• 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 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 30N-

26.95N = +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)

• 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

• 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)

• 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 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.

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