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

1. chrisplusian

2. 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

3. Michele_Laino

here we have to apply this vector equation: ${\mathbf{M = OP \times F}}$ where $$P$$ is the application point of force $$F$$

4. 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.

5. 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 $$F=5$$

6. Michele_Laino

as we can see, only the $$z-$$ component is different from zero

7. 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 $$d$$.

8. 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

9. ganeshie8

|dw:1443888969129:dw|

10. ganeshie8

|dw:1443889094948:dw|

11. ganeshie8

it is right triangle, with one angle = 75, hypotenuse = 3 you can find the opposite side, $$d$$. no big deal

12. chrisplusian

Ok I see it now, I can't beleive that was not obvious. Thank you.

13. ganeshie8

np :)