Two books laid the foundations for our understanding of forces and motion... CHAPTER 4 NEWTON S LAWS. Little bit of history. Forces.

Transcription

1 Two books laid the foundations for our understanding of forces and motion... Little bit of history orces Newton s 1st Law CHAPTER 4 NEWTON S LAWS Newton s 2nd Law mass and weight Newton s 3rd Law action reaction pairs free body diagrams the normal force analysis of day-to-day situations

2 orces A force is best described as an action or influence that when acting on an object, will change its motion. There are two main types of forces: contact forces (push, pull, friction, etc.) direct force non-contact (or field) forces (gravity, magnetic, electric, etc.) action at a distance and long range... but they are treated in exactly the same way. Dimension [M][L] [T] 2 (show later) Unit: Newton (N) 1N weight of a medium size apple, i.e., the downward force exerted on your hand when you hold it.

3 orce is a vector, i.e., with both direction and magnitude and so to find the resultant or the net force when several forces act on a object, they must be added vectorially The net force is R = i i = R R y R θ x = cosθ y = sin θ x Question 4.1: An object on a horizontal table is acted on by three forces, 1 = 3 N, 2 = 4 N and 3 = 3 N, as shown below. What is the net force acting on the object? " 60 " 2 The components of the net force are: Rx = x and Ry = y.

4 y " 3 = 3 N " 1 = 3 N " 2 = 4 N x Define x and y axes. Then, the net force in the x direction is x = i xi and the net force in the y direction is y = x = 1 cos cos60 3 = 2.60 N N 3.00 N = 1.60 N, and y = 1 sin sin 60 yˆ j xˆ i θ " R The angle θ = tan 1 = 1.50 N 3.46 N = 1.96 N. Thus, the net force is " R = (1.60ˆ i 1.96ˆ j ) N, with magnitude " R = ( x ) 2 + ( y ) 2 = (1.60 N) 2 + ( 1.96 N) 2 = 2.53 N. y x i yi. = tan 1 (1.225) = Example of forces: N W W W R Important concept... Two forces: the weight of the ball downward (due to the attraction of the Earth) is balanced by the normal force of the ground on the ball. No net force no motion Two forces: the weight of the ball downward and the force of the foot on the ball. Net resultant force change in motion acceleration One force: the weight of the ball downward. Net force change in motion acceleration... we cannot neglect the effect of the Earth... it always produces a non-contact force

5 Analysis of moving objects led to... A simple example... Newton s 1st Law: A body at rest, or moving with constant velocity, will remain at rest, or moving with constant velocity, unless it is acted on by a net force. (NB: the reference frame is the Earth.) Sometimes called the Law of inertia... Tendancy for a body that s moving to keep moving. Tendancy for a body at rest to remain at rest. If the sum of forces i, i.e., the net force, on an object is zero it remains at rest or continues moving with constant velocity. ( x = 0 and y = 0.) This is called EQUILIBRIUM. Implication to change motion we have to apply a net force... Newton s 2nd Law gives us the relationship between the force and the change in motion. 1 The force on you due to the Earth, i.e., your weight, (downward) is balanced by the force of the seat of the chair on you (upward). No net force. 2 The force on the ice cream cone due to the Earth (downward) is balanced by the force of your hand on the cone (upward). No net force. 3 The total weight of you, the ice cream cone and the chair (downward) is balanced by the force of the ground on the chair legs (upward). No net force.

6 But don t get fooled some common misconceptions... Are you thrown backwards when a car accelerates from rest? NO There is no throwing force; you tend to remain at rest as the seat moves forward DISCUSSION PROBLEM [4.1]: Are you thrown outwards when a car turns a corner? NO There is no throwing force; as the car turns, you tend to continue to move in a straight line; so the side of the car exerts an inwards force on you Why does a wire stretched between two posts always sag in the middle no matter how tightly it is stretched?

