Download P221_2008_week4

• The coefficient of static friction between a 12.5 kg block and table is
0.34 and the coefficient of kinetic friction is 0.22. The block is at rest
on the table when a horizontal force of 30 N is applied to it. A). What
is the force of friction at this point in time? B). The block is now set in
motion (by temporarily applying a larger force), and then the same
30 N force is applied. What is the acceleration in this second case?
For both questions give a brief description of how you arrived at your
answer. (A: 8 correct 19 incorrect; B: 9 correct 18
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incorrect; 28 none)
a)10 N. This is calculated by multiplying the number of
Newtons applied to the stationary block by the coefficient
of friction.
A) F= mg*0.34=41.65 N (as block is not moving we are
using coefficient of static friction)
b) 2.1 m/s^2. This is calculated by multiplying 12.5 by
9.8 and the .22 coefficient of kinetic friction.
fs,max = 0.34*N = 0.34*12.5kg*9.8m/s2 = 41.6N, since this
is greater than the applied force fs=30N opposite the
applied force (since a=0 the forces must cancel !!).
– Once it is moving fk = 0.22*122.5N =26.95N => net force is 30N26.95N = +3.0N so a = 3.0N/12.5kg = 0.24 m/s2
• “The force of friction always opposes motion.” Please
comment briefly on the validity and/or universality of this
statement. (Correct: 5 Incorrect: 24 No answ: 24)
• From Newton's 2nd law, a force must act oppposite of
any given force, so when an object is pushed a force
must push back against it in the direction opposite its
velocity, this force is frictional force.
• Based on Newton's third law we can derive that for
every force acted on an object there is an equal force
being applied on that object in the opposite direction.
• My understanding of friction too is that it always opposes
the motion or attempted motion of one surface across
another. I can't really think of any exception off the top of
my head, so I'd say this statement is valid in most
situations..
• The force of friction always acts opposite to any real or
“virtual” RELATIVE motion of the two surfaces in
contact.
Chapter 6 examples
• Using the terms provided in the reading, explain why the terminal
speed for a skydiver is less when she assumes the spread-eagle
position (figure 6-8) as opposed to a head or foot-down orientation.
Estimate the ratio of speeds (head-first over spread eagle), and
explain how you arrived at your result. (6 correct; 14 tied speed ratio
directly to the area ratio (not the square-root thereof; 9 didn't
quantify and 26 didn't answer; About 6 figured out that the speed
ratio was about 1.5 from the info on p123, few worked this into an
area ratio of 2.2)!
• I would say the ratio is 1:2 because you are essentially doubling
your body's surface area by being in the spread-eagle position.
• v-terminal = sq.rt. [2F-gravity/(drag coefficient*air density*A)], the
terminal velocity will increase when the area decreases (head or
foot-down orientation):
– Using the equation above, the ratio would be A-eagle/A-head.
• 88.9 m/s / 60 m/s in table 6.1 the typical velocity of a sky diver is
60m/s but the man who caught the lady was going 320km/hr
(88.9m/s)
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Consider a Ferris wheel ride, in which the wheel is rotating such that you
are looking away from the wheel when you are going upward (and toward its
axis when going down). If the rotational speed of the wheel is constant,
indicate the direction of the net force (up, down, forward, backward), and the
origin of the greatest contribution to that net force when you are A). at the
bottom B). halfway up going up. C). at the top. and D). halfway down going
down.
• A) Net force = upward B) Net force = upward
C)
Net force = upward D) Net force = ??
• A) forward, center; B) up, center C) backward, center
D) down, center
• A. up. B. backward C. down D. forward The origin of all
the net forces is the support bars of the ferris wheel
pulling the cars inward.
• in A,B,C,and D the net force is always pointing to the
center.
• DIRECTION: 13 correct 9 confused with velocity 29
didn’t answer
• Origin: confused several of you, 9 suggested that the
origin was “centripetal force” What produces the force??
CALM suggestions
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Drag/friction (9)
Uniform Circular motion (6)
Projectile motion (5)
The Big 3 / Free Body Diag. (3 each)
Setting up problems / Tension (2)
EVERYTHING!! (1)
Dot and cross products (3)
Chapter 6 Examples
Chapter 6 Examples
Assume
throughout that
the rotation is at
a constant rate.
Q0: What is the
direction of the net
force on the ball?
More Chapter 4 and 5 Examples
4-
5-
DVB- Review/Summary
(* -> topics that I think we have emphasized the most)
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UNITS, (dimensional analysis and checking your answers)
Newton’s Laws: F=ma ; Free body diagrams*
Interpreting graphs
Kinematics*: Big 3, def’s of a(t),v(t) etc., free fall
Vectors: components*, adding*, products (. & x)
2-D motion*: Projectiles, relative motion,
centripetal acceleration
• Friction and Drag
• To date we’ve had 9 lectures covering new
material, look carefully at each an glean the 2-3
key points, write review questions, …
DVB- Exam details
• NEXT WEDNESDAY (6 Feb.)
• 4 multiple choice questions followed by 2 multipart (4 or 5) problems for a total of 13 individual
questions. Partial credit is available for all.
• If part b uses answer from part a and a is wrong
you can still get full credit for b!!
• Some of the 13 are very straight-forward, a few
are more challenging.
• Show your work and use a PEN not a pencil!!
• Questions at back of chapter and CALM are
pretty good practice in addition to the problems
from the chapters.
DVB’s Formula sheet
• Newton II and III
• K1, K2, K3; g = 9.80 m/s2
• Vector components (if you don’t know trig
well), dot and cross products (2 each).
• Centripetal acceleration
• Friction: Static, kinetic, drag
• Definitions of instantaneous and average
velocity and acceleration.
• VAB = VAC + VCB NOTE: VCB = - VBC
• Any notes I’d need on drawing FBD’s