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Chapter 4 Lecture Notes Chapter 4 Newton’s 1st Law of Motion Aristotle was the first to attempt to classify different types of motion (at least the first that we know about). He stated that there are two types of motion, natural and violent. Natural motion would be dirt falling to the ground (earth returns to the Earth) or a puff of smoke rising into the air. He believed that everything has a natural state ant that it wants to return to that state. Violent motion is imposed on an object by applying a pushing or pulling force. Aristotle believed that all objects wanted to be at rest and this meant that the Earth must be at rest. This was agreed upon for nearly 2000 years. Along comes Copernicus Nicolaus Copernicus was the first to suggest that the Earth is not at rest. He theorized that the reason that the Sun, moon, and planets move in the sky is because the Earth moves around the Sun. Copernicus was unsure of his theory but was finally convinced to publish De Revolutionibus shortly before his death. Galileo Galilei Copernicus had contradicted Aristotle’s theory of motion. While we now know that Copernicus was correct, no one believed him at the time. In the early 17th century, Galileo also started to point out flaws in Aristotle’s theories on motion. For nearly 2000 years people believed that heavier objects fell faster than light ones. Galileo demolished this theory by dropping objects from the Leaning Tower of Pisa. Galileo dropped two stones from the top of the tower and found (as expected) that they landed at basically the same time. He discovered that the pull from the Earth is the same on all objects. Now that Galileo had shown that object fall at the same rate, he started looking at Aristotle’s theory of natural motion. Aristotle looked at nature and stated that all objects would find a way to return to rest. Physics 210 Santiago Canyon College Page 1 Chapter 4 Lecture Notes While Galileo agreed that this is what we see in nature, he wondered why he should be limited to motion we see in nature. When Galileo looked at inclined planes, he noticed that the ball would speed up as it went down the plane and would slow down as it went up one. He also noticed that if he used a very smooth set of inclined planes (Why did they need to be very smooth?) then the ball would return to the same height as when it was released. When the ball reached that same height (on the other side) the ball would come to rest (for an instant) and then start back down the incline. • What if I have a very long horizontal plane that never comes back up to the original height? Galileo said that obviously the ball will continue forever because it will never return to the same height (remember that the ball only stopped when it reached the starting height). The property that the ball will keep moving to the right is called inertia. Obviously, inertia contradicts Aristotle’s theory of natural motion, but it would be until after Galileo’s death that Aristotle’s theories would finally be discarded. Physics 210 Santiago Canyon College Page 2 Chapter 4 Lecture Notes Law of Inertia Isaac Newton refined Galileo’s definition of inertia: Every object continues in its state of rest, or uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it. The is known as Newton’s 1st Law of Motion and was published in Newton’s Principia. Newton’s 2nd Law of Motion A force can be thought of as a push or a pull. Forces can involve contact (pushing a block on the floor) or not (gravity). Force is a vector quantity. This means that we not only care about how big the force is but also in which direction the force is applied. For example: 5N We would say that the force on the block is 5 N î (Note: 1 N = 1 Newton kg ∗ m = 1 2 ). s What is the force on this block? 5N 10 N Fnet = 10 N ˆi + 5 N ˆi = 15 N ˆi Physics 210 Santiago Canyon College Page 3 Chapter 4 Lecture Notes 8N 12 N The net force on the block is: Fnet = 12 N ˆi + −8 N ˆi = 4 N ˆi ( ) Therefore, our diagram above is equivalent to: 4N Example 1: Find the net force in each of the following cases: 7N A. 3N 5N 5N B 10 N 5N C. 8N 7N 3N 6N 5N Solution: Physics 210 Santiago Canyon College Page 4 Chapter 4 Lecture Notes Equilibrium An object is in static equilibrium if: 1. v = 0 2. ∑ F = 0 Example 2: All of the objects shown are in equilibrium. Find the unknown force in each case. A. 12 N 10 N 8N F B. 6N F 15 N 7N C. 2N 1N 2N 4N 2N F 3N Solution: Physics 210 Santiago Canyon College Page 5 Chapter 4 Lecture Notes Normal Forces The force due to gravity on objects is called weight ( w ) . w = mg m g = −9.8 2 ˆj s where So when you are standing on a scale: (1) w Your body pushes down on the scale with a force w . Newton’s 1st law states that any object with a non-zero net force should be moving. Is the person moving? NO!!!!! Therefore the net force must be zero!!!! This means that there must be an upward force acting on the person. This upward force of the scale pushing on the person is called a normal force ( n ) . Physics 210 Santiago Canyon College Page 6 Chapter 4 Lecture Notes n w ∑F = 0 ⇒ n − w = 0 n=w In general, the net force on an object is: F ∑ = ma (2) This is known as Newton’s 2nd Law of Motion. Imagine that a 1 kg block rests on a frictionless table. We now push the block to the right with a force of 5 N for 3 seconds. How fast is the block moving after this period of time? This problem can be solved as follows: 1. Draw a force diagram for the block (also called a Free-Body Diagram). n 5N 1 kg w Physics 210 Santiago Canyon College Page 7 Chapter 4 Lecture Notes 2. Find the net force on the block (in each direction but the net vertical force is zero in this case): ∑F y ∑F x = n − w =0 ⇒ n = w F 5N m =ma x ⇒ a x = = = 5 2 m 1 kg s 3. Now compute the final speed of the block: m m m v f = v i + at = 0 + 5 2 (3 s ) = 15 s s s Example 3: Your car has a mass of 2000 kg. What net force must be km (from rest) in 6 applied to the car to achieve a final speed of 100 hr seconds? How far will the car travel during this time? Solution: Physics 210 Santiago Canyon College Page 8 Chapter 4 Lecture Notes Imagine that a girl is sliding down a frictionless inclined plane. What is her acceleration? 