N (4 sig. figs) 13 Edexcel Page 55 Ex. 3E. 3

24 Motion both up and down a slope When a particle is moving up, the frictional force acts down the slope. When a particle is sliding down, the frictional force acts upwards. On a rough slope, a particle sliding freely will have different accelerations in the upwards and downwards directions Example. A games machine launches a sliding puck up a rough sloping surface at m/s. The surface is angled at 15 to the horizontal and 0.1. Find: (i) The time taken to reach its highest point and the distance travelled up the slope (ii) The time taken to return to its initial position and its speed at this point. R a Resolving perpendicular to the slope: R mg cos15 0 Resolving up the slope: mg sin15 F ma 15 F mg When sliding up: F R 0.1mg cos15 a g sin15 0.1g cos15 g sin15 0.1cos15) m/s v u at, t, t sec. uv 0 s t t m When sliding down: F R 0.1mg cos15 a g sin15 0.1g cos15 g sin15 0.1cos15) m/s (check: less than before) v u as v m/s and t sec

25 Statics of a Particle (equilibrium of coplanar forces) Really no different to the previous examples - just a 0 and 4 forces not 3. See the Four forces handout for lots more examples and alternative methods Statics means the balancing of forces such that the net force is zero and the object does not accelerate. the body is then said to be in equilibrium the sum of the components in any direction will be zero. By particle, we mean that the object is considered to be concentrated at a point. The line of action of each force passes through this point We do not have to worry about whether or not the object rotates. We often have unknowns so we need to solve equations to find them (resolving in different directions). Generally these will be simultaneous equations but if we are cunning we can write them as simple equations in 1 unknown. Example. A particle of weight 16 N is attached to one end of a light string whose other end is fixed. The particle is pulled aside by a horizontal force until the string is at an angle of 30 to the vertical. Find the magnitudes of the horizontal force and the tension in the string. 30 Resolving gives an equation without P: T cos T P 16 T 18.5 N (3 sig. figs) cos30 Resolving gives PTsin N PTsin N (3 sig. figs) 5

26 Example. A load of mass 6kg is supported in equilibrium by two ropes inclined at 60 and 30 to the horizontal. Find in terms of g the tension in each rope. 60 Resolving gives: 1 T sin 60 T sin 30 6g 0 T 1 30 T cos30 T cos60 0 Resolving gives: 1 ( terms so easy to express T in terms of T1): 6g Alternatively: T cos60 T T 1 cos30 cos60 Substituting into the first equation: T1 sin 60 T1 sin 30 6g cos30 3 T Solving simultaneously: T1 6g 13 3g cos60 T T1 13g cos30 o Resolving in the T 1 direction: T1 6gcos30 0, T1 13 3g Strictly speaking one could say T = 13 kg.g with kg. being a force unit equivalent to 9.8 N but this may cause more confusion that clarity! T o Resolving in the T direction: T 6gsin30 0, T 13g T g closed polygon for forces in equilibrium Example. A particle is in equilibrium under the action of the forces shown. Find the magnitude of the forces P and S. y S 30 P 60 8 N 10 N x Resolve horizontally so that P does not occur in the equation: cos60 o - Scos30 o = 0 S N Now resolve vertically P + 8sin60 o - Ssin30 o = P 8 N 3 3 6

27 Example. An 8kg particle sits on a smooth slope in equilibrium under the action of the forces shown. Find the magnitude of the forces P and Q. P 75 N Q Resolving perpendicular to plane so Q does not appear in the equation: 75 Psin30 o 8gcos30 o = g cos30 P 14. N (3sig.figs.) sin g Resolving parallel to plane: Q + Pcos30 o 8gsin30 o = 0 Q = 6.9 N (3sig.figs.) Example A kg particle sits on a rough 0 slope with 0.. Find the applied force X when the particle is about to slide: (a) Down (b) up. P R F We can use one set of equations for both cases, just changing the sign of force F. (a) about to side down, as drawn, F R (b) about to slide up, F resists motion up F R 0 g Resolving vertically so we can find R without knowing P: R cos 0 F sin 0 g 0 (a) R (b) R cos 0 0.sin 0 g, R N cos 0 0.sin 0 g, R.50 N Now we know R (and hence F), resolve horizontally to find P: P F cos 0 R sin 0 0 (a) P R R R (b) sin 0 0. cos 0 sin 0 0.cos N P R sin 0 0.R cos 0 sin 0 0.cos 0 R 11.9 N 7

