Forces Part 1: Newton s Laws

Transcription

1 Forces Part 1: Newton s Laws Last modified: 13/12/2017

2 Forces Introduction Inertia & Newton s First Law Mass & Momentum Change in Momentum & Force Newton s Second Law Example 1 Newton s Third Law Common Forces Weight Reaction Force: Normal Reaction Reaction Force: Attached Objects Example 2 Tension Example 3 Friction Static Friction Example 4 Example 5 Kinetic Friction Example 6 Forces Problems: The Recipe Example 7

3 Introduction In previous lectures, we have looked at the motion of an object in terms of its acceleration, velocity and position, without considering what is causing the motion. Motion is caused by forces acting on an object. In this lecture we willl look at Newton s three Laws of Motion, and apply them to determine the equation of motion of an object. The equation of motion is an expression for the acceleration of an object. As we have seen, we can then (at least theoretically) determine expressions for the velocity and position of the object.

4 Inertia & Newton s First Law Inertia is the name given to the property of an object that resists a change in its motion (either speed or direction). Newton s First Law (also known as the Principle of Inertia ) summarizes this: Newton s First Law of Motion In the absence of any external influences a body s state of motion will remain unchanged. This means that without some sort of external interaction: An object that is not moving will remain stationary. A moving object s velocity vector will remain constant.

5 Mass & Momentum A body s inertia is related to its mass - a large mass will resist a change in motion more than a small mass. As we have already seen, mass is one of the basic dimensional quantities of Physics, and is measured using one of the SI base units. The SI unit of mass is the kilogram (kg) The amount of motion of an object of mass m travelling with velocity v is described by the momentum, p of the object, where: p = mv The SI unit of momentum is kg m s 1 v m p Since velocity is a vector, so too is momentum.

6 Change in Momentum & Force An external influence that acts to change the motion of a body will change the momentum of that body. Since momentum is a vector, then p, the change in momentum, must also be a vector. Whatever is causing this change must be a vector also. Clearly, the time t that this effect is acting for will affect the size of the momentum change. p + p p F The size of the external influence must be given by p t. This external effect is called a force. The SI unit of force is the newton (N): 1 N = 1 kg m s 2

7 Newton s Second Law As we shall soon see, it is common for there to be more than one force acting on an object. The vector sum of these forces is known as the resultant force. Newton s Second Law of Motion connects the resultant force on an object with the change in momentum: Newton s Second Law F resultant = F = dp dt = m dv dt + dm dt v when, as is quite common, there is no change in mass (i.e. dm dt =0): F resultant = F = m dv dt = ma By calculating the resultant force, we can then calculate the acceleration, and thus velocity and position of an object.

8 Example 1 A block of mass 2 kg has two forces F 1 = 5i N and F 2 = 4j N acting on it. What is the acceleration vector of the block? F 2 F 1 F 2 = 4j F 1 = 5i The diagram on the right shows the forces all acting at the centre of the block and is called the free body diagram of the block. The resultant force is: F resultant = F 1 + F 2 = 5i + 4j N and, using Newton s Second Law: a = F resultant m = 5i + 4j 2 = 2.5i + 2j m/s 2

9 Newton s Third Law We know that a force will act to change the momentum of an object by p. Where does this momentum come from? There must be a second object, with an equal and opposite change in momentum p. F 1 p 1 + p There will be a second force acting on this object which is equal in magnitude and opposite in direction to the force on the first object: F 1 = F 2 p 1 p 2 F 2 p 2 p In an interaction, forces transfer momentum between objects. Newton s Third Law For every force acting on a object, there is an equal and opposite force acting on a second object.

10 Forces always come in pairs, acting on different objects. For example: Two blocks colliding The Earth and Moon We may think that since the two forces are equal and opposite, that they will just cancel out when adding forces. This does not happen because the two forces act on different objects. It only makes sense to add forces if they act on the same object!

11 Weight Probably the commonest force we will see is the force due to gravity, or weight force on an object. We know the acceleration of a mass m is g, so the weight force W = mg W m g Mass m always has a positive value, and since the direction of g is downwards, W must also always be directed down. Remenber mass is a scalar, while weight is a vector. It is very common to hear someone incorrectly say for example My weight is 70 kg, when they should be saying either My mass is 70 kg or My weight is 700 N.

12 Reaction Force: Normal Reaction A common situation is where we have one object resting on another. For instance, if a wooden block is supported by a horizontal board. There is a weight force W acting down, R so the board must be pushing up on the block with the reaction force R so that W + R = 0. This reaction force is normal (i.e perpendicular) to the board. W If we now rotate the board, the direction of the reaction R force R also rotates, so that it remains normal to the board. (We should expect the magnitude of the reaction force to also change). Clearly, now W + R 0, and there will be an acceleration of the block. W A flat surface will always apply a reaction force to an object resting on it, in a direction normal to the surface. This is generally called the normal reaction force.

