P202 Lecture 2

1
Chapter 4 examples
(C) What is providing the centripetal force that
keeps the cat in a circular path?
2
Chapter 5 examples
a
N
ffr
q
mg
a). Net force on the rider must be in the
direction of his/her acceleration (Fma), so we
have Fnet 60.0kg3.0m/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 ffr282N N 579 N. This is
a net force of magnitude 640N at an angle of 64o
above the incline (74o above horiz.)
3
Chapter 5 examples ?
F1
a 0
mg
F2
a
In this case we have the two FBDs on the
right a). From the top diagram F1-mg 0 gt the
weight of the landing craft (mg) near this moon
is mg3260 N. b) From the second FBD we get
2200N-mg m(-0.39 m/s2), or 2200N-3260N -1060N
m(-0.39 m/s2). gt m 2718kg 2700 kg. c) We
can find the relevant number for the local
acceleration due to gravity from 3260N
2718kgg gt g 1.20m/s2.
mg
4
4 -
We note that the maximum flight time occurs if
q90o (i.e. if you shoot it straight up, vertical
component of velocity equal to speed vxvo). 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 trajectorys 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 t2vosin(45)/9.8m/s2 and
240m vocos(45)t gt (vo)2 240m(9.8m/s2)/2(si
n(45)cos(45) or vo48.50 m/s. Therefore the
least speed the ball has in the flight for which
the time is half the maximum possible value is
vmin42.0 m/s.
5

• 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/s2, 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.

6

• 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 0.34N 0.3412.5kg9.8m/s2 41.6N,
  since this is greater than the applied force
  fs30N opposite the applied force (since a0 all
  the forces must cancel and these are the only two
  acting in the horizontal direction!!).
• Once it is moving fk 0.22122.5N 26.95N gt 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

7

• 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).

8

• 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).

9
Chapter 6 examples
10

• 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 didnt 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 0.5mx2m1m2 in spread-eagle and
  0.5mx0.5m0.25m2 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

11
CALM suggestions

• Vectors (5)
• Drag/friction (5)
• Relative Motion (4)
• Uniform Circular motion (3)
• Setting up problems, FBDs (3)
• 34 didnt bother to answer??? Ive reopened this
  question until Sunday night.

12

• 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. DONT 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 didnt answer
• Only 5 got the forces responsible correct. I
  think that this is indeed a problem with setting
  up the problem DIAGRAMS!

13

• 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 seats 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
14
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?
15
Chapter 5 examples?
a
N
T
T
a
q
m1g
m2g
Based upon the free body diagrams shown above we
can write the following results from Newtons
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/(m1m2) 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 mkm1gcosq to the FBD for m1, and you find
a0.254m/s2 with T25.8N.
16
DVB- Review/Summary( -gt topics that I think we
have emphasized the most)

• UNITS, (dimensional analysis and checking your
  answers)
• Newtons Laws Fma Free body diagrams
• Interpreting graphs
• Kinematics Big 3, defs 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 weve had 9 lectures covering new
  material, look carefully at each an glean the 2-3
  key points, write review questions,

17
DVB- Exam details

• NEXT WEDNESDAY (11 Feb.)
• 4 multiple choice questions followed by 2
  multi-part (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.

18
DVBs Formula sheet

• Newton II and III
• K1, K2, K3 g 9.80 m/s2
• Vector components (if you dont 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 Id need on drawing FBDs