7 So, to produce a change in motion, there must be a net force... v... to move an object from rest, i.e., to change its velocity, v frictional force opposing motion or slow down a puck or change the direction of a ball. v 1 v 2 v Newton s 2nd Law: In the Principia Newton stated that The change in motion is proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed. Today, we write the 2nd Law as = d p dt, where p = m v, which today is called the momentum; Newton called it the quantity of motion. = d p dt = d(m v ) = m d v dt dt + v dm dt = m a + v dm dt. However, most textbooks ignore the full Newtonian expression and assume that m is constant. So, you will find Newton s 2nd Law usually stated this way: If a net force acts on an object it accelerates (i.e., there is a change in motion). The direction of the acceleration is the same as the net force, i.e., i i = m a.

8 And in component form, x = ma x : y = ma y : z = ma z where m is a constant. Remember, however, these expressions are only true if the mass remains constant. So, this form of the Law does not apply to a rocket that is losing mass by burning fuel. NOTE: The net force has to be an external force, i.e., not inside the object or system, e.g., you can t lift yourself The product m a is not itself a force... it is the net force that is put equal to m a. The acceleration a is parallel to the net force. The dimension of force is the same as mass acceleration [M] [L] [T] 2 = [M][L] [T] 2. The unit of force is the Newton (N). " 3 Question 4.2: An object on a horizontal table is acted on by three forces, 1 = 3 N, 2 = 4 N and 3 = 3 N, as shown above. If the mass of the object is 2.0 kg, and it is initially at rest at the origin, " " 2 (a) what is its acceleration? (b) What is its velocity after 4.0 s? (c) What is its displacement after 4.0 s? ˆ j ˆ i

9 " 3 = 3 N y " 1 = 3 N x " 2 = 4 N (a) rom before (Q4.1) we found the resultant force was: " R = (1.60ˆ i 1.96ˆ j ) N. Using Newton s 2nd Law " R = m a ". a " " = R m = (1.60ˆ i 1.96ˆ j ) N 2.0 kg = (0.80ˆ i 0.98ˆ j ) m/s 2, and the magnitude of the acceleration is " a = (0.80 m/s 2 ) 2 + ( 0.98 m/s 2 ) 2 = 1.27 m/s 2 and it is parallel to " R. (c) r r " = v " t + 1 a t 2 = 1 (0.80ˆ i 0.98ˆ j ) m/s 2 (4.0 s) = (6.40ˆ i 7.84ˆ j ) m. r r " = (6.40 m) 2 + ( 7.84 m) 2 = 10.1 m. (b) " v = " v + " a t = (0.80ˆ i 0.98ˆ j ) m/s s " v = = (3.20ˆ i 3.92ˆ j ) m/s. (3.20 m/s) 2 + ( 3.92 m/s) 2 = 5.06 m/s.

10 The constant, m, in the expression = ma is really a measure of how difficult it is to change the motion of an object. We call it the inertial mass, or simply, the mass. Note: mass is a scalar quantity. To understand that concept, think about how you check the weight of something... DISCUSSION PROBLEM [4.2]: We have just seen that the way we choose between the weights of two objects we move them up-and-down and determine how difficult it is to move them. m 1 m 2 m 1 m 2... you check to see how easy or difficult it is to make them move or the same acceleration... a = 1 m1 = 2 m2, Will this idea work in a weightless environment? i.e., m 1 m2 = 1 2 So, in fact, you are comparing the masses

11 OK... so, a force changes motion. But when you throw a baseball, what keeps it moving to the right after it leaves your hand? The answer is... inertia... remember Newton s 1st Law. Neglecting air resistance, there are no horizontal forces to change its horizontal motion, so it keeps going to the right Note, inertia is NOT a force. w Question 4.3: An object of mass m is traveling at an initial speed v = 25.0 m/s. A net force frictional force of 15.0 N acting on the object in a direction opposing the motion, causes it to come to a stop after a distance of 62.5 m. (a) What is the mass of the object? (b) How long does it take to stop? There is a force downward (its weight), which causes the ball to fall. These effects combine so the ball follows a parabolic trajectory.

