Transcript Friction - Salisbury University

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)
3N
7N
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 a
=m
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
• Fnet = ma
• Fnet = 5*1.5
• Fnet = 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
Fnet = Fapplied - Ffriction
7.5 N = 30 N - 22.5 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
• gM = 1/6 gE
• 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 =  FN
Problem 10 . . . 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 10 . . . Calculate

• Frictional force = 22.5 N (see problem 4.6)
•
f =  FN
• 22.5 =  *50
•  = 0.4
Problem 11 . . . Lugging the Luggage
• Q1. Calculate acceleration if no friction
• Q2. Calculate “ a ” if  = 0.2
40 N
10 kg
Hawaii or
Bust!
60
Solution 11 . . . Hawaii or Bust!
• Q1. F = ma
• 40 cos 60 = 10* a
• a = 2 m/s2
•
•
•
•
Free - body Diagram [FBD]
What’s wrong with this picture?
Q2. F = ma
40 cos 60 - 0.2 (100 - 40 sin 60) = 10* a
a = 0.7 m/s2
Problem 12 . . . Slippery when wet!
• What is “a” in m/s2 if there is no friction?
•
•
•
•
A. 5
B. 7.5
C. 10
D. 15
2 kg
30
Solution 12 . . . Slippery when wet!
• What is “a” in m/s2 if there is no friction?
• F = ma
• mg sin  = ma
m
• a = g sin 
mg sin
• a = 5 m/s2
mg cos
mg

Problem 13 . . . Slippery when wet!
• What is “a” if there is friction ( = 0.2)
•
•
•
•
A. 2.5
B. 3.5
C. 4.5
D. 5.5
2 kg
30
Solution 13 . . . Slippery when wet!
• What is “a” if there is friction ( = 0.2)
• F = mg sin  -  mg cos 
m
• a = g sin  -  g cos 
• a = 3.2 m/s2
mg sin
mg cos
mg

Problem 14 . . . 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
T
T
45
45
Solution 14 . . . The Sun also rises!
• T sin 45 + T sin 45 = 20 N
• T = 14 N
T
T
45
45
20 N
Problem 15 . . . Atwood’s Machine I
• What is the acceleration?
2 kg
3 kg
Solution 15 . . . Atwood’s Machine I
•
•
•
•
F = ma
30 - 20 = 5a
10 = 5a
a = 2 m/s2
2 kg
3 kg
Problem 16 . . . Atwood’s Machine II
• What is the tension in the string?
2 kg
3 kg
Solution 16 . . . Atwood’s Machine II
• What is the tension in the string?
•
•
•
•
30 - T = 3a
30 - T = 3*2
30 - T = 6
T = 24 N
2 kg
3 kg
Problem 17 . . . Who wins?
• A. Calculate the acceleration if there is no friction
• B. Calculate “a” if “” = 0.1
3 kg
2 kg
300
Solution 17 . . . Who wins?
•
•
•
•
•
A.
20-T =2a ….(1)
T – 30 sin 30 =3a …(2)
a =1 m/s2
T = 18 N
•
•
•
•
•
B.
20-T =2a ….(1)
T – 30 sin 30 – 30 cos 30(0.1) =3a …(2)
a = 0.5 m/s2
T = 19 N
Problem 18 . . . 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 18 . . . friction = zippo ;-)
• Zippo …which rhymes with hippo!!!
• Note: f <=  FN
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 19 . . . Tug-of-war
• Big Bubba and Little Suzy pull on a rope in a tugof- war. Who exerts the greater force?
Solution 19 . . . 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 20 . . . 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 20 . . . 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!