28 Alternatively (same example) if you really don t like equations, use a closed vector polygon. You have to be REALLY careful with the arrow directions Really just to give you a deeper understanding! Algebraic solutions are usually easier. For case (a) with F pushing up the slope: R F P R and F must form two sides of a triangle since we know two angles enabling this to be drawn. F R 0.R F R 1 1 tan tan You must draw it (but it does not need to be an accurate scale drawing) g P tan hence P 9.8tan N g arrows nose-to-tail Nb. If P is not horizontal, you will need the sine rule for the final calculation. Steps: Draw thin construction lines for g and P Draw 0 construction line for R Draw dotted line at 11.3 from R line, then F at 0 down from intersection. For case (b), about to slide upwards, the friction force pushes downwards. (i) we could define as negative and use the same equations as above: P tan hence P 9.8tan N g (ii) It may be clearer though to do a fresh diagram. I will replace F pushing up the slope with G pushing down the slope, G R G R P g 0 Again, R and G must form two sides of a triangle since we know two angles enabling this to be drawn. F R 0.R 1F 1 tan tan R You do not need an accurate scale drawing P tan g hence P 9.8tan N Edexcel Page 95. Ex 4A-4D. 8

29 DYNAMICS OF TWO CONNECTED PARTICLES When two accelerating bodies are connected by a taut string (or rigid rod), the tension in the string acts equally (but in opposite directions) on each mass. T T We assume: the string is light, so it has no inertia (otherwise there would be a net force on the string and the tension would be less at one end than at the other) the string is inextensible (so both masses have the same acceleration; the masses do not, for instance, bounce in and out) Example. A car of mass 800kg pulls a trailer of mass 300kg. The force produced by the engine is 4000N. The resistances acting on the car and trailer are 500 N and 00 N respectively. Calculate: (a) The acceleration of the car and trailer. (b) The tension in the towbar. 00 N 300g 300 kg T 500 N 100 kg 100g a 4000 N R R 1 (a) We have two unknowns (T and a) so need a pair of simultaneous equations: F ma for car: 4000 T a ` 3500 T 800a (1) F ma for trailer: T00 300a () Adding the two equations to eliminate T: a 300a a a 3 m/s Alternatively (conceptually more obvious) we can treat the car and trailer as a single particle subject to a drag force of 700 N: a, a, a 3 m/s (b) To find the tension we must use the equation of motion for one mass, e.g. the trailer, T00 300a, T 300a N 9

30 Problems involving pulleys. These are basically the same as car and trailer problems but (i) The direction of motion of the bodies is different so we tend to write a separate F ma equation for each mass rather than imagining a net force along the line of motion. (ii) We can have more complex situations with the string going slack and the masses then falling back. Whilst the string is taut, both have the velocity and acceleration in their direction of motion. We assume the pulley (or peg ) is light (so has no rotational inertia) and smooth (so there is no friction as it rotates or the string slides over it). The tension in the string is then constant throughout its length. Example. A body of mass 3kg rests on a smooth horizontal table. It is connected by a light string, which passes over a smooth pulley at the edge of the table, to another body of mass kg hanging freely. When the system is released, find the acceleration of the bodies and the tension in the string. R a A 3 kg T F ma equation of motion for mass A: T 3a F ma equation of motion for mass B: g T a g T g T 3a a Adding to eliminate T gives: 3g B g 5a, a 0.4g Now substitute into one of the equations of motion, e.g. T 3a 1.g T kg a Note: the direction of positive acceleration a must be consistent for the two masses. Example. Particles of mass 7kg and 5kg are connected by a light string passing over a smooth fixed pulley. The particles hang freely and are released from rest. Find the acceleration of the two particles and the tension in the string. T T 5 7 a 5g 7g Resultant force on pulley = a T 35 g 3 g 6 5kg mass: T 5g 5a 7kg mass: 7g T 7a Add to eliminate T: g 1a 5 35 T 7g 7a 7 g a 7 g g 6 6 a and [ Edexcel Page 64 Ex. 3F.] 30

31 Pulleys with slope and bouncing Without any new theory, we can handle complicated problems involving pulleys, slopes and particles moving under gravity. Example A particle P of mass 3m and Q of mass m are connected by a light inextensible string. 3 P lies on a smooth inclined plane at an angle to the horizontal where sin 5 The string passes over a smooth peg at the top of the incline. Initially the system is held at rest with Q 1 m above a horizontal floor. It is then released. Assuming P does not hit the peg, find: (a) the acceleration of Q, (b) the time that passes until Q hits the floor (c) How far P (in total) moves up the inclined plane before coming to rest First we need to sketch the system P 3m Q m 1 m sin 5 4 cos 5 and define the forces and accelerations for each mass: R a T T 3m P 3mg cos m Q a 3mg 3mg sin mg Equation of motion for P: T 3mg sin 3ma Equation of motion for Q: mg T ma (a) Add to eliminate T: mg 3mg sin 5ma 3 3 Divide by m, put sin : g 3g 5a g 9g 1 g 5a 5 5 a g 5 31

