AAaaaargh I don't understand Aarggghh

[winbacker]

AAaaaargh! I don't understand! Aarggghh!

Homework Statement

OK so I'm trying to understand the solution to a mechanics of materials problem that I'm working on. I understand the entire solution EXCEPT for 3 very tiny and BASIC lines of math which draw upon some BASIC trig property of which I must be unaware.

Homework Equations

Attached (argh!.doc) is the segment of the problem that is driving me crazy.
You will notice that under the picture of the problem are the words "Use member DEF as a free Body".

You then see member DEF (in all its glory) and the forces acting on it.

Now, the hideous lines that are driving me nuts are directly to the right of this diagram.
They are the ones that say

"Sigma F (unreadable symbol) = 0;" (sum of forces in some direction is zero)
(3/5)Dy -(4/5)Dx = 0;
Dy - 4/3Dx = 1200 lb.You can also see on the diagram that he somehow associates a 3-4-5 triangle to the reaction forces Dx and Dy. WTF?! I've taken statics before and I never saw anything like this.

The Attempt at a Solution

Pleeeeease oh pleeease can one of you just explain this little section to me before I defenestrate myself (figuratively). Thanks.
AAAARgh!

 

[PhanthomJay]

Aarggghh, Matey! It is a bit confusing, that FBD. I'd look at it this way: Since the lines of action of the forces F_BE and F_CF both lie along the hypotenuse of a 3-4-5 right triangle, then by summing forces = 0 along an axis parallel to that hypotenuse, then the line of action of the reaction at D must also lie along that axis. Hence, D_y = 4/3D_x.

 

[djeitnstine]

Have you tried investigating the math? one of the angles on a 3-4-5 trangle is ~36.8 degrees. Take the cosine of it and you get ~0.8 or 4/5.

Similar argument for sine of the angle which is 0.6 ~3/5.

Always try to check the math and what it means. And remember "SohCahToa" =)

 

[Integral]

Have you tried investigating the math? one of the angles on a 3-4-5 trangle is ~36.8 degrees. Take the cosine of it and you get ~0.8 or 4/5.

Similar argument for sine of the angle which is 0.6 ~3/5.

Always try to check the math and what it means. And remember "SohCahToa" =)

why not just use the definition of sin and cosin.

$$sin x = \frac O H = \frac 3 5$$

$$cosin x = \frac A H = \frac 4 5$$

 

[winbacker]

I think i see what you're saying PhanthomJay but why then did he orient the triangle at D differently than the other triangle?If the reaction at D was parallel to the other 2 forces, shouldn't both triangles be oriented the same way? So that their hypotenuses are parallel??

 

[PhanthomJay]

I think i see what you're saying PhanthomJay but why then did he orient the triangle at D differently than the other triangle?


If the reaction at D was parallel to the other 2 forces, shouldn't both triangles be oriented the same way? So that their hypotenuses are parallel??

Yes, I agree. I don't understand his orientation, either. EDIT: O,I see what he did, he took an axis pependicular to the chords and summed the components of the Dy and Dx forces along that axis equal to 0, since the chord forces have no components along that chosen axis. I find it simpler to note that since the resultant reaction at D must be along the axis of the chords, then Dx and Dy are trigonometricaly related in acordance with the 3-4-5 triangle geometry. It's a matter of choice as to which method is easier.