Physics 221 Chapter 5

Problem 1... That’s Incredible! Your friend tells you that she once saw a person pull really hard on an object towards the LEFT but the object was moving really fast towards the RIGHT. Explain whether that is ever possible.

Solution 1... That’s incredible! Absolutely! Here is a possible scenario. The brakes fail on a slowly moving car and three people jump out and try to stop it. The car will still keep moving but will hopefully start to slow down enough that it will eventually stop. The moral of the story (other than having your brakes checked regularly) is that one should really focus on the change in the speed (acceleration) and not just the speed itself in order to draw correct conclusions about the nature of the forces acting on an object.

Define the terms “Force ” and “net force” A FORCE is a PUSH or a PULL. Net force is the resultant of all the forces. The net force on the box is 4N (Newtons) 7 N3 N

Problem 2... Play it again, Sam! You see an object moving towards the East. What can you say about the direction of the force acting on it?

Solution 2... Play it again, Sam! One cannot even be certain if there is a force acting on the object, let alone the direction! One needs to observe closely whether the speed or direction are changing. If the speed and direction are not changing then there is no net force acting on the object. If it is moving towards the East but is either speeding up or slowing down then there is certainly a net force acting on it.

Newton’s First Law If the net force on an object is zero A. The object will stay at REST if it was at rest OR B. Keep moving in a straight line with constant speed if it was in motion

Inertia In other words, objects naturally resist attempts to bring about any change in their velocity. Objects at rest want to stay at rest and objects in motion want to keep moving in a straight line with a constant speed. This property of objects is called inertia Note: Mass is a measure of inertia. The more massive an object the more inertia it has!

Problem 3... One more time ;-) You see an object moving. Is there a net force acting on it?

Solution 3... One more time ;-) Depends! One needs to observe closely whether the speed or direction are changing. If the speed or direction are not changing then there is no net force acting on the object. There is a net force if and only if at least one of the following is true : a. Its speed is increasing b. Its speed is decreasing c. Its direction of motion is changing

Problem 4... Playing Newton! You observe an object moving in a circular path. You wonder (like Newton did 300 years ago about the moon) if there is a (net) force acting on it. You put on your scientist’s hat and study the object’s motion carefully. You ascertain that the object is neither getting faster nor slower. It seems to be maintaining a constant speed. You know this because it takes the same amount of time to go around the circular path over and over again. Is there a net force acting on the object?

Solution 4... Playing Newton! Absolutely! In the absence of any external forces, objects will either stay at rest or move in a straight line with constant speed. You didn’t even have to measure the speed of the object in this case to conclude that there must be a force acting on it! Incidentally, the circular path of the heavenly bodies clued Newton into the presence of a universal gravitational force!

In other words... In other words, Newton’s First Law says that: Force causes acceleration Recall that acceleration means increase in speed, decrease in speed, or change in direction. Put another way, acceleration means change in velocity.

Problem 5... Down factory lane! You observe a factory worker push really hard on a crate. The crate seems to be moving in a straight line with constant speed. (The crate doesn’t seem to be getting faster or slower). Can you reconcile this observation with Newton’s First Law of motion?

Solution 5... Down factory lane! The force of FRICTION is opposing the force exerted by the worker. The NET force is zero! And the object is moving ;-)

Newton’s Second Law Acceleration is greater if the force is greater AND Acceleration is smaller if the mass is greater F = ma

A picture is worth a thousand words! F = m a or F = m a

Problem 6 You apply a 30 N force on a 5 kg crate to push it horizontally across a factory floor. The speed of the crate increases at the rate of 1.5 m/s every second. Which statement most accurately describes this experiment? A. The net force on the object is less than 30 N B. The object is not accelerating because of friction C. F = ma does not work if there is friction D. Not enough information is given to determine if the object is accelerating

Solution 6 F net = ma F net = 5*1.5 F net = 7.5 N (correct answer is A) So frictional force must have been 22.5 N Put another way: Net Force = Applied Force - Frictional force F net = F applied - F friction 7.5 N = 30 N N

Problem 7... How heavy is it? The WEIGHT of an object is the force of gravity acting on it. Q1. What are the units of WEIGHT? Q2. What is the WEIGHT of a 5 kg box?

