Lecture 8
Applications of
Newton’s Laws
(Chapter 6)
Announcements
Assignment #4: due tomorrow night, 11:59pm
Midterm Exam #1 scores are up:
Class Average 75.0%
Reading and Review
Will It Budge?
A box of weight 100 N is
at rest on a floor where
μs = 0.4. A rope is
attached to the box and
pulled horizontally with
tension T = 30 N. Which
way does the box move?
a) moves to the left, because the force of static
friction is larger than the applied force
b) moves to the right, because the applied force
is larger than the static friction force
c) the box does not move, because the static
friction force is larger than the applied force
d) the box does not move, because the static
friction force is exactly equal the applied force
e) The answer depends on the value for μk.
Static friction
(s = 0.4 )
m
T
Will It Budge?
A box of weight 100 N is
at rest on a floor where
μs = 0.4. A rope is
attached to the box and
pulled horizontally with
tension T = 30 N. Which
way does the box move?
a) moves to the left, because the force of static
friction is larger than the applied force
b) moves to the right, because the applied force
is larger than the static friction force
c) the box does not move, because the static
friction force is larger than the applied force
d) the box does not move, because the static
friction force is exactly equal the applied force
e) The answer depends on the value for μk.
The static friction force has a
maximum of sN = 40 N. The
tension in the rope is only 30 N. So
Static friction
(s = 0.4 )
m
the pulling force is not big enough
to overcome friction.
Follow-up: What happens if the tension is 35 N? What about 45 N?
T
Tension
T
T
Fs
W
W
Translational
equilibrium?
T
W
T
2.00
kg
W
Tension in the rope?
m1 : x :
y:
m2 : y :
fk
Over the Edge
In which case does block m
a) case (1)
experience a larger acceleration?
b) acceleration is zero
In case (1) there is a 10 kg mass
c) both cases are the same
hanging from a rope and falling.
In case (2) a hand is providing a
d) depends on value of m
constant downward force of 98 N.
e) case (2)
Assume massless ropes.
m
m
10 kg
a
a
F = 98 N
Case (1)
Case (2)
Over the Edge
In which case does block m
a) case (1)
experience a larger acceleration?
b) acceleration is zero
In case (1) there is a 10 kg mass
c) both cases are the same
hanging from a rope and falling.
In case (2) a hand is providing a
d) depends on value of m
constant downward force of 98 N.
e) case (2)
Assume massless ropes.
98 N due to the hand. In
case (1) the tension is
10 kg
a
less than 98 N because
were at rest would the
tension be equal to 98 N.
a
T
the block is accelerating
down. Only if the block
m
m
In case (2) the tension is
Case (1)
a>0 downward
implies T<W
F = 98 N
Case (2)
W
Tension
Force is always along a rope
Ty
T
T
T
T
W
Springs
Hooke’s law for springs states that the
force increases with the amount the spring
is stretched or compressed:
The constant k is called the spring constant.
Springs
Note: we are discussing the force of the
spring on the mass. The force of the spring
on the wall are equal, and opposite.
Springs and Tension
S1
S2
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end,
and M1 is hung below. How far does the combined
spring stretch?
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the
springs are attached
e) L1 + L2
Springs and Tension
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end,
and M1 is hung below. How far does the combined
spring stretch?
S1
S2
Fs=T
W
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the
springs are attached
e) L1 + L2
Springs and Tension
A mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end,
and M1 is hung below. How far does the combined
spring stretch?
S1
S2
Fs=T
W
a) (L1 + L2) / 2
b) L1 or L2, whichever is smaller
c) L1 or L2, whichever is bigger
d) depends on which order the
springs are attached
e) L1 + L2
Spring 1 supports the weight.
Spring 2 supports the weight.
Both feel the same force, and
stretch the same distance as before.
Instantaneous acceleration
Velocity vector is always in the direction of motion;
acceleration vector can points in the direction
velocity is changing:
Circular Motion
An object moving in a circle must have a force acting
on it; otherwise it would move in a straight line.
a
a
If the speed is constant, the
direction of the force and
the acceleration is towards
the center of the circle.
The magnitude of this
centripetal force is given by:
For circular motion problems, it is often convenient to choose coordinate
axes with one pointing along the direction of this centripetal force
Circular Motion
This force may be provided by the tension in a
string, the normal force, or friction, among others.
Examples of centripetal force
when
no friction is
needed to hold
the track!
Circular Motion
An object may be changing its speed as it moves
in a circle; in that case, there is a tangential
acceleration as well:
A hockey puck of mass m is attached to a string that passes
through a hole in the center of a table, as shown in the figure.
The hockey puck moves in a circle of radius r. Tied to the other
end of the string, and hanging vertically beneath the table, is a
mass M. Assuming the tabletop is perfectly smooth, what speed
must the hockey puck have if the mass M is to remain at rest?
A hockey puck of mass m is attached to a string that passes
through a hole in the center of a table, as shown in the figure.
The hockey puck moves in a circle of radius r. Tied to the other
end of the string, and hanging vertically beneath the table, is a
mass M. Assuming the tabletop is perfectly smooth, what speed
must the hockey puck have if the mass M is to remain at rest?
necessary centripetal force:
Only force on puck is tension in the string!
To support mass M, the necessary tension is:
Circular motion and apparent weight
This normal force is the
apparent, or perceived, weight
Key points re: circular motion
1) If
Then
object moving in a circular path
Fc  t  
m v t 
2
r
2) Fc is NOT a separate force; it represents
the sum of the physical forces acting on m
 Fc   fc
Going in Circles I
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel is
at rest, the normal force N exerted by
a) N remains equal to mg
b) N is smaller than mg
your seat is equal to your weight mg.
c) N is larger than mg
How does N change at the top of the
d) none of the above
Ferris wheel when you are in motion?
Going in Circles I
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel is
at rest, the normal force N exerted by
a) N remains equal to mg
b) N is smaller than mg
your seat is equal to your weight mg.
c) N is larger than mg
How does N change at the top of the
d) none of the above
Ferris wheel when you are in motion?
You are in circular motion, so there
has to be a centripetal force pointing
inward. At the top, the only two
forces are mg (down) and N (up), so N
must be smaller than mg.
Follow-up: Where is N larger than mg?
Vertical circular motion
http://www.youtube.com/watch?v=BHu8LAWSKxU
Vertical circular motion
Centripetal acceleration must be
C
vertical
(down)
B
horizontal
A
vertical
(up)
Condition for falling: N=0
at C:
So, as long as:
at the top, then N>0 and pointing down.
(now apparent weight is in the
opposite direction to true weight!)
The Centrifuge
Other common examples:
• spin cycle on washing machine
• salad spinner
• artificial gravity on giant space station in
show on the SciFi channel
Barrel of Fun
A rider in a “barrel of fun”
finds herself stuck with
her back to the wall.
Which diagram correctly
shows the forces acting
on her?
a
b
c
d
e
Barrel of Fun
A rider in a “barrel of fun”
finds herself stuck with
her back to the wall.
Which diagram correctly
shows the forces acting
on her?
a
b
c
d
e
The normal force of the wall on the
rider provides the centripetal force
needed to keep her going around in
a circle. The downward force of
gravity is balanced by the upward
frictional force on her, so she does
not slip vertically.
Follow-up: What happens if the rotation of the ride slows down?