If there is two cables holding up a bar
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If there is two cables holding up a bar

[From: ] [author: ] [Date: 12-03-13] [Hit: ]
15m. from the weaker cable end.......
Let's say that there is a bar with weight 314N and has length 1.97m. The bar is suspended by two cables that can support a max tension of 537N and the other one can support a max tension of 428N. Both the cable are at the ends of the bars.
If I am to place a small mass on the bar, what is the maximum weight I can place on the bar and where?

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Maximum total weight the cables can bear = (537 + 428) = 965N.
Maximum extra weight the system can bear = (965 - 314) = 651N.
The less strong cable is already bearing (314/2) = 157N., so maximum increase on it can only be (428 - 157) = 271N.
The rest must be born by the stronger cable, = (651 - 271) = 380N.
1.97/((380/271)+1) = 0.82m. from the stronger cable end, and therefore (1.97 - 0.82) = 1.15m. from the weaker cable end.
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