The door has a weight of 200lb and a center of gravity at G. Determine the constant force F that must be applied to the door to push it open 12ft to the right in 5 sec. starting from rest. Also, find the vertical reactions at the rollers A and B.
The Attempt at a Solution
200lb --> m = 6.21619 slug
F(x) = 6.21619a
F(y) = N(ab) - 6.21619(32.2) = 0, N(ab) = 200.161318 N
Moment @ G = F(2ft) - N(a)(6ft) + N(b)(6ft) = 0
N(a) + N(b) = 200.161318
Not sure where to go from here