13 Reaction Force: Attached Objects What is different about the block and the board if we now glue the block to the board? R When the board is horizontal, then there is no change in the forces seen previously. The reaction force R is normal to the board and W + R = 0. Now, when the board is rotated to a new direction, we know that because of the glue the block remains stationary, so W + R = 0 will still be true. This must always be the case, even when the board is positioned at different angles: W The angle of the reaction force R relative to the board can take any value.

14 Example 2 A block of mass m is placed on an inclined plane as shown. Use Newton s Laws to determine an expression for a, the magnitude of the acceleration of the block. m θ W R a The free body diagram for the block is shown at left. Clearly the motion must be down the plane. There are two forces acting - first, the weight W = mg, and second, the normal reaction force R. Applying Newton s Second Law gives: ma = R + W

15 This vector equation can be written as separate equations for the x and y directions (just as we did for projectile motion problems): F = ma x = R x + W x and F = ma y = R y + W y x y x mg cos θ mg sin θ θ R a W = mg y Solving these equations will usually be simpler if we choose axes so that the x axis is in the same direction as the acceleration a. In this example, this choice has the bonus that R x = 0 and only one of the vectors, W, must be resolved into components. The two force equations become: F = ma x = R x + W x ma = mg sin θ x F = ma y = R y + W y 0 = R mg cos θ y

16 Cancelling m from both sides of the first of these equations gives our desired result: a = g sin θ The second equation, for the y direction, isn t actually needed. What if we instead used the traditional x and y axes? y x R cos θ mg R sin θ θ a cos θ a R a sin θ In this case, it is necessary to resolve two vectors (a and R) into components, as shown. The two force equations become: F = m(a cos θ) = R sin θ x F = m( a sin θ) = R cos θ mg y

17 Now, both equations contain the variables a and R. To solve for a, we need to eliminate R which involves several lines of algebra and trigonometry to obtain the same answer as before: a = g sin θ The first choice of axes made the problem much simpler. Tips: Always try to choose your x and y axes to minimize the number of vectors needing to be resolved into components. Any choice of axes will work, as long as x and y are perpendicular, but a smart choice can make the algebra easier. Most forces problems, like this one, will reduce down to a set of simultaneous equations in 2 or 3 variables. Thinking carefully about the what is a variable (here: a and R), and what is a parameter (here: m,g,θ) is important.

18 Tension Tension is the force applied to an object by an attached string (or rope, cable, chain, cord, wire... Anything long and flexible). It should be obvious that this tension force will always pull on the object. i.e. A tension force T is always directed away from the object it acts on. Consider a block of mass M being suspended by a rope of mass m, as shown at right. Each end of the rope will apply a tension force (a) T 1 on the block, and (b) T 2 on the roof. What can we determine about the magnitudes of these two tension forces? M T 2 T 1

19 Let us look at a small segment of the rope with mass m. The other sections of the rope will pull down with tension t 1 and up with tension t 2 as shown. There will also be a weight force ( m)g acting down, and since the rope is at rest, applying Newton s Second Law gives: F = 0 = t2 t 1 ( m)g Therefore we have: t 2 = t 1 + ( m)g If we extend this logic to larger and larger sections of the string, we can find the connection between the tensions at the different ends of the string. In general, the two ends of a rope or string will apply different tension forces. M M t 2 m t1 T 2 = T 1 + mg T 1

20 The block is at rest which tells us that T 1 = Mg and thus that T 2 = (M + m)g. If the mass of the string m is much smaller than the mass of the block M, then M + m M, and: T 2 = (M + m)g Mg = T 1 If the mass of the string is negligible (i.e. small enough to ignore, alternatively, the string is massless ) then the tensions at both ends will have the same magnitude (but different directions). In this course, all strings will have negligible mass.

21 Example 3 A string is used to pull a block m up an inclined plane, as shown. Calculate the tension force T required to move the block at a constant speed v. T v m θ The first thing we must realize, is that constant speed a = 0. T W R a = 0 The free body diagram for the block has the same two forces as in the previous example - the weight W = mg, and the normal reaction force R. Additionally there is the tension force T acting up the plane. Newton s Second Law gives: 0 = R + W + T

22 As before we need to choose x and y axes to work with. Any choice will work, but a smart choice will reduce the amount of work we need to do. y T x mg cos θ θ mg sin θ R a = 0 W = mg The best choice is the same as the previous example. Only one force, the weight, needs to be resolved. A different choice of axes would mean we need to resolve two or even three vectors. The two force equations become: F = 0 = mg sin θ T and x F = 0 = R mg cos θ y The first of these gives the required answer: T = mg sin θ

23 Friction In addition to the normal reaction force already discussed, there may also be friction between an object and a surface. The friction force will be parallel to the surface and has two different forms, depending on the motion: Static Friction occurs when an object is stationary relative to the surface, and acts to oppose potential motion. Kinetic Friction occurs when the object is moving relative to the surface and acts to oppose this motion. A surface that has a frictional force is known as a rough surface, while a surface with no (or negligible) friction is said to be smooth.

