chapter 5 - newton's laws

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Chapter 5 (Unit 2) Newton’s Laws

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J. Rushton, Class of 2014

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Page 1: Chapter 5 - Newton's Laws

Chapter 5 (Unit 2)

Newton’s Laws

Page 2: Chapter 5 - Newton's Laws

Newton’s First Law So, we’ve talked about Newton’s first law,

which says that Fnet = 0 when the object is in a “natural state of motion”

What were these two natural states?

What does this “Fnet = 0” imply?

Page 3: Chapter 5 - Newton's Laws

Newton’s Second Law Imagine three children playing in the snow

with a toboggan. Let’s assume: Each friend has the same mass, m Each friend call pull with the same force, F The toboggan has a mass so small that it can be

ignored The snow is frictionless

Page 4: Chapter 5 - Newton's Laws

Newton’s Second Law Imagine you are carrying boxes out to a truck,

and they are pack full with heavy textbooks. What would happen if you picked a box up expecting it to be just as heavy, but it ended up being empty?

This is Newton’s second law!

You put in a lot of force, but it only had a little bit of mass to it accelerated very quickly.

Page 5: Chapter 5 - Newton's Laws

Newton’s Second Law

Page 6: Chapter 5 - Newton's Laws

Newton’s Second LawA man is riding in an elevator. The combined

mass of the man and the elevator is 7.00x102kg. Calculate the magnitude and direction of the elevator’s acceleration if the tension (FT) in the supporting cable is 7.50x103N.

Page 7: Chapter 5 - Newton's Laws

Newton’s Second Law

Page 8: Chapter 5 - Newton's Laws

Newton’s Second Law A sled with a mass of 5.5kg accelerates across

a snowy field because of an applied force of 75N. If the coefficient of kinetic friction between the sled and the snow is 0.33, how fast is the sled accelerating?

Page 9: Chapter 5 - Newton's Laws

Combining Dynamics and KinematicsIf electrons (m=9.1x10-31kg) are accelerated

through an electron gun from rest by an electric force of 5.8x10-15N over a distance of 3.5mm, what is the velocity of a single electron when it exits the gun.

Page 10: Chapter 5 - Newton's Laws

Model ProblemA curler exerts an average force of 9.50N [S] on

a 20.0kg stone. The stone started from rest and was in contact with the girl’s hand for 1.86s. If the ice is frictionless, (a) What is the average acceleration of the stone?

(b) What is the velocity of the stone when the it’s released?

Page 11: Chapter 5 - Newton's Laws

Model Problem

Page 12: Chapter 5 - Newton's Laws

Determining the Net Force There is almost always more than one force

acting on an object, and oftentimes they do not work at 0º or 90º angles.

In order to find out the net force, or resultant force, we must separate all forces into their x and y components before summing them.

An example of this would be if you were pulling something along the ground – it is very uncommon to pull at a 0º angle.

Page 13: Chapter 5 - Newton's Laws

Determining Net Force As we begin to solve problems involving forces

at various angles, you will be working in one-dimension at a time.

You must determine each component of the forces, and then work in each independent direction.

The only time we can combine x and y directions is when finding an overall resultant force (the same way we would find the resultant vector).

Page 14: Chapter 5 - Newton's Laws

Model ProblemTo move a 45kg wooden crate across a wooden

floor (μ=0.20), you tie a rope onto the crate and pull on the rope. While you are pulling the rope with a force of 115N, it makes an angle of 15º with the horizontal. How much time elapses between the time at which the crate just starts to move, and the time at which you are pulling it with a velocity of 1.4m/s?

Page 15: Chapter 5 - Newton's Laws

Model Problem

Page 16: Chapter 5 - Newton's Laws

Model Problem

Page 17: Chapter 5 - Newton's Laws

Determining Net Force So far, we’ve worked with problems where the

direction of the net force is obvious.

However, in many cases it may not be clear.

We will use the same techniques to solve these problems – this is where our free body diagrams will become extremely useful!

