In this chapter you will Use Newton’s laws to solve problems Determine the magnitude and direction of a net force that causes a change in the motion of an object Classify forces according to their cause
Section 4.1 Force & Motion Force - a push or pull acting on an object that can cause the object to speed up, slow down, or change direction –Forces have both magnitude and direction - they are ___. –Forces are divided into contact and field forces Fields: electric, magnetic
Free-body diagrams Draw vectors away from objects Table pushing up on books Books pushing down on table
Determine net force Add forces acting in the same direction Subtract forces acting in opposite directions. 3n E + 2n W 1n E F 1 = 3.0 n E F 2 = 2.0 n E F net = 5.0 n E
Vectors at Right Angles Sin = opp hyp Cos = adj hyp Tan = opp adj
Newton’s 2 nd Law a = F net / 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.
Newton’s 1 st 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 in –No motion or constant motion –Known as equilibrium A net force can result in –Speeding up or slowing down
Problem 1. A rock falls freely from a cliff. Draw vectors and label each. +y V a F net
A skydiver falling towards earth at a constant rate… +y v v v a = 0 F air resistance on diver F net = 0 F Earth’s mass on diver
A rope pulls a box at a constant speed across a horizontal surface. The surface provides a force that resists the box’s motion. +x vvv v F friction on box F pull on box F net = 0
Problem Two 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 10 2 n in the direction of the forces.
4.2 Objectives Describe how the weight and mass of an object are related. Differentiate between actual weight and apparent weight Use Newton’s 2 nd Law in two forms: F = ma F = mg
System Fg V a Known: a = g m Unknown: Fg Fnet = ma Fnet = Fg a = g Therefore: Fg = mg A ball in mid-air in free fall has only the force of gravity acting on it. Air resistance can be neglected.
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 acceleration Newton’s 2 nd law has become: Fg = mg
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.
Newton’s 2 nd 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?
Solution to Susie’s & Sally’s dilemma. In this chapter you will: Find the net Force: 11.0n R + (-10.0n L) F net = 1.0 n R F net = ma a = F net / mw a = 1.0 n / 0.30 kg = 3.33 m/s 2 Right
Apparent Weight The force an object experiences as a result of the contact forces acting on it, giving the object an acceleration
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.
Slowly rising or speeding up while descending. Appar ent weight is less when … Appare nt weight is greater when Speeding up while rising or slowing while going down Fscale Fg Fscale Fg
Going Up? W W app Ground floor Normal feeling v = 0 a = 0 W W app Just starting up Heavy feeling v > 0 a > 0 W W app Between floors Normal feeling v > 0 a = 0 W W app Arriving at top floor Light feeling v > 0 a < 0
W W app Top floor Normal feeling v = 0 a = 0 W W app Arriving at Ground floor Heavy feeling v < 0 a > 0 W W app Between floors Normal feeling v < 0 a = 0 W W app Beginning descent Light feeling v < 0 a < 0 Going Down?
Turn to page 100 Let’s look at Example Problem 2 Refer to 19 &20
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/s 2 ? C. Back on Earth, what do you weigh in pounds? (1 kg = 2.2 kg)
A. What is your mass? m = Fg / g m = 585 n /9.8 m/s 2 m = 59.7 kg
B. What would the scale read on the Moon where g = 1.60 m/s 2 ? Fg = mg moon Fg= (59.7 kg)(1.60m/s 2 ) Fg = 95.5 n
m = 59.7 kg x 2.2 lb 1 kg m = 131 lb Back on Earth…
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.
4.3 Interaction Forces In this section you will : Define Newton’s Third law Explain tension in strings and ropes in terms of Newton’s 3 rd law Define the normal force Determine the value of the normal force by applying Newton’s 2 nd law
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.
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 opposite Numerically, F A on B = - F B on A
Practice Problem 32. Someone please read the problem. Identify the bucket as the system and up as positive. F net = F rope on bucket - F Earth’s mass on bucket = ma a = (F rope on bucket - F Earth’s mass on bucket )/m a = (F rope on bucket - mg)/m a = [450n - (42kg)(9.80m/s 2 )] / 42 kg a = 0.91 m/s 2
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 10 24 kg.
Use Newton’s 2 nd and 3 rd laws to find a Earth F Earth on Ball = m ball a Substitute a = -g F Earth on Ball = m ball (-g) Substitute knowns F Earth on Ball = (0.18kg)(9.8m/s 2 ) F Earth on Ball = 1.8 n
Find Earth’s Acceleration F ball on Earth = - F Earth on ball = - 1.8 n a Earth on ball = F net / m Earth a Earth on ball = 1.8 n / 6.0 x 10 24 kg a Earth on ball = 2.9 x 10 -25 m/s 2 toward the ball
Tension Tension, the specific name for the force exerted by a string or rope is an interaction force.
Normal Force The perpendicular contact force exerted by a surface on another object.