24/06/13 Forces ( F.Robilliard) 1

Transcription

1 R Fr F W 24/06/13 Forces ( F.Robilliard) 1

2 Mass: So far, in our studies of mechanics, we have considered the motion of idealised particles moving geometrically through space. Why a particular particle moves with a particular type of motion, has not yet been considered. The motion of a real body, is determined by the external forces, that act upon it. There is an intrinsic property of a body, however, that determines its precise response to a given applied force the mass (or inertia ) of the body. Mass: Mass is the amount of material composing body. Bodies are made from atoms, and atoms are composed of elementary particles, such as protons, and electrons. These elementary particles have a quantum property which we call mass. This property allows a particle to interact gravitationally with other particles, and also gives the particle the property of inertia. This property is additive. The mass of a body is the total mass of all the elementary particles, of all the atoms, of which the body is composed. SI Unit of mass: kg 1 kg = The total mass of (~5.98x10 26 ) individual protons 24/06/13 Forces ( F.Robilliard) 2

3 Momentum: Next we need to quantify the amount of motion possessed by a body. We need to include both the mass of the body, and its velocity. m v p A body of mass, m, moving with a velocity vector, v, has a momentum vector, p, defined by - p = m v Momentum is a measure of the motion of a body. It is a vector, having the direction of the velocity. SI Units: kg.m s -1 24/06/13 Forces ( F.Robilliard) 3

4 Several Forces: It is common for several forces to act on a body simultaneously. Each individual force will contribute to the overall motion of the body. For example: Fr R W F Block being dragged along a surface: R = upward reaction Force of the surface, on the block W = downward weight force of gravity, on the block. F = applied force dragging the block. Fr = friction force of the surface, on the block. Although we have not yet defined precisely what we mean by force, it is intuitively clear that forces have direction, and are therefore vectors. The individual contributions by individual forces, to the overall motion of a body, will add up, like vectors. We can use the idea of an overall effective force causing an overall effective motion. We can replace the several forces, with a single equivalent force, that has the same total effect on the motion of the body, as all of the individual forces. Such an equivalent force is called a resultant force, and is obtained by vector addition of the several individual forces. 24/06/13 Forces ( F.Robilliard) 4

5 Resultant Force: The vector sum of all the forces that act on a given body, at a given moment, is called the resultant force acting on the body, at that moment, and is represented by ΣF. R Fr F In the diagram ΣF = F + Fr + R + W (vector sum) W It is the resultant force, ΣF, acting on a body, that determines the overall change in motion of that body. 24/06/13 Forces ( F.Robilliard) 5

6 Newton s Laws of Motion: We are now able to define force. This is done in terms of the change in the motion of a body, that the force causes. This idea is due to Isaac Newton and was included in his three laws of motion. Firstly, we will give a general description of the role played by each of these laws. The first law identifies a fundamental spatial property of masses, namely, that they will not change their momentum, unless a force acts upon them. This law was originally discovered by Gallileo, and is known as the law of Inertia. The second law describes the quantitative connection between a force, and the change it produces in the motion of a body. This law effectively defines force. The third law expresses the fundamental property of forces, that they always occur in pairs, and gives the quantitative relationship between the two forces of the pair. This is the law of action and reaction. 24/06/13 Forces ( F.Robilliard) 6

7 Newton s Laws of Motion: Next we give a quantitative statement of each law: Law 1 (N1): Unless a resultant force acts on a body, its momentum vector will not change. Law 2 (N2): The time-rate of change in the momentum of a body, is proportional to the resultant force that acts on that body, and takes place in the direction in which that resultant force acts. Law 3 (N3): If a body A exerts a force on a body B, then B exerts a reaction force back on A, which is equal in magnitude, but opposite in direction to the original action force due to A. This is true at every instant. 24/06/13 Forces ( F.Robilliard) 7

