Transcript P221_2009_week4

Chapter 4 examples
(C) What is providing the centripetal force that keeps the cat in a circular
path?
Chapter 5 examples
a
N
ffr
q
mg
a). Net force on the rider must be in the direction of his/her acceleration (F=ma),
so we have Fnet= 60.0kg*3.0m/s2 = 180 N along the ramp (i.e. 10o above the
horizontal.) Note; in the FBD above, q is equal to the ramp angle (10o).
b). To find the net force from the motorcycle is the vector sum of the Normal and
friction forces from the seat on the rider. Looking at the FBD above: you see that:
ffr-mgsin(10o) = ma and N-mgcos(10o) = 0 so ffr=282N; N= 579 N. This is a net
force of magnitude 640N at an angle of 64o above the incline (74o above horiz.)
Chapter 5 examples
F1
a=0
mg’
F2
a
In this case we have the two FBD’s on the right:
a). From the top diagram: F1-mg= 0 => the weight of the landing craft
(mg’) near this moon is mg’=3260 N.
b) From the second FBD we get: 2200N-mg’ = m*(-0.39 m/s2), or
2200N-3260N -1060N = m*(-0.39 m/s2). => m = 2718kg= 2700 kg.
c) We can find the relevant number for the local acceleration due to
gravity from 3260N = 2718kg*g’ => g’= 1.20m/s2.
mg’
4-
We note that the maximum flight time occurs if q=90o (i.e. if you shoot it
straight up, vertical component of velocity equal to speed vx=vo). To be in
the air half this time you need a vertical velocity component that is half
the speed, or q = 30o. What is the least speed a ball has in flight? Its the
horizontal component of the velocity (which is constant) vocosq = 0.87vo,
since only at the trajectory’s peak, the vertical component is zero. So all
we need to do is figure out vo from the figure.
For level ground the max. range is at 45o. So t=2*vosin(45)/9.8m/s2 and
240m =vocos(45)*t => (vo)2 = 240m*(9.8m/s2)/[2*(sin(45)*cos(45)]; or
vo=48.50 m/s. Therefore the least speed the ball has in the flight for
which the time is half the maximum possible value is: vmin=42.0 m/s.
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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: 17
correct 19 incorrect; B: 12 correct 23 incorrect; 19 no
answer; Only about 5 got both parts!)
a) .34 x 30 N = 10.2 newtons. I used the equation in the
book to figure out how to answer this problem. NO: ms
times the NORMAL force gives fs,max NOT fs!
a). the force of friction at this time is 41.69 N. Since the
block is stil at rest, I multiplied the coefficient of static
friction by the normal force, which is just the mass of the
block times the acceleration of gravity. NO: this is the
Maximum possible value for the force of friction, not fs!
Remember the soda can and the hippo!!
B)the acceleration of the block is 2.4m/s^2, you simply
divide the force (30N) by the mass (12.5kg) NO: Newt. II
refers to the NET FORCE (30N-26.95N in this case)
B) 2.156 m/s/s; set kinetic frictional force equal to mass
multiplied by acceleration. NO: same as the above.
• 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.
• 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 all the forces must cancel and
these are the only two acting in the horizontal
direction!!).
• Once it is moving fk = 0.22*122.5N =26.95N => net force
is 30N-26.95N = +3.05N so a = 3.05N/12.5kg = 0.24
m/s2 in the direction of the applied force
• “The force of friction always opposes motion.” Please
comment briefly on the validity and/or universality of this
statement. (True: 22; not always valid: 9; unclear: 7; no
answer: 18)
• No, the force of friction does not always opposes motion.
Sometimes, there is no friction such as like outside of the
earth.
• This is always true. Friction makes motion slow down
and stop, and it can sometimes prevent motion from
happening.
• Friction only opposes motion parallel to the two surfaces
that are touching, and when there is no motion it
opposes any force acting parallel to the surfaces.
