3 In this chapter you will Use Newton’s laws to solve problemsDetermine the magnitude and direction of a net force that causes a change in the motion of an objectClassify forces according to their cause
4 Section 4.1 Force & MotionForce - a push or pull acting on an object that can cause the object to speed up, slow down, or change directionForces have both magnitude and direction - they are ___.Forces are divided into contact and field forcesFields: electric, magnetic
5 Free-body diagrams Draw vectors away from objects Table pushing up on booksBooks pushing down on table
6 Determine net force Add forces acting in the same direction Subtract forces acting in opposite directions.3n E + 2n W n EF1 = 3.0 n E F2 = 2.0 n E F net = n E
7 Vectors at Right Angles Sin q = opp hypCos q = adj hypTan q = opp adj
8 Newton’s 2nd Law a = Fnet / m The acceleration on an object is equal to the sum of the forces acting on an object divided by the mass of the object.
9 Newton’s 1st Law (Inertia) An object has a tendency to resist a change in its motion unless there is an outside net force acting on it.No net force can result inNo motion or constant motionKnown as equilibriumA net force can result inSpeeding up or slowing down
10 Problem1. A rock falls freely from a cliff. Draw vectors and label each.+yV a Fnet
11 A skydiver falling towards earth at a constant rate… +y F air resistance on diverF net = 0F Earth’s mass on diverva = 0vv
12 A rope pulls a box at a constant speed across a horizontal surface A rope pulls a box at a constant speed across a horizontal surface. The surface provides a force that resists the box’s motion.+xvvvvF friction on boxF pull on boxF net = 0
13 ProblemTwo horizontal forces, 255 n and 184 n, are exerted on a boat. If these forces are applied in the same direction, find the net horizontal force on the boat.F net = 255n + 184n = 4.39 x 102 n in the direction of the forces.
15 4.2 ObjectivesDescribe how the weight and mass of an object are related.Differentiate between actual weight and apparent weightUse Newton’s 2nd Law in two forms:F = maF = mg
16 A ball in mid-air in free fall has only the force of gravity acting on it. Air resistance can be neglected.SystemVaKnown:a = gmUnknown:FgFgFnet = maFnet = Fg a = gTherefore:Fg = mg
17 The only force acting on the falling ball is Fg. Fg is the weight force.Fg is acting down as are the velocity and the accelerationNewton’s 2nd law has become:Fg = mg
18 How a bathroom scale works. When you stand on the scale the spring exerts an upward force on you while are in contact with the scale.You are not accelerating, so the net force acting on you must be zero.The spring force, Fsp upwards must be opposite and equal to your weight Fg that is acting downward.
19 Newton’s 2nd Law Problem Two girls are fighting over a stuffed toy (mass = 0.30 kg). Sally (on left) pulls with a force of 10.0 n and Susie pulls right with a force of 11.0 n. What is the horizontal acceleration of the toy?
21 Solution to Susie’s & Sally’s dilemma. In this chapter you will:Find the net Force:11.0n R + (-10.0n L)Fnet = 1.0 n RFnet = ma a = Fnet / mwa = 1.0 n / 0.30 kg = 3.33 m/s2 Right
22 Apparent WeightThe force an object experiences as a result of the contact forces acting on it, giving the object an acceleration
23 Real and Apparent Weight same when a body is traveling either up or down at a constant rate, in an elevator, for example.Apparent weight < real weight when the elevator is slowing while rising or speeding up while descending.Apparent weight > real weight when speeding up while rising or slowing while going down.
24 Apparent weight isless when…FscaleApparent weight is greater whenFscaleFgFgSlowly rising or speeding up while descending.Speeding up while rising or slowing while going down
30 Problem On Earth, a scale shows that you weigh 585 n. A. What is your mass?B. What would the scale read on the Moon where g = 1.60 m/s2?C. Back on Earth, what do you weigh in pounds? (1 kg = 2.2 kg)
31 A. What is your mass?m = Fg / gm = 585 n /9.8 m/s2m = 59.7 kg
32 B. What would the scale read on the Moon where g = 1.60 m/s2? Fg = mgmoonFg= (59.7 kg)(1.60m/s2)Fg = 95.5 n
34 Drag Force and Terminal Velocity When an object moves through a fluid (liquid or gas), the fluid exerts a drag force opposite to the direction of motion of the object.The force is dependent upon the motion of the object and the properties of the fluid (temperature and viscosity - resistance to flow).As the object’s velocity increases, so does the drag force. The terminal velocity is the maximum velocity reached by the object as it moves through the fluid.
36 4.3 Interaction Forces In this section you will : Define Newton’s Third lawExplain tension in strings and ropes in terms of Newton’s 3rd lawDefine the normal forceDetermine the value of the normal force by applying Newton’s 2nd law
37 Identifying Interactive forces You are on skates and so is your friend. You push on their arm to move them forward and they exert an equal and opposite force on you which causes you to move backwards.These forces are an interaction pair.An interaction pair (or action and reaction) is two forces that have equal magnitude and act in opposite directions.
39 The forces simply exist together or not at all. They result from the contact between the two of you.The two forces act on different objects and are equal and oppositeNumerically, F A on B = - F B on A
40 Practice Problem 32. Someone please read the problem. Identify the bucket as the system and up as positive.Fnet = Frope on bucket - FEarth’s mass on bucket = maa = (Frope on bucket - FEarth’s mass on bucket)/ma = (Frope on bucket - mg)/ma = [450n - (42kg)(9.80m/s 2 )] / 42 kga = 0.91 m/s2
41 When a softball of mass 0.18 kg is dropped, its acceleration toward Earth is g. What is the force on the Earth due to the ball and what is Earth’s resulting acceleration? Earth’s mass is 6.0 x 1024 kg.
42 Use Newton’s 2nd and 3rd laws to find a Earth F Earth on Ball = m ball aSubstitute a = -gF Earth on Ball = m ball (-g)Substitute knownsF Earth on Ball = (0.18kg)(9.8m/s2)F Earth on Ball = 1.8 n
43 Find Earth’s Acceleration F ball on Earth = - F Earth on ball = na Earth on ball = Fnet/ m Eartha Earth on ball = 1.8 n / 6.0 x 1024kga Earth on ball = 2.9 x m/s2 toward the ball
44 TensionTension, the specific name for the force exerted by a string or rope is an interaction force.
45 Normal ForceThe perpendicular contact force exerted by a surface on another object.