8 F = force vector; More Concisely: p = momentum vector N1: N2 : N3: If F = ~ AB ~ 0, ~ then δ p d F ' p $ % " ~ dt & ~ # F = F BA ~ ~ = 0 ~ No cause, no effect! Force causes a proportional rate-of-change in momentum. Never a lonely force; an action causes an equal, but opposite reaction. N1: N1 is a special case of N2. It was included as the first law to emphasise the fundamental property of bodies, that they will not change their momentum unless a force acts upon them. If at rest, they remain at rest; if in motion, they continue their motion with the same velocity, and in the same direction. This is the property of inertia. Prior to Gallileo, it was thought that bodies naturally returned to a state of rest. The role of friction forces had not been understood. 24/06/13 Forces ( F.Robilliard) 8

9 We can simplify N2: Say a resultant force, ΣF, acts on a body, of mass, m, which has a velocity, v, and a momentum of p. p m v ΣF N2 F ~ N2: d dt = k ' p $ % " & ~ # d dt d v ~ = km dt = km a ~ ( ) m v ~ if m = const. where We define the SI unit of force so that k = 1. Thus: where k constant of p mv a acceleration ~ ~ proportionality d v ~ dt F = ma ~ Unit: The SI unit of force is the kg m s -2 = Newton = N 1 N force causes a mass of 1 kg to experience an acceleration of 1 m s /06/13 Forces ( F.Robilliard) 9

10 Find the acceleration of a 2 kg mass, when acted upon by the two forces illustrated. Example: 2 kg y m 3N 4N x θ# ΣF 4N 3N Resultant Force: ΣF = 5 N, at θ = 36.9 deg to x-axis ΣF = ma thus acceleration = a = ΣF/m = 5/2 = 2.5 m s -2 in the direction of the resultant force, namely at θ = 36.9 deg to the +x-axis. Alternatively (using i,j,k notation): Magnitude = a = Direction ΣF = = 4i +3j (N) ΣF = ma 4i +3j = 2 a thus a = 2i + 1.5j m s -2 tan θ = hence = 2.5 m s -2 θ = 36.9 deg Same as before 24/06/13 Forces ( F.Robilliard) 10

11 N3: N3 is often misunderstood. The only way a force can act on a body is by an interaction with another body. There must always be two bodies. The interaction could be by direct contact, or remotely, via a field. If A exerts a force on B (the action force), symmetry would suggest that B would exert reaction force back on A. This can be deduced from the requirement that the motion of a body must be consistent with the motion of its component parts - The action and reaction forces- - are equal in magnitude - are opposite in direction - act on different bodies Examples of action-reaction pairs of forces: Bodies A & B Collide: A B action force of A on B reaction force of B back on A Earth(E)- Moon(M) System: E M gravitational force of E acting on M gravitational force of M acting on E 24/06/13 Forces ( F.Robilliard) 11

12 The Mechanism for Force : We defined force in terms of the change in motion of a single inertial body (a single mass). Forces can only be produced by an interaction between two bodies. Without the second body, the interaction, and therefore the force, cannot exist. This interaction involves momentum being transferred from one of the bodies to the other. Thus the momentum of one body increases, while the momentum of the other body decreases, with time. We define the force acting on each body, in terms of the time-rate of transfer of momentum, to (or from) that body, from (or to) the other body. This idea of momentum transfer unifies the second, and third laws of Newton. Forces can only happen between two bodies. Because there is a mutual transfer of momentum, there must be two, equal, but opposite forces. Each force acts on only one of the two bodies. We will return to this idea, when we consider momentum in more detail, later in this course. Fundamentally, any force involves the transfer of momentum between particles. Different types of force differ only in the mechanism for momentum transfer. 24/06/13 Forces ( F.Robilliard) 12

13 Common Forces: Next, we will describe some of the more common types of force, that occur in problems. Weight: This is the force of gravity on a body. If the acceleration due to gravity (near a planetary surface, say) is g, then the weight, W, of a mass, m, will be g m W = mg from ΣF = m a W = m g Near the earth s surface, g = 9.8 m s -2 = 10 m s -2, approximately Other examples: Body Earth Venus Mars Moon g (m s -2 ) /06/13 Forces ( F.Robilliard) 13

