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Applying the…
Laws
F Of
F
Motion
n, “Normal force”
G3
G2
ΣFy = may
ΣFy = 0
FG1
ΣFy = n - FG1 - FG2 - FG3 = 0
n = FG1 + FG2 + FG3
Extra Practice Problems:
4.11, 4.23, 4.25, 4.29, 4.37, 4.59
Notes and advertisements
Homework for today’s lecture will be
due SUNDAY night.
Use the next few nights to practice
problems!
Wednesday:
We will review and problem solve.
Post to facebook the things you’re
having trouble with!
https://www.facebook.com/groups/366030497095349/
Newton’s laws: a review.
Velocity doesn’t change
unless there’s a force
ΣF = ma
ALWAYS!
If there’s a force,
action = reaction
These are equal and opposite
forces even if you break the
wall or your fist keeps going!
Today
• Gravitational force
• “Contact” and “Normal” force
• Tension (same thing as force!)
• Solving force problems with free body
diagrams
Today we’re going to continue talking about forces: a few special ones, and how to apply them.
Types of forces
• Contact
• Literal contact (things hitting eachother)
• Tension (force in a string)
• Friction
• Field
• Gravity
• Electrical/magnetic
Pushing something, hitting a ball with a bat, kicking the wall are all contact forces.
Pushing/pulling something with a magnet or gravity are field forces.
Let’s talk a little about gravitational forces first.
Types of forces
• Contact
• Literal contact (things hitting eachother)
• Tension (force in a string)
• Friction
• Field
• Gravity
• Electrical/magnetic
Pushing something, hitting a ball with a bat, kicking the wall are all contact forces.
Pushing/pulling something with a magnet or gravity are field forces.
Let’s talk a little about gravitational forces first.
Types of forces
• Contact
• Literal contact (things hitting eachother)
• Tension (force in a string)
• Friction
Next monday!
• Field
• Gravity
• Electrical/magnetic
Physics 102!
Pushing something, hitting a ball with a bat, kicking the wall are all contact forces.
Pushing/pulling something with a magnet or gravity are field forces.
Let’s talk a little about gravitational forces first.
Weight vs. Mass
• “Weight” is an object’s force due to gravity
w = FG = mg
(g = 9.8 m/s2)
Units: N,
kg m
s2
• Mass: amount of matter in an object
(how many kilograms, pounds, etc.)
Weight is an object’s force due to gravity. This is confusing with common terminology, because we say, how much do we weigh? But if you see something as a weight, it
is a force so will always be expressed in units of Newtons, i.e. kg - m per seconds squared. We already know what the acceleration of Earth’s gravity is, so can apply
Newton’s second law to say what’s the force gravity is having on me? It’s my mass, times my acceleration due to gravity, 9.8 m/s^2.
Weight vs. Mass
• “Weight” is an object’s force due to gravity
w = FG = mg
(g = 9.8 m/s2)
Units: N,
kg m
s2
• Mass: amount of matter in an object
(how many kilograms, pounds, etc.)
Earth,
a = 9.8 m/s2
You weigh
less on Mars,
a = 2 m/s2 !
Weight is an object’s force due to gravity. This is confusing with common terminology, because we say, how much do we weigh? But if you see something as a weight, it
is a force so will always be expressed in units of Newtons, i.e. kg - m per seconds squared. We already know what the acceleration of Earth’s gravity is, so can apply
Newton’s second law to say what’s the force gravity is having on me? It’s my mass, times my acceleration due to gravity, 9.8 m/s^2.
+y
+x
Consider a ball in free-fall. Here is a vector representing the force of gravity, it’s acting downwards on the ball. No other forces.
FG = weight = mg
+y
+x
In this case, the only force is that of gravity, pushing down. This is what’s known as a free-body diagram, where I’ve drawn all the forces actually acting on the ball.
FG = weight = mg
Let’s apply Newton’s laws to this ball to
analyze its motion.
+y
+x
Fnet = ma
ΣFx = max ΣFy = may
Let’s note for a moment that force and accelerations are vector quantities. This means we can break it down into x and y components. We know from Newton’s first law
that to change a velocity (and hence get an acceleration) you need a force in that direction. For this ball, we’re not actively pushing it left or right at all. What are the
forces in the x direction? [DO THIS ON THE LIGHT BOARD]
Normal force happens in contact situations. It is
perpendicular to the contact surface
and counteracts the forces applied on an object.
n, “Normal force”
FG (weight)
+y
+x
OK but what happens when the ball sits on a surface? If the surface weren’t there, it’d continue to fall, but since the surface is there, it’s pushing its weight down onto the
ground. That surface has a reaction force, pushing upwards on the ball. This is Newton’s third law in action.
n, “Normal force”
FG (weight)
Let’s apply Newton’s laws to this ball to
analyze its motion.
ΣFy = may = 0 The ball is at rest ΣFy = n - FG = 0
(“in equilibrium”)
n = FG
+y
+x
Normal force equal to mg here!
So let’s analyze its motion like we did last time. As before, there’s no x force here so I’ll stick to y. However, we notice that in fact this ball is at rest, so the y motion is also
zero. In cases like this, we can say this system is in a condition of equilibrium: it’s not moving. So what do we know? From the diagram I drew, I know that there are two y
forces, and I can write these out for the sum. The normal force is equal in magnitude to the gravitational force here!
There are normal forces for all contact forces.
n
n
Fpalm
Book
FG
Fpush
In each of these examples there’s a normal force. In the pushing on a book example, the normal force is equal to the sum of the downward forces: the weight of the book
and the force from the pushing hand. Normal force won’t always be equal to -mg! Since normal is always perpendicular to the surface, sometimes you will need to look at
vector components of the force of gravity; for the homework due Thursday you have one problem that aims to walk you through this process.
