The 1-kg collar C is free to slide along the smooth shaft AB. Determine the accleration of collar C if collar A is subjected to an accleration of 1m/s2 to the left. Consider the shaft to move in the vertical plane.

1) 3.5 m/s2
2) 9.6 m/s2
3) 10.8 m/s2
4) 7.08 m/s2

Equation of motion:
S
Fx' = max'   1(9.81)sin45o = aC/A + 1(-1cos45o)
aC/A = 4.17 m/s2
Relative accleration:
aC = aA + aC/A
aC = -i + (4.17cos45o)i - 4.17sin45oj
aC = (1.95i - 2.95j) m/s2
Thus, the magnitude of the accleration aC is

So, the magnitude of accleration is aC = 3.53 m/s2