14 Common Forces: Tension: This is the force exerted on a body via a rope or string. Strings can pull, but not push. t T B G Block, B, hangs by a string, from girder, G. T is the tension force of the string on the block. t is the tension force of the string on the girder. If the string is massless, T = t t m M T Masses m and M are connected by a string, over a pulley. t is tension force of the string on mass m. T is tension force of the string on mass M. If the string is massless, and the pulley is massless and frictionless, then T = t. 24/06/13 Forces ( F.Robilliard) 14

15 Common Forces: Reaction: S When a body exerts a force on a surface, the surface reacts back on the body, with an equal and opposite reaction force. If there is no mechanism for this reaction force to have a component along the surface (no friction, no hinge etc), then the reaction will be normal to the surface. We call this the normal reaction, R. B R Consider a block, B, resting on a surface, S. r is the action of the block on the surface. (Note: r is not the weight of the block!) R is the normal reaction of the surface, S, back on the block. r By N3: R = r always! Normal reactions occur, where the surfaces in contact are smooth. If the surfaces are rough, we get, in addition, a force component along the surface, which we call friction. 24/06/13 Forces ( F.Robilliard) 15

16 Common Forces: Spring: Let one end of a spring be fixed to a wall and an object is fixed on the other. When an applied force F acts on the object, the object exerts a force F on the spring and the spring exerts a restoring force F on the object. The restoring force is in a direction opposite to the displacement of the object. Hooke's law is a principle of physics that states that the restoring force -F acting on the spring when it has been extended or compressed by some distance of x is proportional to that distance : -F x thus F = -kx where k is the spring constant and is a characteristic of that spring. 24/06/13 Forces ( F.Robilliard) 16

17 Friction: Common Forces: If there is roughness between two surfaces in contact, a contact force along the surface is possible. This force is called friction. The friction force will depend on (1) the nature of the surfaces in contact (2) the normal forces of contact that hold the surfaces together. We identify two situations: (1) static friction is where there is NO relative motion between the surfaces. (2) kinetic friction is where there is relative motion. Friction forces are always directed so that they oppose motion. 24/06/13 Forces ( F.Robilliard) 17

18 f Common Forces - Friction: R F Consider a block of mass, m, resting on a horizontal rough surface. A horizontal force, F, is applied to the block. mg = weight of the block. R = normal reaction of the surface up on the block. mg A friction force, f, opposes the motion of the block. When F = 0 (no dragging force!), f must also = 0, since the block is stationary. Initially, as F is increased, the block remains stationary. Thus f must be equal, but opposite, to F. Eventually, F reaches a value where the block starts to move. At this point, f has reached its maximum value, and F has become slightly greater than f. As F continues to increase, f remains constant, and the block moves, with an increasing acceleration, proportional to (F-f), along the surface. We plot a graph of the friction force, f, against the applied force, F. 24/06/13 Forces ( F.Robilliard) 18

19 Common Forces - Friction: max. static friction force f kinetic friction force In static region: f = F In kinetic region: f = const. static friction kinetic friction F There is a non-simple transition region between the static, and kinetic regions. For any pair of surfaces, there will be a characteristic maximum value of friction force, just before the block slides. This is called the maximum static friction force. As soon as the block starts to slip, the friction force typically relaxes to a slightly smaller constant value, called the kinetic friction force. 24/06/13 Forces ( F.Robilliard) 19

20 Common Forces - Coefficient of Friction: The values of maximum static friction, and kinetic friction, depend on (1) the nature of the contact surfaces (2) the normal reaction force between the surfaces. Experiment shows, for any given pair of surfaces, both the max. static friction force, and the kinetic friction force, are proportional to the reaction force, R, between the surfaces. f max.static R f kinetic R thus f max.static = µ s R thus f kinetic = µ k R where µ s = constant of proportionality = coefficient of maximum static friction. and µ k = constant of proportionality = coefficient of kinetic friction For any pair of surfaces, there will be a characteristic static coefficient, µ s, and a (usually) lower kinetic coefficient of friction, µ k. 24/06/13 Forces ( F.Robilliard) 20

