QUIZ 6 (ANSWERS) (WORKSHEET 2)
QUIZ 6  AND QUIZ 7 ARE MORE CLOSELY  EXAMPLE DRIVEN---HERE WILL BE A CLOSE CORRESPONDENCE,  often with explicit cross indexes,  BETWEEN EXAMPLES  5.1 TO 5.24 AND THE EXERCISE/PROBLEMS.  THESE PROBLEMS HELP US UNDERSTAND  APPLICATIONS OF NEWTON'S LAWS, WHICH IN TURN  OPEN  US   VIA ANALOGY TO  DEEP INNER-WORKINGS OF NATURE. FOR EXAMPLE  CLASSICAL MODELS STARTING WITH THESE PROBLEMS HELP US VISUALIZE  MOLECULAR PROCESSES  OBEYING  NON-INTUITIVE STATISTICS. Sometimes it helps to think of an electron  orbiting in a well defined, classical circular orbit around the nucleus  using this  chapter's definition of centripetal force even though we know  electrons come in waves  or "clouds" with a  probability function  and   UNCERTAINTY PRINCIPLE . 
DISCUSSION QUESTIONS 12, 13, 16, 19, 20, 23, 24, 25, 26, 27, 29, 30
SEC. 5.1:  2, 8 ( Try also #4) , 10, 11 , 14, 15 (Try 13)  , 65 (Try #36); SEC  5.2: 16, 18,  20, 21 (Try #19 , 36, 37),  23, 24;
SEC. 5.3: 28, 31 (Try # 28),  38, 42, 44, 47, 48; (Try #91 and #120, 121)
NOTE: #91 IS EXTREMELY IMPORTANT  IMPORTANT  FOR REASONS THAT CRYSTALLIZE AFTER TODAY.
2. You can also assume in each block  is moving at constant speed along the lines of their motion, which means the acceleration equals zero. This is called DYNAMIC EQUILIBRIUM. In either case AT REST OR WITH CONSTANT SPEED,  with mass m,   EACH BLOCK has acceleration equal to zero , so you can  simply write

sum of forces in y dir = pos - nag = 0 since the  acceleration in y direction = 0,

In  (a) for example ,we have 0 = pos - neg = T - mg.

8.  Form an x-y axes on the plane of you page and let right and upward  on paper be the pos x and pos y directions, respectively. ISOLATE MASS A.
sum of force in x dir = 0 = -TA*cos0 + TB*cos theta, where theta is the angle the B string makes with the pos x-axis.
sum of forces in y dir = 0 = TB*sin theta  - mg
10. DRAW YOUR AXES DIFFERENTLY NOW WITH an x-y axes on the plane of your page and let UP the plane be the pos x direction and let PERPENDICULAR  and away from the plane surface  be the  pos y direction.
sum of forces in x direction = 0 =  T*cos31 - mg*sin 25.
sum of forces in y direction = 0 = T*sin 31 + N - mg*cos 25.
From this info you can answer all questions selectively.
DRAW A FREE BODY DIAGRAM.
11. DYNAMIC EQUILBRIUM MEANS MOTION WITH ZERO ACCELERATION AS IN THIS CASE. 0. DRAW YOUR AXES  NOW WITH an x-y axes on the plane of your page and let DOWN the plane be the pos x direction and let PERPENDICULAR  and away from the plane surface  be the  pos y direction.
(a)   
sum of forces in x direction = 0 =   mg*sin 11 - F
sum of forces in y direction = 0 =  N - mg*cos 11
(b)
sum of forces in x direction = 0 =   mg*sin 11 - F*cos 11
sum of forces in y direction = 0 =  N - mg*cos 11 - F*sin 11
DRAW A FREE BODY DIAGRAM.
14.  See Example 5.4 . To get the tension in the rope connecting A to the wall, you can assume that the that rope supports a total mass MA + MB. Thus TA = (MA + MB)*g*sin alpha, , where alpha is the angle with the horizontal. To find TB, isolate   MB. : Let the x axis be along the incline so that  sum of forces in x direction = 0 =  TB - MB*g*sin alpha. For the normal forces in both cases we have:
sum of forces in y dir = N = M*g*cos alpha 
15. DRAW YOUR AXES DIFFERENTLY NOW WITH an x-y axes on the plane of your page and let UP the plane be the pos x direction and let PERPENDICULAR  and away from the plane surface  be the  pos y direction. APPLY THE FORCES AT THE BALL CENTER.
The tension exerts a force parallel to the horizontal , so it is like #11, part (b).
DRAW A FREE BODY DIAGRAM.
sum of the forces in x direction = 0 = pos - neg =  T*cos35 - mg*sin 35
sum of forces in y direction = 0 =  N - mg*cos 35 - T*sin 35
65.
(a) See example 5.5 but add friction and assume DYNAMIC EQUILIBRIUM---in both cases the acceleration equals zero for both masses. Isolate each mass. For mass 1 assume: UP the plane is the pos x direction and let PERPENDICULAR  and away from the plane surface  be the  pos y direction. For mass 2 assume vertically down on the page is the positive y direction. ISOLATE:

