ANSWERS
Quiz 4 - Chapter 4 
Ch. 4:  1, 3, 5, 6, 7, 10, 12, 15, 16, 17, 19, 20, 23, 25, 26, 28, 31, 32, 52
QUIZ 4 "TURN INS' :  5 [Re-do assuming box on table weighs 15.0(kg)], 10, 12, 19 [Re-do assuming box on table weighs 40.0(N)],  26, 28, 31, 52
10/17/2010 02:46:27 AM.   Note the correction to problem 26 hint below. 
5. 
Assume the box on the table has mass 15.0 kg. 
magnitude of weight = m*g, where m = 15.0 kg. Now,  the net force on the 15.0-kg box is zero. Let up be the positive y-direction. 

(a) 
Assume the 10.0-kg mass is NOT on top of the 15.0 - kg mass. 
Sum of Fy = pos = neg .
m*ay = N - m*g , where N = magnitude of the upward directed normal force from the ground  and m = 15.0 kg.  In other words,
0 =  N - m*g . Solve for N. 
(b)
Isolate the 15.0-kg mass and write the following equation. Assume the 10.0-kg mass is on top of the 15.0 - kg mass. 
Sum of Fy = pos - neg .
m*ay = N - m*g - N', where N = magnitude of the upward directed normal force from the ground and N' = magnitude of  downward directed normal force from the 10.0-kg mass on top. In other words,
0 =  N - m*g - N'. 
We want to solve for N. Thus we need another equation.  Isolate the 10.0-kg mass on top and write:
Sum of Fy = pos = neg .
0 = N' - m'*g , where m' = 10.0 kg. Solve for N' and plug into the first equation to get N. Your answer will be larger than part (a) ' s.   
12. The bucket  acceleration must be up  since T is greater  than mg. Let up be positive:

ma = pos - neg =T - mg where T is given. Find a. 

Alternatively  since the motion is down, you could say that down is positive, in which case you would get a negative acceleration indicating that the bucket is slowing down as it descends:
ma = pos - neg = mg - T.  Solve for a.  
15. The elevator acceleration must be down since N is less than mg. Down is positive:
ma = pos - neg = mg - N, where N = 0.75mg. Find a. 
16. ma = pos - neg, where the positive direction is up. 
ma = T - mg, The values of T and mg are given, so find a. 
17. 
(a) Let down be positive direction: ma = pos - neg = mg - mg/4. Find a.
(b) ma = pos - neg = mg - R, where  a = 0. Solve for the magnitude R of the air resistance force. 
19. Call the box on the ground object (A) and the hanging mass object (B). Note that = mAg = 40.0 (N). 

Isolate A:

(I) mAa = T + N -  mAg -
 
Isolate B: 

(II) mBa =  mBg - T. 

Note:  a = 0.

Solve (I) and (II) for N in the 3 cases given: mBg = 30.0 (N), 60.0 (N), or 90.0 (N).  Note that if N is negative, the system is in motion and you would have to solve for the acceleration a.  In the event N is negative when a  is set to zero,  set N = 0 and solve the following system for a:

(I) mAa = T -  mAg -
 
Isolate B: 

(II) mBa =  mBg - T. 
20. 

(a) Just before leaving the ground, the player has a net force directed upward causing him to accelerate upward  and to leave the ground.

 May = pos - neg

May = N - Mg, where N >Mg-

(b) 

In the air, N = 0, and the only force acting on the player if gravity.
 May = pos - neg

May = 0 - Mg so that ay  = -g if up is positive (pos). . 
23.  Isolate Arlene.  Her weight acts downward. Upward is the force of the tight rope on her feet.  

Arlene is stretching the tightrope, which behaves like a spring.  The rope is under tension  because Arlene is pushing down on it and causing it it to stretch. Thus,  all the molecules in the rope are being pulled apart from each other because she's pushing downward. 

The tension in the rope pulls on the building roof where it is attached at  two points on the roof; that tension is directed along the rope toward Arlene.  The vertical components of the tension  push up on Arlene, preventing  her from falling. The components each have the value Tsin10, where T = tension magnitude.  

May = pos - neg
0 = Tsin10 +Tsin 10 - Mg, where M = Arlene's mass.    
Solve for T. 
25. Call the top bucket object A and bottom B:

 Isolate A:
mAa = T1 - mAg  - T2 
Isolate B: mBa = T2 - mBg

Solve these equations, symbolically for T1 and T2.

(a) Solve T2 and T1 when a = 0
(b) Solve for  T1 and T2 when a = 1.6 m/s2. 
26. 
(1) ma = FcosØ  -  µN .
(2) 0 = mg + FsinØ   - N .  
For parts (a) to (c) assume that the acceleration a = 0 in equation (1) . Solve (1) and (2)  for N and for the friction coefficient  µ .
(d) Use the kinematical  information to solve for acceleration a.   Solve (1)  for  F with the non-zero  acceleration a you compute from information given in part (d)  .

NOTE: It is not really necessary to find  µ to do the problem since all the problem wants is the entire friction  force of magnitude fk .  Thus you could write equation (1) as ma = FcosØ  -  fk  and simply solve for fk.  HOWEVER ON A TEST I COULD ASK YOU FOR  µ. 

ALSO NOTE THAT THE PROBLEM ASKS YOU TO ASSUME IN PART (D) THAT THE FRICTION FORCE IS THE SAME. But the friction force will change. THAT'S BECAUSE the friction force depends on the normal force  N and in reality  N would change since F changes. To see this, solve equation (2) to get  N = mg + FsinØ .  So you are making an approximation ignoring  changes in N and  friction force. However,  this approximation seems to work since the acceleration is relatively small. 
28. 
 (1) ma = T1  - T2 
(2) ma = T2.  
Plug in ma for T2  in  (1) to get  T1 = 2ma.
Find the ratio of the two tensions.  
31. 
mAax = T - µmAg
mBay = mBg  - T

ay =  ax = a .

Solve for a   and T. From these you find all requested information. 
32.


TsinØ  = max .
TcosØ  = mg

Divide these equations to get tanØ. Solve for the x-component of acceleration  and find equation the angle.  
52. 

(a) max = mgsin22 - µmgcos22 
(b) Use  Ch. 2 methods with  equation  2.11 (a) to (d).   

More hints later !