Diagram is attached.
The mast OA is supported by 3 cables. If cable AB is subjected to a tension of 500N, determine the tension in cables AC and AD and the vertical force F which the mast exerts along its axis on the collar at A.
a = 6m
b = 3m
c = 6m
d = 3m
e = 2m
f = 1.5m
g = 2m
For this assignment, I'm pretty sure the prof just wants us to consider sum of Fx, Fy and Fz, and to take no moments.
The Attempt at a Solution
I have a complete solution, but apparently it's not the right answer according to the teacher assistant (I don't know what the right answer is).
so vector AB = 2i + 3j - 6k
AC = -1.5i + 2j - 6k
AD = -3i -6j - 6k
and I get that vector Fab = (142.85i + 214.285j - 428.55k)N
Fac = (-0.2308(Fac)i + 0.30769(Fac)j - 0.923076(Fac)k)N (where the Fac's on the right side of the equation represent the magnitude of the force *vector* Fac)
Fad = (-0.333(Fad)i - 0.666(Fad)j - 0.666(Fad)k)N
and of course, F = (Fk)N
where i, j and k are all unit vectors in the direction of the x, y and z axis, respectively.
From that, I have that:
sum of Fx = 0
Equation 1: 142.85i - 0.2308(Fac)i - 0.333(Fad)i = 0
sum of Fy = 0
Equation 2: 214.285j + 0.30769(Fac)j - 0.666(Fad)j = 0
sum of Fz = 0
Equation 3: -428.55k - 0.923076(Fac)k - 0.666(Fad)k + F(k)= 0
That's 3 equations in 3 unknowns. I then rearranged equation 1 to get Fac in terms of Fad. I plugged this Fac equality into equation 2, to get Fad alone.
My answer for Fad is 910.2 N. Apparently that's wrong. Plugging that in elsewhere, I get that Fac = -694.2 N and that F = 393.9 N, which are both wrong according to the marker.
Any suggestions/corrections? Did I not approach the question properly or assume too much?
17.9 KB Views: 1,781