College Physics 140 Chapter 4: Force and Newton s Laws of Motion


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1 College Physics 140 Chapter 4: Force and Newton s Laws of Motion We will be investigating what makes you move (forces) and how that accelerates objects.
2 Chapter 4: Forces and Newton s Laws of Motion Forces Newton s Three Laws of Motion The Gravitational Force Contact Forces (normal, friction, tension) Application of Newton s Second Law Apparent Weight Air Resistance Fundamental Forces
3 Forces Isaac Newton was the first to discover that the laws that govern motions on the Earth also applied to celestial bodies. Over the next few chapters, we will study how bodies interact with one another. Simply, a force is a push or pull on an object.
4 Forces How can a force be measured? One way is with a spring scale. By hanging masses on a spring we find that the spring stretch applied force. The net force is the vector sum of all the forces acting on a body. This means force is a vector quantity. F net = ΣF = F 1 + F 2 + F 3 + The units of force are Newtons (N).
5 Inertia and Equilibrium: Newton s First Law of Motion Newton s 1 st Law (The Law of Inertia): If no force acts on an object, then its speed and direction of motion do not change. Inertia is a measure of an object s resistance to changes in its motion. If the object is at rest, it remains at rest (speed = 0). If the object is in motion, it continues to move in a straight line with the same speed. No force is required to keep a body in straight line motion when effects such as friction are negligible. Therefore, turning requires an acceleration. An object is in translational equilibrium if the net force on it is zero.
6 Net Force, Mass, and Acceleration: Newton s Second Law of Motion Newton s 2nd Law: The acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the body s mass. Mathematically: a = F net m or F net = ma Some forces you will deal with: Gravitational, Normal, Friction, Centripetal, Tension,...
7 Net Force, Mass, and Acceleration: Newton s Second Law of Motion An object s mass is a measure of its inertia. The more mass, the more force is required to obtain a given acceleration. The net force is just the vector sum of all of the forces acting on the body, often wri]en as ΣF. If a = 0, then ΣF = 0. This body can have: Speed = 0 which is called static equilibrium, or Speed 0, but constant, which is called dynamic equilibrium. F=ma For Newton s Second Law to be valid it must be applied in an inertial reference frame. An inertial reference frame is one where Newton s First law is valid. A non accelerating reference frame!
8 Interaction Pairs: Newton s Third Law of Motion Newton s 3 rd Law: When 2 bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction. Or, forces come in pairs. Mathematically: F21 = F 12.
9 Interaction Pairs: Newton s Third Law of Motion Example: Consider a box resting on a table. F 1 (a) F 1 is the force of the Earth on the box, what is the interaction partner of this force? The table! F 2 (b) F 2 is the force of the box on the table, what is the interaction partner of this force? The Floor!
10 External forces: Any force on a system from a body outside of the system. F Pulling a box across the floor Force between bodies of a system. F ext Pulling 2 boxes across the floor where the two boxes are a]ached to each other by a rope.
11 Gravitational Forces Gravity is the force between two masses. Gravity is a long range or field force. No contact is needed between the bodies. The force of gravity is always a]ractive! The force between two objects F G = GMm r 2 G = 6.67 x m 3 /kgs 2 r is the distance between the two masses M 1 and M 2 and G = Nm 2 /kg 2. M 1 F 21 F 12 M 2 F21 = F 12. r Because the Earth is so large the force is really large. As you move away from the Earth, the force gets smaller, but not zero!
12 Gravitational Forces! Let M 1 = mass of the Earth. F = GM E # " r 2 $ &M 2 % Here F = the force the Earth exerts on mass M 2. This is the force known as weight, w.! w = GM E # 2 " r E $ &M 2 = gm 2. % M E = kg r E = 6400 km where g = GM E r E 2 = 9.8 N/kg = 9.8 m/s 2 Near the surface of the Earth
13 Gravitational Forces Example: What is the weight of a 100 kg astronaut on the surface of the Earth (force of the Earth on the astronaut)? How about in low Earth orbit? This is an orbit about 300 km above the surface of the Earth. On Earth: w = mg = 980 N! In low Earth orbit: w = mg(r o ) = m GM E # " R E + h ( ) 2 $ & = 890 N % Their weight is reduced by about 10%. The astronaut is NOT weightless!
