SF= (0+f_k-wsin60)i + (N-wcos60)j = m(-ai +0j)
1.    f_k-wsin60 = m(-a)
2.    N-wcos60 = 0  so   N = mgcos60      and f_k =N*m_k= mgm_scos60
3.    put f_k = into 1 and solve for a
if you have positive a: assumption was correct
 
 
case b)  (0,N),  (-f_k,0),  (-wsin60, -wcos60)       a = -ai +0j
SF= (0 - f_k-wsin60)i + (N-wcos60)j = m(-ai +0j)
1.    -f_k-wsin60 = m(-a)
2.    N-wcos60 = 0  so   N = mgcos60      and f_k =N*m_k= mgm_kcos60
3.    put f_k = into 1 and solve for a
if you have positive a: assumption was correct

33
m_A = 10kg,  incline is at angle Q = 30^o to horizontal,  m_k= 0.20
What is the mass of block B if block A slides down the incline at constant speed.
in coordinate system shown:  four forces on first FBD are: (0,N), (-T,0), (-f_k,0), (w_Asin30,-w_Acos30)  and acceleration is a = (0,0)
So Newton's Law is SF = (0-T-f_k+w_Asin30)i + (N+0+0-w_Acos30)j =  m_A(0i+0j)
two forces on second FBD are +Ti and -w_Bi(second component is always 0)
So Newton's Law is SF = T - w_B = m_B * 0
There are 3 equations from the above:
1.   0-T-f_k+w_Asin30 = 0   and we know that f_k= Nm_k  so  0-T- Nm_k+w_Asin30 = 0
2.   N+0+0-w_Acos30 = 0
3.    T - w_B = 0
There are three unknown variables T, w_B, and N and three equations.  Find T from 3), N from 2) plug into 1 and solve for  w_B.     w_B = m_Bg
 
37

Coefficient of static friction between the above two blocks can be calculated.
 F_smax = Nm_s  where N is Normal force = 40N and F_smax = 12N so m_s= 0.3
FBDs show the forces acting on both blocks A and B
From Newton's Law:
  1.     SF = (F_A-f_s)i + (N_A-50-40)j = m_A(ai + 0j)  because we can expect acceleration to be only horizontal.
   2.    SF = f_si + (N_B-40)j = m_B(ai + 0j)
or we can write Newton's Law for the whole system:
3.    SF = F_Ai + (N-90)j = (m_A+ m_B)(ai + 0j)

From equation 2:  to have maximum acceleration a you must use maximum friction force fs so
f_smax = m_Ba_max= 12N we can calculate maximum acceleration a_max.  a_max= 3 m/s²
Then it is reasonable to use equation 3)  to find F_Amax.   F_A = (m_A+ m_B) a_max = 9 * 3 = 27N
You don't have to use equation 1 at all.
 
57

From the above FBD for mass M and y-axis pointing up:   +T - Mg = 0  so T = Mg
From the second FBD (in  r  direction)    +T = m*a = m v²/_r
 so   Mg = mv²/_r
and v² = Mgr/m