24 Static Friction A block is resting on a rough surface. If we apply a small force F, the block does not move. There must be a frictional force f s which is equal and opposite to F. f s F R W Static friction f s acts to stop potential motion. If we increase the applied force F, the opposing static friction f s must also increase, until at some point the block starts to move. The static friction f s has a maximum magnitude, f s,max. This maximum value is related to the magnitude of the normal reaction force: f s,max = µ s R where µ s is the coefficient of static friction between the block and the surface. Note that µ s is a dimensionless number, typically less than 1.

25 Example 4 A block of mass m = 100 kg is placed on a surface with a coefficient of static friction µ s = 0.5. A rope is attached to the block and we attempt to move the block by pulling horizontally on the rope. Calculate the minimum tension force T required to move the block. y a = 0 x f s mg R T The free body diagram for this block is shown at left. For small values of the force T, the friction f s will balance it out and the block will not move. The minimum value of T for movement to occur is when T min = f s,max = µ s R Since the forces in the y direction add to zero, R = mg and: T min = µ s R = µ s mg = = 500 N

26 Example 5 Re-do the previous problem but with the rope lifted so that it is now pulling at an angle of θ = 45 above the surface. The free body diagram for this block is only slightly modified from the previous example. y x f s a = 0 R mg θ T cos θ T T sin θ Again we apply Newton s Second Law in both x and y directions: F = 0 = T cos θ f s x F = 0 = R + T sin θ mg y The minimum value of T T min cos θ = f s,max = µ s R for movement to occur is when

27 The two equations now become: 0 = T min cos θ µ s R 0 = R + T min sin θ mg and to find T min we must eliminate R: T min = 0 = T min cos θ µ s (mg T min sin θ) µmg cos θ + µ sin θ Substitute the known values θ = 45, µ = 0.5 and m = 100 kg: T min = 471 N Which is less than the value required to pull horizontally! This is because the reaction R and hence the friction f s is smaller in this case.

28 Kinetic Friction Now we consider a block moving on a rough surface. The kinetic friction f k is directed in the opposite direction to the motion. f k R W v Kinetic friction f k acts to oppose actual motion. For most surfaces, this friction is (approximately) independent of velocity and the magnitude of f k is related to the magnitude of the normal reaction force: f k = µ k R where µ k is the coefficient of kinetic friction between the block and the surface. Like µ s, µ k is a dimensionless number and typically µ k < µ s.

29 Example 6 A 5 kg mass rests on a horizontal surface with coefficient of kinetic friction µ k = 0.5. Find the acceleration a when a force of F = 50 N acts horizontally on the mass. The free body diagram for this mass looks very similar to that seen in Example 4, except in this case there is a non-zero acceleration. y x f k a R mg F Applying Newton s 2nd Law in the x and y directions: F = ma = F f k = F µ k R x F = 0 = R mg y

30 From the y direction equation R = mg which we substitute into the x equation: ma = F µ k R = F µ k mg a = F m µ kg Finally we can substitute the given values, F = 50 N, m = 5 kg and µ k = 0.5 to obtain the answer: a = = 5 m/s2 5

31 A plot of frictional force f on a block resting on a rough surface, against the applied force F, will look like: f s,max = µ sr f k = µ k R f Static friction Kinetic friction f s = F No Motion Block is Moving Before the block moves, the (static) friction is variable, up to the maximum value. As the block starts to move, the (kinetic) friction will be constant. F

32 Forces Problems Recipe THINK, then draw in direction of acceleration a Draw a free body diagram, including all forces on the object mass? weight force down. rope? tension force away. contact with surface? normal reaction force + (maybe) friction. Choose directions of x and y axes to minimize number of vectors needing to be broken into components. (Usually this means x in the direction of a) Resolve any vectors not already in x or y directions. Use Newton s 2nd Law in both x and y directions: F = ma =... F = 0 =... x Solve these equations, usually for a. Substitute numerical values of the variables, if given. y

33 Example 7 Fluffy the cat is still enjoying the ride on the turntable seen in the last lecture. If the coefficient of static friction between Fluffy and the turntable is µ s, determine an expression for the maximum angular velocity ω that the turntable can reach before Fluffy slips. rotation r Fluffy is moving, BUT so is the turntable. There is no RELATIVE movement between Fluffy and the turntable, so the friction acting must be static friction.

34 What direction is the friction? Relative to the turntable, Fluffy s inertia is acting to send Fluffy flying off in a tangential direction. The friction is preventing this, by acting toward the centre of the circular path (i.e. the central point of the turntable). After remembering from our previous lecture on rotational motion that the direction of the acceleration must be toward the centre of the path (and also that it is connected to the angular speed ω by a = ω 2 r), the free body diagram is straightforward. x y f s mg a R Applying Newton s 2nd Law in the x and y directions: F = ma = mω 2 r = f s x F = 0 = R mg y

35 Using the x equation, we find an expression for ω: fs ω = mr Clearly, as we increase the angular speed, the friction must also increase until it reaches its maximum value f s,max = µ s R ω max = fs,max mr = µs R mr µs g = r Where we have used R = mg from the y equation above. Increasing the angular speed past this point will cause poor Fluffy to be sent flying off the turntable..