14 Gravitational Forces Mass vs. Weight We perceive the force of gravity as our weight. Don t confuse this with mass. Mass is an intrinsic property of ma]er, but weight will change depending on gravity. GMm FG = 2 r Since the moon has 1/6 the mass of the Earth, it has 1/6 the force of gravity. This means you will weight 1/6 your weight on earth. If you weight 120 lbs. on earth, then you will weight 20 lbs. on the moon!
15 Contact Forces Contact forces: these forces arise because of an interaction between the atoms in the surfaces in contact. Normal force: this force acts in the direction perpendicular to the contact surface. N Normal force of the ground on the box w N Normal force of the ramp on the box w
16 Contact Forces Example: Consider a box on a table. y FBD for box N x w Apply Newton s 2 nd law F y = N w = 0 So that N = w = mg This just says the magnitude of the normal force equals the magnitude of the weight; they are not Newton s third law interaction partners.
17 Contact Forces Friction: a contact force parallel to the contact surfaces. Static friction acts to prevent objects from sliding. The force of static friction is modeled as fs µ sn. where µ s is the coefficient of static friction and N is the normal force. Kinetic friction acts to make sliding objects slow down. The force of kinetic friction is modeled as f = µ k k N. where µ k is the coefficient of kinetic friction and N is the normal force.
18 Contact Forces Example (text problem 4.95): A box full of books rests on a wooden floor. The normal force the floor exerts on the box is 250 N. (a) You push horizontally on the box with a force of 120 N, but it refuses to budge. What can you say about the coefficient of friction between the box and the floor? FBD for N y box (1) F y = N w = 0 f s w F x Apply Newton s 2 nd Law (2) F x = F f s = 0
19 Contact Forces Example (text problem 4.95): A box full of books rests on a wooden floor. The normal force the floor exerts on the box is 250 N. (a) You push horizontally on the box with a force of 120 N, but it refuses to budge. What can you say about the coefficient of friction between the box and the floor? From (2): F = f s = µ s N µ s = F N = 0.48 This is the minimum value of µ s, so µ s > (b) If you must push horizontally on the box with 150 N force to start it sliding, what is the coefficient of static friction? Again from (2): F = f s = µ s N µ s = F N = 0.60
20 Contact Forces Example (text problem 4.95): A box full of books rests on a wooden floor. The normal force the floor exerts on the box is 250 N. (c) Once the box is sliding, you only have to push with a force of 120 N to keep it sliding. What is the coefficient of kinetic friction? y FBD for box N F x Apply Newton s 2 nd Law (1) F y = N w = 0 (2) F x = F f k = 0 f k w From 2: F = f k = µ k N µ k = F N = 120 N 250 N = 0.48
21 Contact Forces Consider a box of mass m that is at rest on an incline. Its FBD is: F RB y There is one long range force acting on the box: gravity. w x There is one contact force acting on the box from the ramp. If the net force acting on the box is zero, then the contact force from the ramp must have the same magnitude as the weight force, but be in the opposite direction.
22 Contact Forces The force F RB can be resolved into components that are perpendicular and parallel to the ramp. F RB y N The perpendicular component is what we call the normal force. f s w x The parallel component is the static friction force.
23 Example: Let the box on the ramp have a mass 2.5 kg. If the angle between the incline and the horizontal is 25, what are the magnitudes of the weight force, normal force, and static friction force acting on the box? F RB f s θ y N Apply Newton s 2 nd Law F y = N wcosθ = 0 F x = wsinθ f s = 0 w The forces are: x w = mg = 2.5 kg ( )( 9.8 N/kg) = 24.5 N N = wcosθ = ( 24.5 N)cos25 = 22.2 N f s = wsinθ = ( 24.5 N)sin25 =10.4 N
24 Tension This is the force transmi]ed through a rope from one end to the other. An ideal cord has zero mass, does not stretch, and the tension is the same throughout the cord.
25 Free Body Diagrams: Forces Must be drawn for problems when forces are involved. Must be large so that they are readable. Draw an idealization of the body in question (a dot, a box, ). You will need one free body diagram for each body in the problem that will provide useful information for you to solve the given problem. Indicate only the forces acting on the body. Label the forces appropriately. Do not include the forces that this body exerts on any other body. A coordinate system is a must. Do not include fictitious forces. Remember that ma is itself not a force! You may indicate the direction of the body s acceleration or direction of motion if you wish, but it must be done well off to the side of the free body diagram.
