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Mathematics 9 Online
OpenStudy (anonymous):

Please help me out. :)

OpenStudy (anonymous):

OpenStudy (mathmate):

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OpenStudy (mathmate):

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.

ganeshie8 (ganeshie8):

still looking for help ?

OpenStudy (anonymous):

i dont know

OpenStudy (anonymous):

yes? :)

ganeshie8 (ganeshie8):

Since all the three forces are in same direction, simply adding their magnitudes gives you the resultant force : \[\large 100+F+300 = 600\] solve \(F\)

rvc (rvc):

now its easy

OpenStudy (anonymous):

F = 200. :)

ganeshie8 (ganeshie8):

yes! plug that in the second equation given by @mathmate and solve x : \[\large 100*45+F*(45-x)-300*25=0\]

OpenStudy (anonymous):

x = 30. :)

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