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Please help me out. :)
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|dw:1404650498528:dw|
You can find the magnitude of F by summing forces: \[\Sigma F = Resultant\] The location of F can be found by summing moments: If you take moments about C (where the resultant passes) \[\Sigma F*d =0\] This means \[100*45+F*(45-x)-300*25=0\] You can then solve for x.
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yes? :)
Since all the three forces are in same direction, simply adding their magnitudes gives you the resultant force : \[\large 100+F+300 = 600\] solve \(F\)
now its easy
F = 200. :)
yes! plug that in the second equation given by @mathmate and solve x : \[\large 100*45+F*(45-x)-300*25=0\]
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x = 30. :)
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