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2015-07-14
1
© 2012 Pearson Education, Inc. Slide 1-1
PreClass Notes: Chapter 5, Sections 5.1-5.3
• From Essential University Physics 3rd Edition
• by Richard Wolfson, Middlebury College
• ©2016 by Pearson Education, Inc.
• Narration and extra little notes by Jason Harlow,
University of Toronto
• This video is meant for University of Toronto
students taking PHY131.
© 2012 Pearson Education, Inc. Slide 1-2
Outline
• 5.1 Problem Solving with
Newton’s Second Law
• 5.2 Objects Connected by
Ropes and Pulleys
• 5.3 Circular Motion
“Why does an airplane tip when
it’s turning?” – R.Wolfson
Image from http://www.decodedscience.com/side-effect-of-rolling-an-airplane-aircraft-yaw/7209
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© 2012 Pearson Education, Inc. Slide 1-3
• Interpret the problem to make sure Newton’s second law is
the relevant concept. Identify which objects are of interest
and think about the forces on each object. Identify
connections between objects and the constraints on their
motion.
• Draw a free-body diagram for each object. Develop your
solution plan by writing Newton’s second law in
components for each object.
• Execute your plan and solve the equations. Remember
that the constraints are like equations.
• Assess your solution to see whether it makes sense.
Think about units, special cases.
Problem Solving Strategy 5.1
© 2012 Pearson Education, Inc. Slide 1-4
Example 5.1
A skier of mass m = 65 kg glides down a
slope at an angle of θ = 32°, as shown.
Find (a) the skier’s acceleration and (b) the
force the snow exerts on the skier. The snow
is so slippery that you can neglect friction.
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© 2012 Pearson Education, Inc. Slide 1-5
Example 5.1
© 2012 Pearson Education, Inc. Slide 1-6
Example 5.1
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© 2012 Pearson Education, Inc. Slide 1-7
Got it?
• A ball of mass m is suspended by a string from the
ceiling inside an elevator. If the elevator is moving
upward with a constant speed, the tension in the
string
A. is greater than mg.
B. is equal to mg.
C. is less than mg.
D. depends on the speed of the elevator.
© 2012 Pearson Education, Inc. Slide 1-8
Tension
Figure (a) shows a heavy
safe hanging from a rope
The combined pulling force
of billions of stretched
molecular springs is called
tension
Tension pulls equally in both
directions
Figure (b) is a very thin cross
section through the rope
This small piece is in
equilibrium, so it must be
pulled equally from both sides
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© 2012 Pearson Education, Inc. Slide 1-9
Example 5.2
To protect her 17 kg pack
from bears, a camper hangs it
from ropes between two trees,
as shown. What’s the tension
in each rope?
© 2012 Pearson Education, Inc. Slide 1-10
Example 5.2
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© 2012 Pearson Education, Inc. Slide 1-11
The Massless String Approximation
Often in problems the mass of the string or rope is much
less than the masses of the objects that it connects.
In such cases, we can adopt the following massless string
approximation:
© 2012 Pearson Education, Inc. Slide 1-12
Got it?
• A rope is tied to a hook that is attached to a wall. If
you pull the rope with a 1-N force, the force
exerted by the hook on the rope
A. is greater than 1 N.
B. is less than 1 N.
C. is equal to 1 N.
D. cannot be determined from the information given.
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© 2012 Pearson Education, Inc. Slide 1-13
Pulleys
Block B drags block A across a frictionless table as it falls
The string and the pulley are both massless
There is no friction where the pulley turns on its axle
Therefore, TA on S = TB on S
© 2012 Pearson Education, Inc. Slide 1-14
Acceleration Constraints
If two objects A and B move together, their accelerations are
constrained to be equal: aA = aB
This equation is called an acceleration constraint
Consider a car being towed by a truck
In this case, the
acceleration constraint is
aCx = aTx = ax
Because the
accelerations of both
objects are equal, we can
drop the subscripts C and
T and call both of them ax
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© 2012 Pearson Education, Inc. Slide 1-15
Acceleration Constraints
Sometimes the acceleration
of A and B may have
different signs
Consider the blocks A and
B in the figure
The string constrains the
two objects to accelerate
together
But, as A moves to the right in the +x direction, B moves
down in the −y direction
In this case, the acceleration constraint is aAx = −aBy
© 2012 Pearson Education, Inc. Slide 1-16
Example 5.4
A 73 kg climber finds himself dangling
over the edge of an ice cliff, as shown.
Fortunately, he’s roped to a 940 kg rock
located 51 m from the edge of the cliff.
Unfortunately, the ice is frictionless, and
the climber accelerates downward.
What’s his acceleration?
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© 2012 Pearson Education, Inc. Slide 1-17
Example 5.4
© 2012 Pearson Education, Inc. Slide 1-18
Example 5.4
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© 2012 Pearson Education, Inc. Slide 1-19
Dynamics of Uniform Circular Motion
Without such a force,
the object would move
off in a straight line
tangent to the circle.
The car would end
up in the ditch!
An object in uniform circular motion is not traveling at a
constant velocity in a straight line.
Consequently, the particle must have a net force acting
on it
© 2012 Pearson Education, Inc. Slide 1-20
Example 5.6
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© 2012 Pearson Education, Inc. Slide 1-21
Example 5.6
© 2012 Pearson Education, Inc. Slide 1-22
Example 5.6
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© 2012 Pearson Education, Inc. Slide 1-23
Example 5.6
© 2012 Pearson Education, Inc. Slide 1-24
Example 5.7
The “Dragon Fire” roller coaster at Canada’s Wonderland features a
double loop section. One of the loops is shown, and the radius of
curvature at the top is 6.3 m. What’s the required speed for a roller
coaster at the top of the loop if the normal force from the track is to be
zero (neither pushing nor pulling)?
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© 2012 Pearson Education, Inc. Slide 1-25
Example 5.7
© 2012 Pearson Education, Inc. Slide 1-26
Example 5.7