lecture 8 applications of newton’s laws (chapter 6)
TRANSCRIPT
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) 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.
Tm
Static friction (s= 0.4)
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?
The static friction force has a
maximum of sN = 40 N. The
tension in the rope is only 30 N.
So the pulling force is not big
enough to overcome friction.
Will It Budge?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.
Follow-up: What happens if the tension is 35 N? What about 45 N?
Tm
Static friction (s= 0.4)
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?
TensionT
Fs
W
W
T
2.00 kg
Tension in the rope?
Translationalequilibrium?
W
W
TT
y :m1 : x :
m2 : y :
fk
Over the Edge
m
10 kg a
m
a
F = 98 N
Case (1) Case (2)
a) case (1)
b) acceleration is zero
c) both cases are the same
d) depends on value of m
e) case (2)
In which case does block m
experience a larger acceleration?
In case (1) there is a 10 kg mass
hanging from a rope and falling.
In case (2) a hand is providing a
constant downward force of 98
N. Assume massless ropes.
In case (2) the tension is
98 N due to the hand.
In case (1) the tension is
less than 98 N because
the block is accelerating
down. Only if the block
were at rest would the
tension be equal to 98
N.
Over the Edge
m
10 kg a
m
a
F = 98 N
Case (1) Case (2)
a) case (1)
b) acceleration is zero
c) both cases are the same
d) depends on value of m
e) case (2)
In which case does block m
experience a larger acceleration?
In case (1) there is a 10 kg mass
hanging from a rope and falling.
In case (2) a hand is providing a
constant downward force of 98
N. Assume massless ropes.
W
T
a>0 downwardimplies T<W
TensionForce is always along a rope
W
TTTy
TT
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.
SpringsNote: 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 TensionA 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
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2
Springs and TensionA 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
W
Fs=T
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2
Springs and TensionA 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?
Spring 1 supports the weight.Spring 2 supports the weight.Both feel the same force, and stretch the same distance as before.
S1
S2
W
Fs=T
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2
Instantaneous acceleration
Velocity vector is always in the direction of motion; acceleration vector can points in the direction velocity is changing:
From
Lectu
re
3 Fr
om Le
cture
3
Circular MotionAn object moving in a circle must have a force
acting on it; otherwise it would move in a straight line.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
aa
Circular Motion
This force may be provided by the tension in a string, the normal force, or friction, among others.
Examples of centripetal force
whenno 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 object moving in a circular path
Then
2) is NOT a separate force; it represents
the sum of the physical forces acting on m
2c
m v tF t r
cF
c cF f
Going in Circles IGoing in Circles I
a) N remains equal to mg
b) N is smaller than mg
c) N is larger than mg
d) none of the above
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel
is at rest, the normal force N exerted
by your seat is equal to your weight
mg. How does N change at the top of
the Ferris wheel when you are in
motion?
Going in Circles IGoing in Circles I
a) N remains equal to mg
b) N is smaller than mg
c) N is larger than mg
d) none of the above
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel
is at rest, the normal force N exerted
by your seat is equal to your weight
mg. How does N change at the top of
the 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
A
B
C vertical (down)
vertical (up)
horizontal
Centripetal acceleration must be
Condition for falling: N=0
at C:
(now apparent weight is in the opposite direction to true weight!)
So, as long as:
at the top, then N>0 and pointing down.
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 FunBarrel 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.
Barrel of FunBarrel 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?
Follow-up: What happens if the rotation of the ride slows down?
a b c d e