• The statement is false because without friction we
wouldn't be able to walk--(motion).
• “The force of friction always opposes motion.” Please
comment briefly on the validity and/or universality of this
statement. (True: 22 not always valid: 9; unclear: 7; no
answer 18)
• No, the force of friction does not always opposes motion.
Sometimes, there is no friction such as like outside of the
earth. (I guess the question was not clear)
• This is always true. Friction makes motion slow down and
stop, and it can sometimes prevent motion from
happening. (but see the answers below)
• Friction only opposes motion parallel to the two surfaces
that are touching, and when there is no motion it opposes
any force acting parallel to the surfaces.
• The statement is false because without friction we wouldn't
be able to walk--(motion).
• The force of friction always acts opposite to any real or
“virtual” RELATIVE motion of the two surfaces in contact.
In so doing, it can generate motion (e.g walking).
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. (essentially all respondents
figured out that spread-eagle gives slower terminal speed; Area ratio
only about 6 were more or less correct; 11 made an error; 3 were
confused; 33 didn’t answer!??)
• Terminal velocity is inversly related to the cross-sectional area, and I
assume this area is about 3 times greater in the spread-eagle
position than the head or foot-down orientation, so I think the
skydiver's terminal velocity would be 3 times greater going head or
foot down vs spread eagle. This was the most common mistake, the
inverse relationship is with the SQUARE ROOT of the AREA!!
• To estimate the ratio of the speeds, assume the body is roughly
0.5mx2m=1m2 in spread-eagle and 0.5mx0.5m=0.25m2 head-on.
Ratio of the areas is roughly a factor of 4, ratio of the terminal speed
is then a factor of 2 since vt ~ (A)1/2
CALM suggestions
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Vectors (5)
Drag/friction (5)
Relative Motion (4)
Uniform Circular motion (3)
Setting up problems, FBD’s (3)
34 didn’t bother to answer??? I’ve
reopened this question until Sunday night.
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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.
• the centripetal force is uniform without, thus the contributing
forces are the same throughout Uniform in magnitude, not in
direction and the forces responsible do change!
• A)The net force is going forward; and the greatest force is the
centripedal force. B)The net force is going up; and the greatest
force is the normal force. DON’T confuse forces with
velocities! REMEMBER the centripetal force has to come
FROM SOMETHING, so “centripetal force” cannot be the
answer to a question asking for its origin!!
• DIRECTION: 11 correct 10 confused with velocity 29 didn’t
answer
• Only 5 got the forces responsible correct. I think that this is
indeed a problem with “setting up the problem”: DIAGRAMS!
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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.
• DRAW a diagram as below:
• Solution: A) UP, Normal force from seat B)
BACK, friction of the seat C) DOWN, gravity D).
FORWARD, friction (or possibly normal force
from the seat’s back)
Net force must be in
the direction of the
acceleration in each
case, draw a FBD to
see what force must
be the largest to get a
net force in the given
direction for each
position
a
a
a
a
Chapter 5 examples
(d) What are the answers to parts a and c if there is friction between m1
and the ramp with mk = 0.10?
Chapter 5 examples
a
N
T
T
a
q
m1 g
m2 g
Based upon the free body diagrams shown above we can write the following
results from Newton’s second law:
T-m1gsinq = m1 a for the first mass (taking the x axis for this mass to be up
and to the right along the ramp). T – m2g = m2a for the second mass. Note that
T is the same in both cases because the rope is assumed to be massless, and
the a is the same in both because the two blocks move together. Subtracting
the two equations we get a = (m2-m1sinq)g/(m1+m2) = 0.735 m/s2 (positive
sign means m2 goes down, m1 goes up the ramp) plugging this back in we get
T= 20.8 N. If mk = 0.1 between m1 and the ramp, you add a force –mk*m1gcosq
to the FBD for m1, and you find a=0.254m/s2 with T=25.8N.
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 (11 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