Maximum weight a bar and cables can hold before breaking

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Alison A. said:
So I am finding all the forces from the moments first then applying those to the force equations?
No, you use the one moment equation I indicated,

ΣMz = FAB(12) - FAC(18.46) = 0 to find FAB in terms of FAC

Then, you can substitute FAB back into the force equations to find the other forces in terms of W.
 
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SteamKing said:
No, you use the one moment equation I indicated,

ΣMz = FAB(12) - FAC(18.46) = 0 to find FAB in terms of FAC

Then, you can substitute FAB back into the force equations to find the other forces in terms of W.
Alright I found FAB to be (4615/5539)W and FAC to be (3000/5539)W
 
Alison A. said:
Alright I found FAB to be (4615/5539)W and FAC to be (3000/5539)W

And from there force equations I found FOA to be -0.6983W
 
WOOOOOOOOOOOOOO:oldsurprised: I got the right answer. Wow that was a lot of work, I think I've gone through about 10 pages of paper. :bow:Thank you so much for sticking with me even when it seemed like I couldn't grasp the most simple concepts . Could you help me find the last part to my other problem you've been answering? That is my last problem... then I'm finally done.