Solution 7... How heavy is it? A1. Weight is a FORCE -- so units are NEWTONS (N) A2. What is the WEIGHT of a 5 kg box? F = ma F = 5*9.8 F = 50 N NOTE: W = mg is a special form of F = ma

Problem 8... Moon Walk An object weighs 50 N on Earth. Its weight on the moon will be A. zero N B. 50 N C. Substantially less than 50 N D. Substantially more than 50 N

Solution 8... Moon Walk C. Substantially less than 50 N g M = 1/6 g E Weight on the moon will be 5*9.8/6 = 8.3 N

Problem 9... Mass on the moon Object P weighs the same on the Moon as object Q does on Earth. Identify the correct statement: A. Mass of P is more than the mass of Q B. Mass of P is less than the mass of Q C. Mass of P on the Moon is more than its mass on Earth D. Weight of P on the Moon is more than the weight of Q on Earth

Solution 9... Mass on the moon A. Mass of P is more than the mass of Q Mass of P must be 6 times the mass of Q! NOTES : Mass never changes. Mass is mass is mass. (Unless you break the object in two!) WEIGHT changes if acceleration due to gravity changes.

Friction The Force of Friction: 1. Always OPPOSES the motion 2. Depends on the surfaces in contact 3. Depends on the NORMAL force (how hard the object is pushed against the surface) friction = coefficient of friction * normal force f =  F N

Problem Calculate  You apply a 30 N force on a 5 kg crate to push it horizontally across a factory floor. The speed of the crate increases at the rate of 1.5 m/s every second. What is the coefficient of friction?

Solution Calculate  Frictional force = 22.5 N (see problem 4.6) f =  F N 22.5 =  *50  = 0.4

Problem Lugging the Luggage Q1. Calculate acceleration if no friction Q2. Calculate “ a ” if  = N Hawaii or Bust! 10 kg

Solution Hawaii or Bust! Q1. F = ma 40 cos 60 = 10* a a = 2 m/s 2 What’s wrong with this picture? Q2. F = ma 40 cos ( sin 60) = 10* a a = 0.7 m/s 2 Free - body Diagram [FBD]

Problem Slippery when wet! What is “a” in m/s 2 if there is no friction? A. 5 B. 7.5 C. 10 D kg

Solution Slippery when wet! What is “a” in m/s 2 if there is no friction? F = ma mg sin  = ma a = g sin  a = 5 m/s 2  m mg mg sin  mg cos 

Problem Slippery when wet! What is “a” if there is friction (  = 0.2) A. 2.5 B. 3.5 C. 4.5 D kg

Solution Slippery when wet! What is “a” if there is friction (  = 0.2) F = mg sin  -  mg cos  a = g sin  -  g cos  a = 3.2 m/s 2  m mg mg sin  mg cos 

Problem The Sun also rises! What is the tension in the string (T) if the mass of the Van Gogh painting is 2 kg and the price is 37 million dollars? A. 10 N B. 15 N C. 20 N D. 40 N 45 T T

Solution The Sun also rises! T sin 45 + T sin 45 = 20 N T = 14 N 45 T T 20 N

Problem Atwood’s Machine I What is the acceleration? 2 kg 3 kg

Solution Atwood’s Machine I F = ma = 5a 10 = 5a a = 2 m/s 2 2 kg 3 kg

Problem Atwood’s Machine II What is the tension in the string? 2 kg 3 kg

Solution Atwood’s Machine II What is the tension in the string? 30 - T = 3a 30 - T = 3* T = 6 T = 24 N 2 kg 3 kg

Problem Who wins? A. Calculate the acceleration if there is no friction B. Calculate “a” if “  ” = kg 2 kg

Solution Who wins? A. 20-T =2a ….(1) T – 30 sin 30 =3a …(2) a =1 m/s 2 T = 18 N B. 20-T =2a ….(1) T – 30 sin 30 – 30 cos 30(0.1) =3a …(2) a = 0.5 m/s 2 T = 19 N

Problem friction = ? A 3 kg crate sits on the factory floor. The coefficient of friction = 0.3. What is the frictional force on the crate?

Solution friction = zippo ;-) Zippo …which rhymes with hippo!!! Note: f <=  F N

Newton’s Third Law Action and reaction are equal and opposite Action: You stand on a skateboard and throw a ball Reaction: The ball pushes you back!

Problem Tug-of-war Big Bubba and Little Suzy pull on a rope in a tug- of- war. Who exerts the greater force?

Solution Tug-of-war They exert the SAME force on each other. You cannot pull without being pulled! It takes two to tango! Action equals reaction.

Problem Rear-ended! A big truck slams into a Neon. Which statement is most nearly correct: A. Force of truck on neon equals the force of Neon on truck B. Force of truck on neon far exceeds the force of Neon on truck C. Force of truck on Neon is much less than the force of Neon on truck D. Neither experiences any force because the equal and opposite forces cancel each other out

Solution Rear-ended! A big truck slams into a Neon. Which statement is most nearly correct: A. Force of truck on neon equals the force of Neon on truck B. Force of truck on neon far exceeds the force of Neon on truck C. Force of truck on Neon is much less than the force of Neon on truck D. Neither experiences any force because the equal and opposite forces cancel each other out

That’s all Folks!