Page 18: Chapter 5 - Newton's Laws

Model ProblemThree children are each pulling on their older

sibling, who has a mass of 65kg. The forces exerted by each child are listed below. Use a scale diagram to determine the resultant acceleration of the older sibling.

Page 19: Chapter 5 - Newton's Laws

Model Problem

Page 20: Chapter 5 - Newton's Laws

Model Problem

Page 21: Chapter 5 - Newton's Laws

Newton’s Third Law For Newton’s first and second laws, we

focused on individual objects and all of the forces acting on it.

We used the net force to determine the change, or lack of change, in the motion of that object.

Now, we will work with two objects that are interacting together – we must consider the forces acting between them!

Page 22: Chapter 5 - Newton's Laws

Newton’s Third Law

Page 23: Chapter 5 - Newton's Laws

Newton’s Third Law So, Newton’s third law states that forces

always act in pairs.

An object cannot experience a force without also exerting an equal and opposite force.

There are always two forces acting and two objects involved.

Page 24: Chapter 5 - Newton's Laws

Newton’s Third Law Compare the forces in the following:

If both situations cause the blue football player to stop, what does that mean?

Page 25: Chapter 5 - Newton's Laws

Applying Newton’s Third Law If we look at a tractor trailer that has three

sections (pictured below), what forces are on each section?

Is the hitch holding the second trailer subjected to the same force as the hitch attaching the first trailer to the truck?

Page 26: Chapter 5 - Newton's Laws

Model ProblemA tractor-trailer pulling two trailers starts from

rest and accelerated to a speed of 16.2km/hr is 15s on a straight, level section of highway. The mass of the truck itself (T) is 5450kg. The mass of the first trailer (A) is 31,500kg and the mass of the second trailer (B) is 19,600kg. What magnitude of force must the truck generate in order to accelerate the entire vehicle? What magnitude of force must each of the trailer hitches withstand while the vehicle is accelerating?

Page 27: Chapter 5 - Newton's Laws

Model Problem

Page 28: Chapter 5 - Newton's Laws

Model Problem

Page 29: Chapter 5 - Newton's Laws

Free Fall Have you ever been on an amusement park

ride that lets you fall with almost no support for a short period of time?

Free fall is a condition in which gravity is the only force acting on you.

Imagine you were standing on a scale in an elevator. If the cable was to snap and there were no safety devices (and air friction was negligible), what would you weight read on the scale?

Page 30: Chapter 5 - Newton's Laws

Free Fall When your weight on a scale can be read as

zero (also called your “apparent weight”), is it often called “weightlessness”.

Your mass has not changed, but you feel weightless because nothing is pushing up on you, preventing you from acceleration at the value of the acceleration due to gravity.

Page 31: Chapter 5 - Newton's Laws

Free Fall When we are close to the Earth’s surface,

weightlessness is rarely experienced, due to the resistance of the atmosphere.

As an object collides with molecules of gases and particles in the air, the collisions act as a force opposing the force of gravity.

What is this force most often called?

Page 32: Chapter 5 - Newton's Laws

Free Fall It is important to remember that air resistance

(or air friction) is quite different from the surface friction that we’ve talked about.

When an object moves through a fluid such as air, the force of friction increases as the velocity of the object increases.

While this is true, a falling object eventually reaches a velocity at which the force of friction is equal to the force of gravity.

Page 33: Chapter 5 - Newton's Laws

Free Fall At this point, the net force is equal to zero so

the object is no longer accelerating (it is at constant velocity).

This velocity is called terminal velocity, and it can be affected by the shape and orientation of an object.

Page 34: Chapter 5 - Newton's Laws

Inclined Planes If you are skiing down a hill, but haven’t put

any effort into your motion (no applied force), then what causes you to accelerate down the hill?

Why will this acceleration will always be smaller than that of a free-falling object?