21 Common Forces - Coefficient of Friction: Some typical values for µ: µ s = max. static friction coefficient µ k = kinetic friction coefficient Surfaces µ s µ k steel on steel copper on steel ski on wet snow f C B A F teflon on teflon human joints Curves for different pairs of surfaces A, B, and C. 24/06/13 Forces ( F.Robilliard) 21

22 Example: What horizontal force, F, would be needed to drag a steel block, of mass 5.0 kg, along a horizontal steel surface, at a constant velocity? µr R m mg F F = applied force on block R = reaction force of surface on block µr = friction force of surface on block (note: µ is the kinetic value µ k ) mg = weight of block Since there is no acceleration vertically, nor horizontally: Vertically: R = mg...(1) Horizontally: F = µr...(2) (1) --> (2): F = µ mg = 0.57 x 5.0 x 9.8 = 27.9 N An applied force of 27.9 N is required. 24/06/13 Forces ( F.Robilliard) 22

23 Force Diagrams: When addressing force problems, the first, and probably most important, step is to draw a diagram showing all the individual forces, that act individually on each body in the problem. These are force diagrams. There are clearly two steps: (1) divide the system up into individual bodies (2) for each chosen body, draw a diagram showing all the forces that act on that particular body. Example: Pulley System: T mg m M T Mg We break the system up into two individual masses - m and M. Because the string is massless and the pulley is massless and frictionless, the same string tension, T, acts on each block. (The pulley s only effect is to change the direction of the tension force in the string.) The motion of each mass is individually determined by the tension and the weight, that act upon it. 24/06/13 Forces ( F.Robilliard) 23 mg m T M T Mg

24 Force Diagrams: The Box in the Lift: Assume the lift has acceleration, a, upward. Let - m = mass of box M = mass of cage of lift T = tension in the lift cable R = reaction of lift floor on box = action force of box on lift floor. a m M We can break the system up in three ways. Box: Cage: T Box & cage: R T R mg m Mg (M + m) g 24/06/13 Forces ( F.Robilliard) 24

25 Force Diagrams: The Engine and Carriages: a R m m M T t M Mg r m T mg r m t F Assume train is accelerating forward. We can break the train up into three bodies engine, first carriage, last carriage. Let: M = mass of engine m = mass of either carriage F = traction force due to wheels pulling engine forward R = normal upward reaction of rails on engine r = normal upward reaction of rails on either carriage T = tension in coupling between engine and first carriage t = tension in coupling between the two carriages mg 24/06/13 Forces ( F.Robilliard) 25

26 Force Diagrams: The Block sliding on a Wheeled Wedge: m The Block: µr r M Assume frictionless wheels. The Wedge: r R Mg µr mg m = mass of block M = mass of wedge r = normal reaction of wedge surface on block R = normal reaction of horizontal surface up on wheels µ = coefficient of friction between block and wedge Relative to the wedge, the block will accelerate down the wedge s inclined face. The wedge will accelerate horizontally to the left, relative to the horizontal surface. 24/06/13 Forces ( F.Robilliard) 26

27 Solution of Force Problems: For systems of masses, the general strategy is to divide and conquer divide the problem up into convenient masses, apply N2 to each, thereby obtaining a set of equations, and finally, solve the equations for the unknowns. Steps: (1) Divide the system into convenient component masses. (2) Draw a force diagram for each mass (3) Label acceleration for each mass (4) Resolve all forces (and acceleration) into mutually perpendicular directions for each mass (5) Assuming consistent positive directions, write ΣF = ma, for each mass, in each direction (6) Solve all equations, for all masses, for the unknowns. (7) Interpret your answers 24/06/13 Forces ( F.Robilliard) 27