32 (b) Q falls 1 m, Q is initially at rest. 1 1 g s ut at hence 0t t 1, 5 50 t, t.6 seconds. 9.8 (c) We must calculate the speed v of Q at the instant when it hits the ground before finding how far it carries on sliding up the slope. After the acceleration so far, the energy equation is: 9.8 g v u as After Q has hit the ground, the string is slack and T = 0. The equation of motion for P: T 3mg sin 3ma then becomes 0 3mg sin 3ma which simplifies to 3g a gsin 5 v 0 Sliding up (slowing down) Using the energy equation a second time as it slows from (our previous v is our new u ): g u to 5 v 0 : u g 3g v u as s 0, multiply by g gives 5 30s 0 0 g Hence 1 s m The total distance P moves is then 1 1 m

33 MOMENTUM AND IMPULSE Inertia describes the general way in which a mass requires an impulse (force time) to achieve a change in velocity. Momentum quantifies the impulse that must have acted to achieve a given velocity. The momentum of a body is the product of its mass and its velocity. momentum vector = mv If the units of mass and velocity are kg and m/s, then the units of momentum are kg.m/s. Edexcel tend to write this (completely equivalently) as Newton-seconds (Ns). The impulse, I, of a constant force, F, is defined as Ft, where t is the time for which the force is acting. I Ft but F ma hence I mat and v u at hence v u at I m v u I mv mu Impulse = change in momentum Force = rate of change of momentum This is called the Impulse - Momentum Principle. The unit of impulse is the Newton-second (Ns). Example. Find the magnitude of the impulse given to a body of mass kg when its speed changes from 6 m/s to 15 m/s in the same direction. I mv mu Ns Example. A squash ball of mass 5g hits the wall of a court at a speed of 40 m/s and rebounds at 3 m/s. Find the impulse of the ball. 0.05g 40ms -1 3ms g Edexcel Page 68 Ex. 3G. Take the direction towards the wall as positive. I mv mu Ns (Negative, i.e. in a direction away from the wall). 33

34 Conservation of momentum. When a collision occurs between two bodies, they exert equal and opposite forces on each other (Newton's third Law). These forces will occur for the same length of time and since Impulse = Force x Time it follows that the two resulting impulses will be equal in magnitude and opposite in direction. Therefore, equal and opposite impulses produce equal and opposite changes in momentum. If there is a gain in momentum of one body, then there must be an equal loss in the momentum of the other body. The total momentum before impact must be the same as the total momentum after impact. This is known as the Principle of Conservation of Linear Momentum. Consider two objects with masses of m 1 and m with velocities u 1 and u before contact and velocities v 1 and v afterwards. m 1 u 1 + m v = m 1 v 1 + m v In dealing with examples on collision, it is advisable to draw two diagrams, one for before the collision and one for afterwards. In some instances, after colliding, the bodies do not rebound from one another, but move as a single body. They are said to coalesce. Example. Two bodies collide on a horizontal surface. Find the speed v of the lighter body after impact. Before After 6ms -1 4ms -1 v 1ms -1 kg 5kg kg 5kg Taking velocities to the right as positive m 1 u 1 + m v = m 1 v 1 + m v (6 x ) + (5 x (-4)) = -v + (5 x (-1)) 1-0 = -v -5 v = 1.5ms -1 Example. A body of mass kg moving on a smooth horizontal surface at 3ms -1 collides with a second body of mass 1kg which is at rest. After the collision, the bodies coalesce. Find the common speed of the bodies after impact. Before After 3ms -1 at rest v kg 1kg Taking velocities to the right as positive. m 1 u 1 + m u = v(m 1 + m ) ( x 3) + (1 x 0 ) = v( + 1) v = ms -1 3kg 34

35 Jerk in a string. If a string is jerked, equal and opposite impulsive tensions act in the string. Therefore, equal and opposite impulses act on the objects to which the two ends of the string are attached. m 1 m I I F The two ends of the taut string have equal velocity components in the direction of the string. The total momentum of the system remains constant, although the momentum of each individual object is changed in the direction of the string. Perpendicular to the string, no impulse acts and the momentum of each object is unchanged. Example. Two particles, P and Q, with masses 3kg and 6kg respectively, are connected by a light inextensible string. Initially, they are at rest on a smooth table with the string slack. Q is projected directly away from P with a speed of 3ms -1. Find their common speed, and the impulse in the string when it becomes taut. at rest 3ms -1 Before P Q 3kg 6kg After P vms -1 vms -1 Q Using the principle of conservation of linear momentum: (3 x 0) + (6 x 3) = 3v + 6v v = ms -1 Consider particle P: I = (3 x ) - (3 x 0) I = 6Ns Edexcel Page 7 Ex. 3H. Page 73 Mixed Ex. 3I. 35