12 v = 25.0 m/s v = 0 x = 15.0 N x (x x ) = 62.5 m +ve direction Define the positive direction to the right. irst, we need to find the (negative) acceleration. (a) We have: v 2 = v 2 + 2a(x x ) a = v2 v 2 (25.0 m)2 = 2(x x ) m = 5.00 m/s2. Note that opposes the motion, i.e., it acts in the ve direction (same as the acceleration). Using Newton s 2nd Law, = ma, i.e., m = a = 15.0 N 2 = 3.00 kg m/s Question 4.4: Two forces, 1 and 2, act on an object on a frictionless surface, as shown. In which case is the acceleration greatest? 2 = 3 N 1 = 12 N 2 = 4 N 1 = 8 N 6 kg 4 kg A B 2 = 2 N 1 = 8 N 2 = 2 N 1 = 8 N 3 kg 2 kg C D (b) We have v = v + at, i.e., t = v v a = m 2 = 5.00 s m/s

13 2 = 3 N 6 kg 1 = 12 N 2 = 4 N 4 kg 1 = 8 N A B 2 = 2 N 3 kg 1 = 8 N 2 = 2 N 2 kg 1 = 8 N What s the difference between mass and weight? C D The net force on a block in each case is 1 2 =. By Newton s 2nd law, that force produces an acceleration given by = ma, i.e., a = m. MASS: inertial property larger mass larger force required to accelerate. WEIGHT: the force exerted by the Earth on an object. a A = 9 N 6 kg = 1.5 m/s2 : a B = 4 N 4 kg = 1 m/s2 a C = 6 N 3 kg = 2 m/s2 : a D = 6 N 2 kg = 3 m/s2 So, the answer is D. Example: Bowling ball difficult to throw (mass) difficult to lift (weight)

14 If you analyze a situation involving forces you ll find that forces ALWAYS occur in pairs: BT TB B Who pulls harder, i.e., who applies the greater force, in the scenario shown here? A AB BA B Books on table (down) Table on books (up) AB BA B (Spring) AB BA 50 BA 0 A B Pulling force A (to the left) is applied on B ( AB ). Pulling force B (to the right) is applied on A ( BA ). AW WA A Pulling force A (to the left) is applied on the wall ( AW ). The wall applies a force (to the right) on A ( WA ). 50 AB Time (s) Ha They both pull equal amounts... the force exerted by A on B is equal and opposite to the force exerted by B on A, i.e., AB = BA and AB = BA at all times This leads us to...

15 Examples of action/reaction pairs: BT TB Newton s third Law: If an object A exerts a force on an object B (an action), object B exerts an equal and opposite force on A (the reaction), i.e., Note: AB = AB = BA ( BA ) the two forces act on different objects, sometimes called an action/reaction pair. B B B = B Books on table (down) Table on books (up) BT = TB E E E = E the Law applies whether the objects are at rest, moving with constant velocity or accelerating. That s weird How come it seems to hurt your eye more than it hurts your finger? But we have to be careful in identifying action/reaction pairs correctly (later).

16 Often, Newton s laws appear in combination look... Centripetal force (Action) Car on you Reaction ( Centrifugal force ) You on car irst Law: Every body perseveres in its state of rest, or of uniform motion in a right [straight] line, unless it is compelled to change that state by forces impressed thereon. Third Law: To every action there is always opposed an equal reaction: or the mutual action of two bodies on each other are always equal, and directed to contrary parts. Principia Mathematica (1687). So, if the Earth applies a force on an apple (its weight), the apple must apply an equal and opposite force on the Earth Does that mean the Earth moves towards an apple when we drop the apple? Let s see... We know EA = AE, and the 2nd Law tells us: EA = m A a A and AE = m E a E. m A a A = m E a E a E = m Aa A m E. But m E kg, m A 0.1 kg and a A = g a E m/s 2. If the apple takes 1 s to reach Earth... WOW... that s small Δy A = 1 2 a At m. Δy E = 1 2 a Et m. EA AE