28 Example: A 5 kg mass rests on a horizontal surface, with which its kinetic friction coefficient is 0.5. Find its acceleration, a, when a horizontal force of 50 N acts upon it. µr a m R F +y +x N2: ΣF = ma x direction: +F µr = ma.(1) y-direction: +R mg = 0.(2) m = 5 kg F = 50 N µ = 0.5 mg Solve (1) & (2) for a: (2): R = mg..(3) (3)->(1): F µmg = ma thus: a = F/m µg = 50/5 (0.5)(10) = 5 m s -2 24/06/13 Forces ( F.Robilliard) 28

29 Example: M µ# θ# M = 10 kg m = 5 kg µ = 0.2 tan θ = 3/4 Find the acceleration of the masses, and the tension in the string, after the system is released from rest. Since M >> m, mass m will slide up the incline. + a T M µ# T a R + µr + Draw force diagrams for each mass. Label acceleration of system. Choose positive directions. Mg mg θ# 24/06/13 Forces ( F.Robilliard) 29

30 + T a M Mg µ# T mg cosθ# a θ # mg R µr + mg sinθ# Example: θ# + Find a: Resolve weight, mg, along incline, and perpendicular to incline. N2: ΣF = ma For M (vertically): +Mg T = Ma.(1) For m (perpendicular to incline): For m (parallel to incline): +R mg cosθ = 0.(2) +T mg sinθ - µr = ma.(3) (2)->(3): +T mg sinθ µ(mg cosθ) = ma.(4) (1) + (4): +Mg mg sinθ µmg cosθ = Ma + ma Thus : a = M m sinθ ( + µ cosθ) ( M + m) g...(5) 24/06/13 Forces ( F.Robilliard) 30

31 Find T: Example: a = +Mg T = Ma.(1) M m sinθ ( + µ cosθ) ( M + m) g...(5) From (1): T = M (g - a)..(6) (5)->(6) gives: Numerical values: From (5): a M m sin θ = & 10 5$ = % = 4.05 ( + µ cos θ) ( M + m) 3 5 ( ) m s #! " (8) g T T & = M$ g % & = Mg$ 1 % Mm = M m sin θ ( + µ cos θ) ( M + m) ( + µ cos θ) ( M + m) M m sin θ ( ) [ 1 + sin θ + µ cos θ] M + m g....(7) 24/06/13 Forces ( F.Robilliard) 31 # g! " #! " (8)->(6): T = 10 ( ) =57.5 N

32 Forces in Rotation: O a r v t m Say that a mass, m, moves about a circle of radius, r, at constant angular velocity, ω. We know from our study of circular motion, that the mass has a centripetal acceleration, a, at every instant, t, given by: ω = const a = ω 2 r = v r 2 where v = the tangential velocity. By N2, this acceleration must be caused by some centripetal force, F, given by: F = ma = m ω 2 r = m v r 2 at every instant of time. Various physical forces, such as friction, tension, gravitational forces, magnetic forces, etc, can play the role of the centripetal force, in particular systems. 24/06/13 Forces ( F.Robilliard) 32

33 Example: A car drives in ever decreasing circles, on a flat horizontal surface, at a constant speed of 100 km/hr. The coefficient of maximum static friction, µ, between the car s tyres and the road surface is 0.8. At what radius, will the car slip? µr v 2 /r R m The static friction force µr must play the role of centripetal force. It will be directed toward the centre of the circle in order to oppose slipping. mg As r is decreased, the static friction force will increase, until it reaches its maximum static value. Friction can not increase beyond this value, and when r decreases further, the car will slip. N2 gives But thus Hence v µr = m r R = mg v µmg = m r 2 2 since there is no vertical acceleration 2 v ' r = = µg % & = 96.5 m at the point of 3 $ " # slipping 24/06/13 Forces ( F.Robilliard) 33

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35 R Fr F W 24/06/13 Forces ( F.Robilliard) 35