Page 35: Chapter 5 - Newton's Laws

Inclined Planes The key to analyzing the motion of objects on

inclined planes is the chose a coordinate system that simplifies the procedure.

Since most of the motion will be along the plane, it is easiest to call the surface of the plane the x-axis.

Page 36: Chapter 5 - Newton's Laws

Inclined Planes The force of gravity affects the motion on an

inclined plane, but the force vector is at an angle to the plane.

Thus, you must split the gravitational force into its x and y components (parallel and perpendicular to the plane).

Page 37: Chapter 5 - Newton's Laws

Model Problem You are holding an 85kg trunk at the top of a

ramp that slopes from a moving van to the ground, making an angle of 35º with the ground. You lose your grip and the trunk begins to slide.

(a) If the coefficient of friction between the trunk and the ramp is 0.42, what is the acceleration of the trunk?

(b) If the trunk slides 1.3m before reaching the bottom of the ramp, for what time interval did it slide?

Page 38: Chapter 5 - Newton's Laws

Model Problem

Page 39: Chapter 5 - Newton's Laws

Model Problem

Page 40: Chapter 5 - Newton's Laws

Model Problem

Page 41: Chapter 5 - Newton's Laws

Momentum and Newton’s Laws When a racecar driver walks away from a crash

unscratched, it’s most often not because of good luck. It’s because because of the practical applications of Newton’s laws, and engineers design the car and its safety equipment to protect the driver from injury.

There are a few scenarios where it is difficult to describe the forces acting on an object. This is when they collisions, explosions, or recoil.

For these scenarios, it is easier to observe the motion of objects before and after, and then use Newton’s laws.

Page 42: Chapter 5 - Newton's Laws

Momentum and Newton’s Laws What do you think about when you think of

“momentum”?

Momentum is the product of an object’s mass and its velocity. Since it is the product of a vector and a scalar, momentum is a vector. The direction of the momentum is the same as the direction of the velocity.

Page 43: Chapter 5 - Newton's Laws

Momentum and Newton’s Laws

Page 44: Chapter 5 - Newton's Laws

Model Problem Determine the momentum of a 0.300kg

hockey puck travelling across the ice at a velocity of 5.55m/s [N].

Page 45: Chapter 5 - Newton's Laws

Impulse Originally, Newton expressed his second law

by stating that the change in an object’s motion (rate of change of momentum) is proportional to the force impressed on it.

Expressed Mathematically, Newton’s second law can be written as:

Page 46: Chapter 5 - Newton's Laws

Impulse We can show that this expression is

equivalent to the equation we’re used to for Newton’s second law:

Page 47: Chapter 5 - Newton's Laws

Impulse Impulse is the product of the force exerted on

an object and the time interval over which the force acts.

Impulse is a vector quantity because it is a vector multiplied by a scalar, and its direction is the same as that of the force that causes it.

Page 48: Chapter 5 - Newton's Laws

Impulse

Page 49: Chapter 5 - Newton's Laws

Model Problem If a golf club exerts an average force of

5.25x103N[W] on a golf ball over a time interval of 5.45x10-4s, what is the impulse of the interaction?

Page 50: Chapter 5 - Newton's Laws

Impulse-Momentum Theorem The impulse is really equal to the change in

the momentum of an object.

This relationship is called the impulse-momentum theorem.

Page 51: Chapter 5 - Newton's Laws

Impulse-Momentum Theorem

Page 52: Chapter 5 - Newton's Laws

Model Problem A student practices her tennis volleys by

hitting a tennis ball against a wall.(a) If the 0.600kg ball travels 48m/s before hitting the wall and then bounces directly backward at 35m/s, what is the impulse of the interaction?(b) If the duration of the interaction is 25ms, what is the average force exerted on the ball by the wall?

Page 53: Chapter 5 - Newton's Laws

Model Problem

Page 54: Chapter 5 - Newton's Laws

Chapter 5 and Unit Review

Page 208-209 #23-27, 30-45

Page 212-213 #29-47