17 Take something as simple as a book or cups on a table on the Earth (i.e., the floor)... ~ Let s identify ALL action-reaction pairs ~ DISCUSSION PROBLEM [4.3]: You are holding a baseball in your hand. The reaction force to the weight of the baseball is... orces acting ON the book: TB and EB (= w B ) These are not an action-reaction pair Why?? orces acting ON the table: BT and ET (= w T ) orces acting ON the Earth: BE and TE or action-reaction pairs: 12 = 21. A: the gravitational force of the Earth on the ball, B: the gravitational force of the ball on the Earth, C: the force of your hand on the ball, D: the force of the ball on your hand, or E: the gravitational force of the Earth on your hand.

18 Does Newton s 3rd Law apply if objects are moving? You betcha W R S You know all about the birds and bees, but now it s time to learn about BD s Believe me, dear, crucial to solving force problems is the free-body diagram. Identify all the forces associated with the rope: (Take L R as positive direction) RS (rope on sled)* + SR (sled on rope)* WR (woman on rope)# + RW (rope on woman)# But RS = SR and WR = RW. The net force on the rope WR SR. So, if WR > SR the rope and everything attached to it, moves to the right, but the equations above still hold A free-body diagram (BD) shows all of the forces acting ON an object. or example, the forces acting on a ball that s been kicked... N W W W

19 The Normal orce DISCUSSION PROBLEM [4.4]: Which of the free-body diagrams below best represents the forces acting on an object sliding down a frictionless incline? v In mechanics, the normal force N (sometimes n or N) is the component of the contact force exerted on an object, perpendicular to the surface of contact by, for example, the surface of a floor or incline, preventing the object from penetrating the surface. In the case of an object N mg resting on a horizontal surface (e.g., a table top, the ground), the normal force is equal and opposite to the weight of the object (Newton s 3rd Law), i.e., N = mg. A B C D When an object rests on an incline, the normal force is N mgcosθ mg mgsin θ θ perpendicular to the plane the object rests on. In that case, the normal force is equal and opposite to the component of the weight perpendicular to the surface, i.e., N = mg cosθ.

20 There is a force acting parallel to the incline, i.e., the component of the weight parallel to the surface, mg sin θ. N N STRATEGY for analyzing force problems: θ mg mgcosθ mgsin θ R θ mg mgcosθ mgsin θ Draw free body diagram (identify the forces acting ON an object) Newton s third Law Therefore, when an object sits on a frictionless incline, the resultant (net) force, R = mg sin θ produces motion down the incline. By Newton s 2nd Law the acceleration of the object down the slope is given by a = R m a result I showed in Chapter 2. = gsin θ, Equilibrium (i.e., at rest or constant velocity) Newton s first Law Separate the force components in two (perpendicular) directions. Motion (i.e., accelerating) Newton s second Law

21 2 f 1 Question 4.5: Two forces, 1 and 2, act on an object on a surface, as shown. If the objects are all moving with a constant velocity, in which case is the frictional force greatest? 2 = 3 N 1 = 12 N 2 = 4 N 1 = 8 N 6 kg 4 kg A B 2 = 2 N 1 = 8 N 2 = 2 N 1 3 kg 2 kg = 8 N C D Draw the BD for the blocks. Since the blocks are not accelerating, Newton s 2nd Law tells us the net force on each block is zero, i.e., 1 2 f = 0 f = 1 2. ( 1 2 ) is largest for block A ( = 9 N to the right). Since the net force must be zero, the frictional force of the ground on block A must be 9 N (to the left). 2 = 3 N 1 = 12 N 2 = 4 N 1 = 8 N 6 kg 4 kg f A B 2 = 2 N 1 = 8 N 2 = 2 N 1 = 8 N 3 kg 2 kg C D