MASS 2:  sum of forces in y direction = 0 = pos - neg = m2*g - T.
MASS 1:  Sum of forces in x direction = 0 = T -  m1*g*sin alpha - uk*m1*g*cos alpha.
sum of forces in y direction = 0 = N -  m1*g*cos alpha.
DRAW A FREE BODY DIAGRAM.
16. See example 5.9. Let the power supply play the same role as the woman. Note  the difference. In this problem the rocket moves up and accelerates upward ( speeds up ) whereas in the example, the elevator slows down while moving down, which means the acceleration also points up. So the acceleration is the same but the motions are  different between the example and this problem.   If up is positive for the motion, you'd write after isolating the woman:
For part (b), m*a = pos - neg = N - mg, where m is the woman's mass and N is the magnitude of the normal force on her feet. Note:
For part (a), (m + M)*a = F - (m + M)*g  where F = 1720 (N) and M + m is the sum of the elevator's and woman's mass.
18.   See in-class quiz with three  mass-less strings connecting 4 blocks together as they moved with common acceleration under the influence of an external force of magnitude F = 102 (N). Formally,  in this problem you'll have three equations and three unknowns, acceleration a, and TA and TB. ISOLATE EACH MASS !
20.
(a) See class notes: Sum of the forces in the x dir = m*a = pos - neg = m*g*sin theta.  DRAW YOUR AXES  NOW WITH an x-y axes on the plane of your page and let DOWN the plane be the pos x direction and let PERPENDICULAR  and away from the plane surface  be the  pos y direction. You can  acceleration a from Ch. 2 theory.
(b)  Add friction and the equation becomes: Sum of the forces in the x dir = m*a = pos - neg = m*g*sin theta. - f, where f = 10.0 (N).
21.  See example 5.12.
23. Nice problem,  harkening back to example 5.8 for a string with mass.  Check out this link to  an earlier ICQ3-14 AND ALSO #54, CH. 4, QUIZ 5.      Let upward be the +y direction.  Let a lifting clamp labeled A  meet the   top of the chain at point A.  Label the chain S just like I did last in icq3-14-11, and the boulder B. Use the same notation as icq3-14 but remember each object-- boulder B and the chain S--have a downward gravitational force on them. For example the string  equation reads
ms*a = FSA - FSB  - ms*g. The boulder equation reads:
mB*a = FBS  - mB*g. Note FSA = the tension at the top of chain S = force of the clamp A on  chain. Note: FBS = the tension in  string at the bottom. Note :  FBS=  FSB from Newton's 3rd Law.

Now, if loading is done in dynamic equilibrium at constant speed directed upward, a = 0. Thus: FSA =  FSB  + ms*g  and   FBS  = mB*g. You can now find FSA and FBS .   Note: Theoretically the tension at the midpoint of chain is   (FSA + FSB )/2.

For the problem at hand, let's consider the case when loading is not in equilibrium and  upward acceleration a is greater than zero.
Return to equations:

ms*a = FSA - FSB  - ms*g. 
mB*a = FBS  - mB*g.
Set FSA  =   2.50*ms*g.  Solve for a.
(b) Ch. 2.
24. The main thing is finding out  if the magnitude of the scale force reading is larger or smaller than the student's normal weight mg. The scale force reading is equal to  magnitude N of the normal force on the student.   If N >mg, the elevator is accelerating upward. If N < mg, the elevator is accelerating downward.

Assume the motion  and thus the positive direction are upward  without loss of generality.
In that case: sum of forces =  m*a =  pos - neg = N - mg. Thus,
N = m*a + mg.  
If acceleration a is positive, the acceleration is upward, thus a > 0 and N >mg. If a is negative, acceleration is downward, thus a < 0 and N < mg . 

(a) N = 450 (N).  Find a. 
(b) N = 670 (N).  Find a
(c) If N = 0  what is a? Should the student be worried? Explain.
(d) To find the tension T in the cable, write sum of force s=  pos - neg for the entire system of elevator and student:
(m + M)*a = T - (m + M)*g, where M is the elevator's mass. Solve for T in parts (a) and (c).
28. See section 5.3 and the  figure 5.19 narrative.
31.
(a) fsMAX = 313 (N);  fk = 208 (N).
(b) m*a = pos - neg = F - fk ,  where a = 1.10 m/s^2.
(c) Repeat if  g = 1.62 m/s^2.
38. Example 5.18,  Use CH. 2 methods friction with constant acceleration, found from Newton's 2ND Law: 
m*a = - pos - nag, where neg = 0 in the case when the  net  horizontal force is due to rolling motion only.
42. Example 5.16. See  WORKSHEET 2 discussion for problem 3, GRP2.
44. Example  5.15 and  WORKSHEET 2 discussion for problem 1, GRP2.
47. example 5.19 (sec. 5.3)
48. Let the positive y direction be up. Let a be the y component of acceleration. In both cases set v = vterm/2, where vterm is the terminal speed of section 5.3  The magnitude of the air resistance force is R= b*v2, proportional to the speed squared.
(a)  sum of forces in y dir = m*a = pos - neg = 0 - mg - R, where R= b*v2.
(b) sum of forces in y dir = m*a = pos - neg = R - mg