26 Example (text problem 4.77): A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is a]ached to the wall. The rope fastened to the wall makes a right angle with the wall. Neglect the masses of the rope and the pulley. Find the tension in the rope from which the pulley hangs and the angle θ. y FDB for the mass M T x w Apply Newton s 2 nd Law to the mass M. F y = T w = 0 T = w = Mg
27 Example (text problem 4.77): A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is a]ached to the wall. The rope fastened to the wall makes a right angle with the wall. Neglect the masses of the rope and the pulley. Find the tension in the rope from which the pulley hangs and the angle θ. FBD for the pulley: Apply Newton s 2 nd Law: y F x = F cosθ T = 0 T θ F x F y = F sinθ T = 0 T = F cosθ = F sinθ This statement is true only when θ = 45 and T F = 2 T = 2 Mg
28 Example: Find the tension in the cord connecting the two blocks as shown. A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m 1 = 3.00 kg and m 2 = 1.00 kg. block 2 block 1 F Assume that the rope stays taut so that both blocks have the same acceleration.
29 Example: Find the tension in the cord connecting the two blocks as shown. A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m 1 = 3.00 kg and m 2 = 1.00 kg. FBD for block 1: FBD for block 2: y y N 1 N 2 T T F x x w 2 F x Apply Newton s 2 nd Law to each block: = T = m 2 a F y = N 2 w 2 = 0 F x w 1 = F T = m 1 a F y = N 1 w 1 = 0
30 Example: Find the tension in the cord connecting the two blocks as shown. A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m 1 = 3.00 kg and m 2 = 1.00 kg. F T = m 1 a T = m 2 a (1) (2) These two equations contain the unknowns: a and T. To solve for T, a must be eliminated. Solve for a in (2) and substitute in (1). " F T = m 1 a = m 1 $ # " F = m 1 $ # T m 2 F T = 1+ m = " % 1 $ ' # & m 2 T m 2 % ' & % " '+T = 1+ m % 1 $ 'T & # & m 2 10 N 1+ 3 kg = 2.5 N " % $ ' # 1 kg &
31 Apparent Weight Stand on a bathroom scale. y FBD for the person: N w x Apply Newton s 2 nd Law: F y = N w = ma y N mg = ma y The normal force is the force the scale exerts on you. By Newton s 3 rd Law this is also the force (magnitude only) you exert on the scale. A scale will read the normal force. N = m g + a y ( ) is what the scale reads. When a y = 0, N = mg. The scale reads your true weight. When a y 0, N > mg or N < mg.
32 Example (text problem 4.128): A woman of mass 51 kg is standing in an elevator. If the elevator pushes up on her feet with 408 newtons of force, what is the acceleration of the elevator? FBD for woman: y N Apply Newton s 2 nd Law: F y w Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s 2 Unknown: a y x = N w = ma y N mg = ma y (1) Solving (1) for a y : a y = N mg m = 1.8 m/s2 The elevator could be (1) traveling upward with decreasing speed, or (2) traveling downward with increasing speed.
33 Air Resistance A stone is dropped from the edge of a cliff; if air resistance cannot be ignored, the FBD for the stone is: y F d w x Apply Newton s Second Law Fy = Fd w = ma Where F d is the magnitude of the drag force on the stone. This force is directed opposite the object s velocity
34 Assume that F d = bv 2 Air Resistance b is a parameter that depends on the size and shape of the object. Since F d v 2, can the object be in equilibrium? F y = F d w = ma bv 2 mg = 0 yes, when v = v t = mg b
35 Example: A paratrooper with a fully loaded pack has a mass of 120 kg. The force due to air resistance has a magnitude of F d = bv 2 where b = 0.14 N s 2 /m 2. (a) If the paratrooper falls with a speed of 64 m/s, what is the force of air resistance on the paratrooper? FBD: F d = bv 2 = ( 0.14 N s 2 /m 2 )( 64 m/s) 2 = 570 N (b) What is the paratrooper s acceleration? y Apply Newton s Second Law and solve for a. F d x F y = F d w = ma a = F d mg m w (c) What is the paratrooper s terminal speed? F y = F d w = ma = 0 v t = bv t 2 mg = 0 mg b = 92 m/s = 5.1 m/s2
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