22 DISCUSSION PROBLEM [4.5]: A B You and a friend have identical cars. You want to rip an old pair of Levi s apart; there are two possible arrangements, as shown above. Which is more likely to do the job, i.e., in which is the tension in the jeans greater? A: A B: B C: They re the same Question 4.6: To prevent a box from sliding down a frictionless inclined plane, physics student Anna pushes on the box horizontally with just enough force so that the box is stationary. If the mass of the box is 2.00 kg and the slope of the incline is 35, what is the magnitude of the force she exerts? Anna 35

23 Draw the free body diagram for the box: N A mgcosθ y x The components of the weight force along x and along y are ( mg sin θ) and ( mg cosθ), respectively. The components of the force Anna applies ( A ) along x and y A sin θ A θ A cosθ θ θ mg mgsin θ are A cosθ and A sin θ, respectively. Summing the forces in the x-direction, we have = A cosθ mg sin θ, i.e., A = mg tanθ. A = ( 2.00 kg) 9.81 m/s 2 x = 0 ( ) ( ) = 13.7 N. Question 4.7: The system shown below is in equilibrium. ind the three tensions T 1, T 2 and T 3 in the string, and the unknown mass, M T 3 T 1 M T2 6 kg

24 T 2 mg irst draw the free body diagram for the 6 kg mass; the two forces are the weight (mg) and T 2. Since the system is in equilibrium, T 2 = mg = 6.00 kg 9.81 m/s 2 = 58.9 N. M 60 T 3 T 1 T T 3 T 1 M T2 6 kg 6 kg The tensions all pass through a single point (circled); draw the free body diagram at that point. Then, we have T T 3 T 2 i.e., T 1 + T 3 = h = 0 and v = 0. (h) T 1 sin 30 + T 3 sin 30 = 0. T 1 = T 3 (v) T 1 cos30 + T 3 cos30 T 2 = 0, T 2 cos30 = 58.9 N = 68.0 N T 1 = T 3 = 34.0 N. Since the pulley only changes the direction of T 1, it does not affect its magnitude. T 1 = Mg, i.e., M = T 1 g = 34.0 N 2 = 3.47 kg m/s

25 Sometimes, we may not know the angles as in this problem... Identify all of the forces acting through Isobel s hand and draw the vector force diagram. Question 4.8: Isobel Newton hangs motionless by one hand from a clothes line, as shown. If the clothes line is about to break, which side is most likely to break? 1 2 w w 1 2 Since she is not moving, the net force on her hand is zero, i.e., net = w = 0. So the vector diagram shown on the right must be closed. Clearly, 1 > 2 and so the part of the line on the right has the greater tension and so is more likely to break A: The left side. B: The right side. C: Equal chance of either side breaking. Hence, even if the angles are not known, you can use the fact that the vector diagram of forces must be a closed figure to solve a problem.

26 Question 4.9: A 2 kg mass hangs from a spring scale, calibrated in Newtons, that hangs from the ceiling of an elevator. What does the scale read when the elevator is (a) moving up with a constant velocity of 30 m/s, (b) moving down with a constant velocity of 30 m/s, and (c) is ascending at 20 m/s and gaining speed at the rate of 3 m/s 2? (d) rom t = 0 to t = 5 s, the elevator moves up at 10 m/s. Its velocity is then reduced uniformly to zero in the next 4 s, so that it comes to rest at t = 9 s. What does the scale read during the time interval 0 < t < 9 s? N (Upward force due to spring balance, i.e., apparent weight. a v w = mg (orce due to Earth) Draw the free body diagram for the 2 kg block. In parts (a) and (b) the block is not accelerating. a v = 0 so v = 0. Take up direction as positive: N +( mg ) = 0, i.e., N = mg = (2 kg)(9.81 m/s 2 ) = 19.8 N... (2 kg). (c) Since the block is accelerating upward we have: v = ma v ( a v > 0). Then N +( mg ) = ma v i.e., N = m(g + a v ) = (2 kg)(12.81 m/s 2 ) = 25.6 N... (2.61 kg). (d) When 0 < t 5 s, a v = 0, so N = 19.8 N.

