QUIZ 5 (ANSWERS)
SAMPLE TEST 2
READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE. 
CH. 5 
CH. 5 - Multiple choice  (TBA) and exercise/problems  equilibrium: 2 (try #15 ) , 3, 5( see #12), 6, 12, 14, 16 (see #29); accelerometers:  23, 25, 26; general no friction: 22, 27, 28 (try #29, done in class),  29*, 66, 29; friction: 34,  35, 36, 37, 39, 46, 47; drag forces: 51, 52; elastic forces (Hooke's Law) : 59, 60, 61.     
TURN IN :CH. 5~2*, 5, 6,  23, 26, 28, 39, 46, 52, 60 *  *DONE IN CLASS. 
DISCUSSIONS PROVIDED. 
SELECTED DISCUSSIONS TO exercises/problems BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES.
error log #1 --See corrections to #6 below  regarding the  forces perpendicular to the incline.
error log #2 --See corrections to #26 below  regarding the leftward motion in which the car slows down; the cable would swing left in that case.
5. SEE EXAMPLE 5.2 FOR A GEOMETRICAL 2D (2 dimensional) system.   Resolve the vector components along horizontal and vertical axes.

(a) sum of forces in x direction = pos - neg = 0 =  Tcos 10 - T*cos 10. 

The above equation provides insight but very little help in actually finding T . So use the y-component equations:
sum of forces in y direction = 0 = pos - neg = Tsin10 + Tsin10 - mg.  

Solve for T.    
(b) Set T = 25000 (N) and solve 2*T*sin theta = mg for theta.

6. See example 5.3 for a simplified version of problem; This time, in contrast to the previous problem (#5), we resolve our components along x and y axes that are parallel and perpendicular to the inclined plane, respectively.
Let the positive x direction be UP the incline.
SUM of forces in x direction = pos - neg = 0 = T*cos 31   - mg*sin 25.
SUM of forces in y direction = pos - neg = 0 =  N + T*sin31 - mg*cos 25.

Use these equations to answer parts (b) and (c) to find T AND N;  use the definition of normal force, exerted by the surface on the car.   

23.  See example 5.5 for the spirit of this problem.
 Let ay be the y-component of acceleration .

(a) Let up be the positive y direction, the assumed direction of motion.
sum of forces in y direction = m*ay = pos - neg = Fs - mg, where Fs is the magnitude of the spring force on the fish; note the value Fs is the scale reading ! The scale reading is  Fs =  60.0 (N). 
Now, you are given   ay  =  + 2.45 m/s2. Solve for m and find mg.
(b) Let DOWN  be the positive y direction, the assumed direction of motion.  In the previous part, we  we see that mg < 60.0 (N). This must be true otherwise the fish would not  be speeding up as it moves upward.   If the fish accelerates downward, i.e., speeds up as it moves down, we have a different situation. In that case, mg > Fs.   Assuming the object moves down, we have sum of forces in y direction =  m*ay = pos - neg = mg  -   Fs .  You are given  Fs   = 35.0 (N). Solve for the acceleration component.  
(c) ZERO in which case the fish would be in free fall. 
26.  See example 5.5 for a direct application of this problem to a real world situation. As the car moves right, speeding up,  the acceleration points right and the cable swings  left. Meanwhile, as the car moves left, the cable would swing  right if the speed was increasing. But the speed decreases during leftward motion, which means the acceleration points right and thus the cable swings left as in part (a) . 

Note in part (c) the cable will hang vertically with no deflection since the acceleration is ZERO.
28.  See class notes on #29 we worked on the board. Let a be the magnitude of the common acceleration of the boxes.

Isolate the 373-N mass: Let this mass be labeled M.
Resolve forces along the vertical and horizontal axes.
sum of forces in x direction = M*a = pos - neg = T  - 0 = T
M*a = T ( equation I)
Sum force forces in y direction = = 0 = N - Mg (equation II).  This equation is not needed since there is no friction. 
Isolate the 22-kg mass: Let this mass be labeled m.
Resolve forces along the vertical  axis. Let down be the positive y direction
sum of forces in y direction = m*a = pos - neg = mg - T

Solve (I) and (III) for a and for T.
(c) Use equation 2.13. 

39. (a) sum of forces in x direction = pos = neg = 0 since the velocity is constant as the box moves along the horizontal.
0 = F - fk, where F is the magnitude of  person's push force and fk is the friction force magnitude. Note:  = uk*N, where uk is the coefficient of kinetic friction and N is the normal force magnitude.  

sum of the forces in the y direction = pos - neg = 0 = N - mg; thus N = mg. 
Find F.
(b)  Yo, in this case the only horizontal force acting on the slowing down box is kinetic friction and the ACCELERATION IS NOT ZERO. 

From part (a), N = mg as before, but now we have a different situation in the x-direction:

sum of forces in x direction = m*ax = pos - neg = 0 - fk,  since there is no positively directed force !

Find  acceleration  ax  , which will be negative.   
46. (a) Let up the incline be the positive x direction.

Going up the incline, sum of forces in the x direction = m* ax = pos - neg = 0 - fk - mg*sin 40 since all the  x-directed -force components  point down incline; there is no "pos" force along x axis ! Note:  = uk*N, where uk is the coefficient of kinetic friction and N is the normal force magnitude.  
Note: sum of forces in y direction = pos - neg = N - mg*cos 40; thus N = mg*cos40.    
Find the acceleration.
(b) Let down the incline be the positive x direction. Going down the incline, sum of forces in the x direction =
 m* ax = pos - neg =mg*sin 40  - fk  since  the  x-directed-weight  component points down incline and friction points in the opposite direction (up). Note:  = uk*N, where uk is the coefficient of kinetic friction and N is the normal force magnitude.  
Note: sum of forces in y direction = pos - neg = N - mg*cos 40; thus N = mg*cos40.    
Find the acceleration. NOTE: IF we chose to keep up the incline as positive, the acceleration would have the same magnitude but opposite sign.  
52.  See example 5.13: Let down be positive;  sum of forces  in y direction = m*a = mg - FD,   where a is the magnitude of the downward acceleration and FD   is the drag force magnitude.  In the formula for FD set  vy = vT / 2 , where the terminal velocity vT is derived in the example.