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anonymous
 one year ago
I need help with geometry/ trig problem.... see attachment please
anonymous
 one year ago
I need help with geometry/ trig problem.... see attachment please

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not sure how they are finding "d". All they say is "through trigonometry". I didn't do well on my last test and it wasn't because I didn't know the concepts of this class (statics) it was mainly because I couldn't do things like this so I actually need an explanation if you are able. Thanks in advance

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here we have to apply this vector equation: \[{\mathbf{M = OP \times F}}\] where \(P\) is the application point of force \(F\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am actually ok with the ideas in this section.... I see what your trying to say, that the cross product of the position vector and the force vector is the moment about the point. I don't have a problem with that. In this problem they want you to use the M=Fd where d is the moment arm. I can't figure out how to dissect the diagram in a way that I can find d. So I am trying to take a step back and understand how they did it. All they said was "the moment arm d in the figure can be found from trigonometry. I am good with trig, but I don't see how they are doing it.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can compute such vector, developing the determinant of the subsequent matrix, along the first row: \[\left\ {\begin{array}{*{20}{c}} {{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\ {d\cos 30}&{d\cos 60}&0 \\ {F\cos 45}&{  F\sin 45}&0 \end{array}} \right\\] where \(d=3\) and \(F=5\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1as we can see, only the \(z\) component is different from zero

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2They are doing two things in the diagram : 1) extend the force vector 2) draw a perpendicular from point O to the force extended force vector; the length of this perpendicular is \(d\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 my thought was that they give me an angle 75 degrees and it is on a right triangle so the other angle between the 3m pole is 15 degrees. Is it valid to use the law of sines when there is a right triangle? If so I thought I could find d that way. Somehow that is not what I think they did. I tried drawing this out and using simple concepts to find something that helped but I couldn't come up with anything. If you could show me visually how they did this I would greatly appreciate it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443888969129:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443889094948:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2it is right triangle, with one angle = 75, hypotenuse = 3 you can find the opposite side, \(d\). no big deal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok I see it now, I can't beleive that was not obvious. Thank you.
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