27 This is ME... Draw the free body diagram for the forces on me: N (Upward force from scale) w = mg (force due to Earth) When 5 s < t 9s, a v = Δv (0 10 m/s) = = 2.5 m/s 2. Δt 4 s Since N +( mg ) = ma v... but a v < 0. N = m(g + a v ) = (2 kg)(7.31 m/s 2 ) = 14.6 N... (1.49 kg) v v = N mg = ma v. My apparent weight is the force I exert on the scale 4.0 a v (m/s 2 ) N = m(g + a v ). acceleration upward 2.0 acceleration downward 4.0 So, if you know your true weight, you can find the acceleration of an elevator Apparent weight (pounds)

28 Note that the apparent weight of someone in an elevator if the cable breaks is zero. Draw the BD for the forces acting on the person... N (Upward force from scale) w = mg (force due to Earth) Yikes rom the free-body diagram N mg = ma v, so apparent weight ( ) = m( g + ( g)) N = m g + a v = 0 Conceptually what s happening is... the floor and weighing scales are falling at the same rate as the person inside the car so there s no net force on the weighing scale. Question 4.10: Two objects are connected by a massless string, as shown above. The incline and pulleys are frictionless. (a) Show that the acceleration of the objects is a = (m 2 m 1 sin θ)g. (m 1 + m 2 ) (b) ind the acceleration and tension in the string when m 1 = m 2 = 5 kg and θ = 30. This is what is popularly known as weightlessness except, of course, the person still has real weight

29 (a) Treat each object separately. Note that T, the tension, is common to both free-body diagrams as the pulley only changes the direction of T. Assume m 2 descends (i.e., m 1 ascends the slope). [#1] along x: T + ( m 1 gsin θ) = m 1 a... (i) [#2] along y: T + ( m 2 g) = m 2 a (ii) Subtract (ii) from (i): N T m 1 g (m 2 m 1 sin θ)g = (m 1 + m 2 )a a = (m 2 m 1 sin θ)g. (m 1 + m 2 ) Note: if m 2 > m 1 sin θ then a > 0, if m 2 < m 1 sin θ then a < 0, if m 2 = m 1 sin θ then a = 0. y x y T m 2 g x y N T m 1 g (b) With m 1 = m 2 = 5 kg and θ = 30, ( a = m 2 m 1 sin θ)g ( m 1 + m 2 ) ( ) (5 kg + 5 kg) ( 5 kg ( 5 kg 0.50 ) ) 9.81 m/s2 = = 2.45 m/s 2. Since a > 0, m 1 moves up the slope. Also, using equation (ii), T = m 2 (g a) = (5 kg)(7.36 m/s 2 ) = 36.8 N. x y T m 2 g x

30 N 1 N m 1 g m 2 g Draw the free-body diagrams for each object. It is the contact force of m 1 on m 2 that causes m 2 to accelerate. Question 4.11: Two blocks are in contact on a frictionless, horizontal surface. A horizontal force is applied to one of them, as shown. ind the acceleration and the contact force for (a) general values of, m 1 and m 2, and (b) when = 3.2 N, m 1 = 2 kg and m 2 = 6 kg. (a) Consider only horizontal forces (why??)... # = m 1 a. # = m 2 a. But 12 = 21. a = The contact force is: 12 (= 21 ) = m 1 + m 2. m 2 m 1 + m 2. (b) a = = 3.2 m 1 + m = m/s2. 12 = = 2.4 N

31 (a) m 1 < m 2 Do the results change if the force is applied from the right? Conceptually, the magnitude of the acceleration will be the same because the total mass (m 1 + m 2 ) and are the same. But, the contact force will be smaller since a smaller force will be required to produce the same acceleration for m 1. We get the same result analytically. (b) N 1 N 2 We have 21 m 1 g 12 m 2 g and 12 = m 2 a, 21 = m 1 a. Eliminating a, and putting 12 = 21, we find m 21 = 1 (m 1 + m 2 ), i.e., the contact force of m 2 on m 1, 21, depends on the mass of the second block, m 1, which in this case is smaller (2 kg) than in the problem (6 kg).