Question

Body A in the figure weighs 92 N, and body B weighs 71 N. The coefficients of friction between A and the incline are μs = 0.51 and μk = 0.22. Angle θ is 39°. Let the positive direction of an x axis be up the incline. What is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

Answer 1

(1) Considering the FBD of mass b
T = w
where w is the weight of block B = 71 N
Now the FBD of block A
f_s = T - WSin39
f_s = 71 - 92Sin39 = 13.1 N
Now we know that the maximum value of friction will be
f_smax = u_s*(WCos39) = 0.51*92*Cos39 = 36.463 N
since the maximum value of friction is more than the applied value hence the block A will remain at rest.
hence the acceeleration will be = 0 m/s²
(b) Now when the block A is moving up incline
Considering the FBD of block B
w -T = ma
71 -T = (71/9.81)a
71-T = 7.237a
T = 71 -7.237a ------------(1)
Now considering the FBD of block A
T - WSin39 - f_K = Ma
T = Ma + WSin39 +u_K*WCos39
T = 9.378a + 57.897 + 15.729
T = 9.378a + 73.626 --------(2)
Now equating 1 and 2
9.378a + 73.626 = 71 -7.237a   
a = - 1.226 m/s²
-ve sign indicate that the acceleration will be in opposite direction that means the speed will be decreasing.
(c) Considering the FBD of block B
w-T= ma
T = 71 -7.237 a -----------(1)
Now considering the FBD of block A
T - WSin39 + f_K = Ma
T = Ma + WSin39 -u_K*WCos39
T = 9.378a + 57.897 - 15.729
T = 9.378a + 42.258 --------(2)
Now equating 1 and 2
9.378a + 42.258 = 71 -7.237 a
a = 13.424 m/s²