chrisplusian
  • chrisplusian
I need help with geometry/ trig problem.... see attachment please
Mathematics
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katieb
  • katieb
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chrisplusian
  • chrisplusian
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chrisplusian
  • chrisplusian
Not 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
  • Michele_Laino
here we have to apply this vector equation: \[{\mathbf{M = OP \times F}}\] where \(P\) is the application point of force \(F\)

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chrisplusian
  • chrisplusian
I 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
  • Michele_Laino
we 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
  • Michele_Laino
as we can see, only the \(z-\) component is different from zero
ganeshie8
  • ganeshie8
They 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 \(d\).
chrisplusian
  • chrisplusian
@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
  • ganeshie8
|dw:1443888969129:dw|
ganeshie8
  • ganeshie8
|dw:1443889094948:dw|
ganeshie8
  • ganeshie8
it is right triangle, with one angle = 75, hypotenuse = 3 you can find the opposite side, \(d\). no big deal
chrisplusian
  • chrisplusian
Ok I see it now, I can't beleive that was not obvious. Thank you.
ganeshie8
  • ganeshie8
np :)

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