36 EXTENDED BODY PROBLEMS (MOMENTS) A moment is the bending effect due to a force acting in a transverse direction at some distance from a point. moment d F Moment about point A M Fd where F is perpendicular to d. A F d sin If the force and distance are not perpendicular, the moment is M Fd sin moment A d The units of moment are Newton metres (Nm). A positive moment, like a positive angle, is anti-clockwise as seen in the x-y plane. A moment is really a vector property o The vector direction is the axis of attempted rotation (z-axis here, perpendicular to x-y plane). Extension to aid conceptual understanding 1. We call it a moment if we are dealing with -dimensional geometry (forces in an (x,y) plane) and/or the effect is to bend a beam. Remember: tension Tension tries to make something longer Compression tries to make something shorter compression A moment acting on a beam tries to snap it: tension on one side, compression on the other. moment moment. A moment is much like a torque. The distinction is simply that is the effect is to bend a beam, we call it a moment; if the effect is to twist a shaft, we call it a torque. (e.g. a Ferrari 360 has a V8 engine producing 373 Nm of torque ). If I try to tighten a bolt using a spanner, the force I exert on the spanner produces a moment in the spanner and a torque on the bolt. moment force torque 36

37 3. Work done (energy) is also measured in Newton metres (1 Nm = 1 Joule) but that is different. work = force distance moved in the force s direction, moment = force distance perpendicular to the force (with or without movement). Example. Find the moment of the force F about the point P. P 5.1 m 35.8 N F Moment of F about P =.8 5.1sin35 = 8.19 Nm (3sig figs) Edexcel Page 118 Ex. 5A. Resultant moment. When several coplanar forces act on a body, the resultant moment about a point in the plane is the sum of the moments of the individual forces about that point. If all these forces balance (so there is no resultant force, hence no acceleration), the resultant moment is a constant that does not depend on the position of the point P. If there is a resultant moment the object will experience angular acceleration (e.g. flywheel of an engine as speed (rpm) increases). In M1 and M we do not have angular acceleration and bodies are in equilibrium because both the resultant force and resultant moment are zero. o (so we can do a moment calculation about any point we fancy!) Example. Calculate the sum of the moments of the forces about P acting on a lamina. 6 N X 1 N 30 8 m.7 m P 9 m 5 N Z Y Sum of the moments about P = ( 6 x.7 ) + ( 5 x 9 ) (1sin30 o x 8) Edexcel Page 11 Ex. 5B. = 13. Nm 37

38 Particles and bodies in equilibrium. When a particle or body is in equilibrium under the action of a set of coplanar forces: The vector sum of the forces is zero (no net force in any direction, the object does not accelerate) the algebraic sum of the moments of the forces about any point is zero (there is no rotational acceleration, it does not start to rotate). These two conditions let us define a set of simultaneous equations for any problem. For a - dimensional problem (forces in the x-y plane only) we can define 3 equations: Net force in x-direction = 0 Net force in y-direction = 0 Net moment due to all forces in this plane = 0 So we can solve problems with 3 unknowns (M). Moment questions in M1 tend to have forces all acting in one direction giving simpler questions needing just equations for unknowns. Example. A uniform bar is 3m long and has a mass of kg. Masses of 3kg and 4kg are suspended from the ends A and D respectively. At which point must the bar be suspended so that it hangs horizontally in equilibrium? T x A 3g B g C 4g D 3 m 1.5 m Taking moments clockwise about D: Vertical force balance: T (3g g 4 g) 0, T 9g Tx g 1.5 3g 3 Tx 1g 0, Tx 1g Hence 1g 1g 4 x m T 9g 3 The net moment = 0 regardless of which points we take moments about. We could for instance have taken moments anti-clockwise about C. This is nice because we then do not need to find force T (T passes through C so generates no moment regardless of how big it is). 3g 3 x g 1.5 x 4gx 0 Perhaps divide all terms by common factor g? 93x 3 x 4x 1 9x x 9 3 m Edexcel Page 15 Ex. 5C. Page 18 Ex. 5D. Mixed exercise 5E. Page