1. Draw the free body diagram: 2. Find the sum of forces in x and y: ∑F = y n − w y = n − mg cos θ= 0 ⇒ n= mg cos θ ∑ F= x θ ma x ⇒ a= w= mg sin= g sin θ x x Now that we have the acceleration, let’s find the time that it takes for the girl to reach the bottom. If the length of the hill is d, then the time is: 1 1 x f = x i + v ix t + a x t 2 ⇒ d = ( g sin θ ) t 2 2 2 Physics 210 Santiago Canyon College Page 9 Chapter 4 Lecture Notes t= 2d gsin θ And the speed of sled at the bottom of the hill is: 2 2 + 2a x d v= v 0x x = vx = 2a x d 2dg sin θ Example 4: A lamp of mass m = 25.0 kg is hanging from the wires as shown below. Find the tensions in each of the wires. Solution: Physics 210 Santiago Canyon College Page 10 Chapter 4 Lecture Notes Example 5: The system above is in equilibrium. What is the value of m? If the mass m is doubled, what is the acceleration of the system? Solution: Physics 210 Santiago Canyon College Page 11 Chapter 4 Lecture Notes Example 6: Two 100 kg boxes are dragged along a frictionless surface m with a constant acceleration of 1.0 2 , as shown below. Each rope has a s mass of 1.0 kg. Find the force F and the tension in the ropes at points A, B, and C. Solution: Physics 210 Santiago Canyon College Page 12 Chapter 4 Lecture Notes Example 7: For the diagram below, find the acceleration of either mass and the tension in the string (this is called an Atwood’s Machine). Solution: Physics 210 Santiago Canyon College Page 13 Chapter 4 Lecture Notes Apparent Weight Have you ever noticed that you feel heavier when an elevator starts moving upward? In order to explain this we first need to note that scales read the normal force ( n ) . This is because the normal force is the reaction of the scale to your weight pushing on it. Now imagine that we have the following situation: Elevator Scale If we draw a free-body diagram for the scale: n w Now we apply Newton’s 2nd law to the scale: ∑F y Physics 210 Santiago Canyon College =n − w =ma y n − mg = ma y Page 14 Chapter 4 Lecture Notes ( = n m ay + g ) If a y = 0 (constant velocity or at rest): ( (3) ) n= m a y + g = m (0 + g )= mg If a y > 0 then n > w (elevator is accelerating upward) If a y < 0 then n < w (elevator is accelerating downward) Example 8: A 50.0 kg woman stands in an elevator which accelerates m upward at 2.00 2 . What is her apparent weight? s Solution: Physics 210 Santiago Canyon College Page 15 Chapter 4 Lecture Notes Example 9: A 70.0 kg man is in an elevator that is accelerating m downward at 3.00 2 . He decides that he needs to go back up to his s m office and rides back up. The elevator accelerates upward at 2.0 2 . s What is the difference in the man’s apparent weight during the two parts of his trip? Solution: Physics 210 Santiago Canyon College Page 16 Chapter 4 Lecture Notes Action and Reaction So far, we have thought of forces as either pushes or pulls. Now we want to look at them in a different way. Imagine that a book is sitting on a table. We know that the table exerts a normal force on the book because the weight of the book is pushing on it. In other words, the normal force is in reaction to the weight of the object pushing down on the table. We say that the weight is an action force and that the normal force is a reaction force. Some examples of action-reaction pairs are: Newton’s 3rd Law of Motion Newton’s 3rd law states: Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first. Physics 210 Santiago Canyon College Page 17 Chapter 4 Lecture Notes This is more commonly stated as: Every action has an equal and opposite reaction. Example 10: A farmer wants the mule shown above to pull the items in the cart from one side of the farm to another. The mule states that he can’t do it because Newton’s 3rd law states that every action has an equal and opposite reaction so the cart won’t move. Where did the mule go wrong? Solution: Physics 210 Santiago Canyon College Page 18 Chapter 4 Lecture Notes We have stated that every action has an equal and opposite reaction. Now let’s look at the case of the rock falling toward the Earth: m W=mg Earth (mass = M) Our action force is: ( ) w= mg − ˆj By Newton’s 3rd law, our reaction force is: Freaction = mg ˆj Does the Earth actually move because of the rock? YES! It just does not move very far. To see why the Earth barely moves, let’s apply Newton’s 2nd law to the Earth: Freaction = mg = Ma Solving for the acceleration of the Earth: a= m g M m 1 . This means that the acceleration of the Earth is M so small that it barely moves at all! Since m M, Physics 210 Santiago Canyon College Page 19 Chapter 4 Lecture Notes Example 11: A 10 gram bullet is fired from a 5 kg gun. As the bullet is m fired the gun accelerates backward at 6 2 . What is the acceleration of s the bullet? If the force on the bullet is applied for 0.1 s, how fast will it be moving when it exits the gun? Solution: Physics 210 Santiago Canyon College Page 20