What are the forces Fu and Fv given the Resultant(R)?

[Bradracer18]

Homework Statement

Two forces applied to a bracket, Fu and Fv. The Resultant(R) of the two forces has a magnitude of 725 lb and a direction shown in my pic. Determine the magnitude of the two forces(Fu and Fv).

Homework Equations

So far, I've just used law of Sines, but I'm sure law of Cosines...pythagorean theorm...etc.

The Attempt at a Solution

Ok basically, I've got the answers in the back of the book...but I'd like to know how to get the answers. To tell you the truth, a picture(or a good word explanation) would be very helpful...when applying the degrees used(so I know what "corner" they are coming from.

I drew my picture(see the attachment)...and I've found many of the angles, using the law of sines/pythagorean.

Then, for Fv I did this(and it seems to be the right answer too...I think I understand what I did, kinda...but I guess not good enough to do Fu).

Here's what I did:

Fv = (sin(75.9638) * 725lb) / sin(132.274) = 950.559lb


Thanks in advance for the help!

I've done a few more on my own, thanks to the last help on here I had...but, I still need more instruction I guess. This one confuses me, because you are given the resultant instead of the forces, and one of the forces points in a different direction.
BradView attachment vector1.bmp

 

[Bradracer18]

If you can't see the pic...I'll try to explain it in words.

 

[Bradracer18]

Can anyone help...or do I need to explain more?

 

[cristo]

I cant' see your picture; I tried saving it and zooming, but I can't see the values of angles, or your working.

However, I get the general idea. How many of the angles do you know? Do you know the angles of each of the Fu and Fv to the horizontal? Do you know the angle of the resultant to the horizontal?

Also, I'm guessing that the forces are applied at the same point on the bracket, and that the bracket is not moving. Am I correct?

 

[radou]

You might want to post a better picture, anything can hardly be seen from this one. Btw, I think I saw this same question a week ago, try to search for the topic.

Edit: oops, Cristo reacted faster.

 

[Bradracer18]

I tried the search...with no luck(maybe I'm not good at it either, not sure what to search).

Anyways...I'll try explaining the problem...but you might have to refer back to my origional picture...to help understand.Fv is the first force, being 14.0362 deg from the x axis. It is also labeled with a triangle which is (in x,y form)...(4,1). This force goes towards the orgin.

Fy the second force is 63.4349 degrees from the x axis. It was labeled with the triangle (1,2). This force goes "up" towards the pos Y axis.

The Resultant(R) is shown as 33.6901 deg from the NEG x axis. Labeled with the triangle (3,2). The resultant goes off in the direction towards the neg x axis.

Hope this helps explain it better. The forces going in different directions is confusing to me. Thanks!