Free-Body Diagrams (FBDs)
n
Book
Fpush
FG
If it doesn’t have units of
Newtons (kg m/s2), then it
doesn’t belong on a FBD!
(No velocities,
accelerations, or even m*a)
• Label all of the forces acting on (not by) an
object.
• Only forces acting on an object will affect its
motion (acceleration).
For instance, we don’t label the gravitational force of the book pulling on the Earth.
Draw a Free-Body Diagram
Box (mass m) at rest on a surface with a string
lightly pulling up on it, but not lifting it.
T = Tension (N)
+y
+x
What forces are acting on the box (there are 3)?
Which way are they pointing?
(What’s the normal force here in terms of m, g, and T?)
I wanted to introduce the concept here of tension, which is just another word to describe the force in a string that’s pulling on something. In this case it’s pulling up on the
box. I’d like you to draw this image and make a FBD for the box. If you think you’ve got the FBD down, compare with your neighbor and try to calculate the value of the
normal force in terms of m, g, and T.
Draw a Free-Body Diagram
Box (mass m) at rest on a surface with a string
lightly pulling up on it, but not lifting it.
+y
+x
What is the normal force in terms of m, g, and T?
A. n = 0 N
D. n = T - mg
B. n = -mg
E. n = mg - T
Q35
C. n = mg
Answer: E.
Think about it like this: the normal force balances out the other forces acting on an object! So Fg and T are counteracting eachother, and the normal force is pointing up.
Two ways to think about it! Conceptually (balance the forces) or with Newton’s laws. Try both to confirm you’ve done it right!
Draw a Free-Body Diagram
Box (mass m) at rest on a surface with a string
lightly pulling up on it, but not lifting it.
n
+y
+x
FG (weight of box)
What is the normal force in terms of m, g, and T?
A. n = 0 N
D. n = T - mg
B. n = -mg
E. n = mg - T
Q35
C. n = mg
Answer: E.
Think about it like this: the normal force balances out the other forces acting on an object! So Fg and T are counteracting eachother, and the normal force is pointing up.
Two ways to think about it! Conceptually (balance the forces) or with Newton’s laws. Try both to confirm you’ve done it right!
Draw a Free-Body Diagram
Box (mass m) at rest on a surface with a string
lightly pulling up on it, but not lifting it.
n
T
+y
+x
FG (weight of box)
What is the normal force in terms of m, g, and T?
A. n = 0 N
D. n = T - mg
B. n = -mg
E. n = mg - T
Q35
C. n = mg
Answer: E.
Think about it like this: the normal force balances out the other forces acting on an object! So Fg and T are counteracting eachother, and the normal force is pointing up.
Two ways to think about it! Conceptually (balance the forces) or with Newton’s laws. Try both to confirm you’ve done it right!
Draw a Free-Body Diagram
+y
+x
What happens if I pull with a tension greater than mg?
Q36
A.
B.
C.
D.
Here’s a question to help you build intuition about systems like this. Answer: A.
The force upwards will be greater than the force downwards.
Box will lift.
System remains in equilibrium.
Table will break.
String will break.
Draw a Free-Body Diagram
n
+y
+x
FG (weight of box)
What happens if I pull with a tension greater than mg?
Q36
A.
B.
C.
D.
Here’s a question to help you build intuition about systems like this. Answer: A.
The force upwards will be greater than the force downwards.
Box will lift.
System remains in equilibrium.
Table will break.
String will break.
Draw a Free-Body Diagram
n
T
+y
+x
FG (weight of box)
What happens if I pull with a tension greater than mg?
Q36
A.
B.
C.
D.
Here’s a question to help you build intuition about systems like this. Answer: A.
The force upwards will be greater than the force downwards.
Box will lift.
System remains in equilibrium.
Table will break.
String will break.
Application: Hospitals
Some are designed to pull the leg,
some are designed to support.
You’ll see a problem on your homework for one of these contraptions.
Tension note: pulleys
T1
T1
T1
T1
T1
We will do a more complicated analysis of this system on Friday!
Tension note: pulleys
T1
T1
T1
T1
T1
As long as the rope is free to move (e.g. on a
pulley), the tension in the rope is the same at
all locations on the rope.
We will do a more complicated analysis of this system on Friday!
Two blocks are arranged as shown and kept
at rest by holding the 1-kg block in place.
The tension force in the string is closest to:
A.
B.
C.
D.
E.
Answer:C
10 N
11 N
Q37
12 N
10 N at the left end; 12 N at the right
Smoothly varying from 10 N by the left block to 12 N
by the right block
An elevator is being lowered at a constant
speed by a steel cable attached to an electric
motor. There is no air resistance, nor is there
any friction between the elevator and the
walls of the elevator shaft.
The upward force exerted on the elevator by
the cable is
Motor
Cable
v
Elevator
A. greater than the downward force of gravity.
B. equal to the force of gravity.
C. less than the force of gravity.
D. any of the above, depending on the speed
Q38
of the elevator.
Answer: B
A.
B.
C.
D.
E.
2000 N
1500 N
1000 N
750 N
500 N
Q39
I mentioned before that every string “segment” of a pulley system will have the same tension. By rigging up this piano so that its weight (ma) is supported by two strings, the puller has effectively divided by two the tension that he
must maintain in the string (hence, he has to pull with half the force that he would if he were pulling the piano straight up, or with only one string attached to it).
Note, Constant rate=no acceleration, which means no net force on piano.
½ of the weight, Answer E.
Get studying!
See you Wednesday!