Statics: Magnitude of vectors HELP

[mrpburke]

Statics: Magnitude of vectors HELP!

Could someone just tell me how to start this problem. I can't seem to think of anything right now.

 

[mrpburke]

Anybody?

I found the answer...68.2kN but I have no idea how to get that.

 

[jtbell]

Call the vector sum of the three forces {\vec F}_{net}, so you have the vector equation

{\vec F}_{net} = {\vec F}_A + {\vec F}_B + {\vec F}_C

Write down the corresponding x- and y- component equations, using the magnitudes and angles of the forces. For example, the x-component of {\vec F}_A is |{\vec F}_A} | \cos \theta_A. You're given that the three forces have the same magnitude, so use |{\vec F}_A} | for all of them, but of course the angles are different. And {\vec F}_{net} has its own magnitude (which you're given) and angle (which is unknown).

Once you have those two equations, you should have enough information to solve for |{\vec F}_A} |.

 

[FluxCapacitator]

As with all vector problems, I'd start by finding the components. To find the components, use basic geometry to find the angles.

Then you could write one equation saying 200= the sum of all the components
and a few equations equating the magnitudes.

If that didn't help you, say so, and I'll try to explain it more explicitly.

 

[jtbell]

Write down the corresponding x- and y- component equations, using the magnitudes and angles of the forces.

I'd better point out that whenever you have a problem involving magnitudes and angles of forces (or any other vectors, for that matter), this is usually a good way to start.

 

[mrpburke]

Could someone work that out a bit more. I'm just not seeing it:(

 

[FredGarvin]

Each force, A B and C can be broken down into their x and y components...You are given the triangle for each vector. Can you get that far to describe each of the three vectos?

 

[mrpburke]

I have A, B, and C broken down to x and y components. Now what?

 

[mrpburke]

I keep getting 47.09kN for the magnitude of Fa. I think the answer in the book is wrong.

 

[jtbell]

Well, I don't think anybody here has psychic "remote vision," so we can't tell you where your mistake is unless you show us your work!

My calculation agrees with the book's answer of 68.2 kN, by the way.

 

[mrpburke]

How are you getting that answer?

 

[jtbell]

I did it the way I described earlier. The x- and y- equations gave me two equations in two unknowns: the magnitude of the individual forces, | {\vec F}_A |, and the direction (angle) of the net force.

 

[mrpburke]

I'm workin on it.

 

[mrpburke]

So set up into 2 equations is it...

200cos@ = |Fa|1.88
200sin@ = |Fa|2.45
?

when that's worked out it comes to 64.7 which isn't quite right.

 

[mrpburke]

I got 67.4kN. I'll scan it and post up the work tomorrow morning so someone can tell me what I did wrong.

 

[mrpburke]

Here's what I have. Somebody PLEASE show me what I'm doing wrong!

 

[jtbell]

When calculating components, always measure the angle from the x-axis (in which case the x-component uses the cosine and the y-component uses the sine) or from the y-axis (in which case you switch the sine and cosine). It looks like your angles are the interior angles of those triangles in the diagram, which is OK for one of them but not for the other two.

 

[mrpburke]

Holy crap! I could of swore I've done this before but I think I finally got it right.

Lemme work it out neatly and post it up!

 

[jtbell]

Yeah, just fixing those two angles should do it.

 

[mrpburke]

I think that's going to be it!

Thanks!

 

[mrpburke]

Yeah, just fixing those two angles should do it.

I've been all over that just never did it quite like that. I don't know why I didnt even think of that.


gahh...I hate it when it's something simple like that