Chapter 7 : Do listed problems.  
Simulations
Quiz 7
CQ 13 14 15 Problems 10 12 23 36 38 39
40 41 42
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CQ13, 14, 15: See class notes and ICQ's
10. Important problems dealing with Newton's third law and contact forces as action reaction pairs.
Let M1, M2 and M3 be the three masses of 1 kg, 2 kg and 3 kg respectively.  Isolate each to get

M1a = F - FA
M2 a = FA - FB
M3a = FB  
FA and FB are the contact forces.  Just add the equations to cancel them and solve for a. Then go back and find the two contact forces.  
12. See ICQ4-9-08 and ICQ4-11-08 which asks you to find the tension T in the rope: MAg = T since the system is at rest. MA = 60 kg.  From that,  find the normal force of magnitude N on MB,  which is sitting on the ground.
23.
MBa = T2 - fkB  -  T1
MAa = T1 - fkA .

In each case fk = µN .

Find a and T2. You are given T1  .
36.  See ICQ4-2 or ICQ4-7, depending on the section which derives the accleration a for fig. 7.14 when the blocks are not equal in mass.
MAa = T - MAg.
MBa = MBg  -  T.
Solve for acceleration a symbolically.
Find acceleration a numerically from Chapter 2 kinematics. Then use the symbolic formula for acceleration a to find mass A. 
38. Yo, you have three equations and three unknowns ! Label the masses from left to right as A, B and C.

MAa = TA - MAg
MBa = TB - TA  -  µN
MCa = MCg  -  TB .

Add equations to find the common acceleration a.  Then go back and substitute into each to find the two tensions.

39.
(a) If the system does not move then a = 0.

(i) First, we assume the friction force on mass m points down incline.
For mass m:
ma = 0 = T - mgsin20 - fsmax   (I)
For mass M = 2.0 kg,
Ma = 0 = Mg - T.
Thus T = Mg (II)
Sub. into (I)
0 = Mg - mgsin30 - µsN,  where N = mgcos20.  Find m.

(ii) Second,   try the problem when the friction force on m points up the incline:
for mass m,
ma = 0 = mgsin20 - T - fsmax   (I)
For mass M,
Ma = 0 = T- Mg.
Thus T = Mg (II)
Sub. into (I)
ma = 0 = mgsin20 - Mg - fsmax   ,  where N = mgcos20.  Find m. 

From the above we have two choices of m.  Case (ii) gives the largest mass m.

(b)
For case (i), the mass m will move up the incline once "nudged". Thus,

ma = T - mgsin20 - fk  , where fk  =  µkN and  N = mgcos20.
Ma   = Mg - T

Find a and T.

For case (ii),  the mass m will move down incline once, "nudged".  Thus
ma  = mgsin20 - T - fk  , where fk  =  µkN and  N = mgcos20.
Ma = T- Mg.

Find a and T. 

ERROR LOG #1 : Update on #39. As discovered in class, case (ii) produces a negative mass m, which is unphysical.  Thus, I told you to re-do the problem with ANGLE  45 DEGREES. This would be turned in as Q7 '. For this angle,  you would find the mass in both cases (i) and (ii) above.  Case (i) would produce m = mMIN and case (ii) would produce m = mMAX.   These results give a RANGE of values  of  m in which the system would be at rest. In other words, a = 0 when,
  mMIN  <  m <  mMAX .
More hints later !!!