finding magnitude of force P if resultant of 3 force is vertical.

Question

adunb8 · Answer

i have no idea how the answer became to be that.. anyone help?

anonymous · Answer

First of all do you understand the question?

adunb8 · Answer

the question itself is confusing =/

anonymous · Answer

attachment is something different and the question is different .??

adunb8 · Answer

oh wait i put wrong attachment sorry

anonymous · Answer

LOL

adunb8 · Answer

attachment

adunb8 · Answer

@andras i got the right one

anonymous · Answer

You even have the solution. 
What is the problem?

adunb8 · Answer

i dont know how the answer came to be that i need like steps...

anonymous · Answer

First he calculates all the forces in the x direction. That is $$R _{x}$$
It has 3 parts as there are 3 forces. 
P is easy as it has only x direction, so its x component is the same as itself. 
For the other 2 vectors you need to understand the magical world of trigonometry, do you need explanation for that?

adunb8 · Answer

it would be great if i can get that magical world trig.

anonymous · Answer

OK lets try, it is a pit hard to draw nicely here

anonymous · Answer

|dw:1389608982847:dw|

adunb8 · Answer

cool i understood this part!

anonymous · Answer

That x is the x component of that 200lb force. How can we find x? With sin function http://www.mathsisfun.com/algebra/trigonometry.html There you can read a bit more about sin/cos/tan in a triangle. Can you answer me this: In that drawing what would sin40 equal to?

adunb8 · Answer

would it be 200lbsin40 = Fx (x component?)

anonymous · Answer

Yes that is right, it comes from this
$$\sin 40 =\frac{ x }{ 200 }$$

adunb8 · Answer

i see! how would i draw 400lb part? its confusing.

anonymous · Answer

So you found the x component of one of the vectors.

anonymous · Answer

|dw:1389609535932:dw|

anonymous · Answer

Just keep in mind that in this case I labeled the x component of that vector as y. AND it is pointing to the negative direction

adunb8 · Answer

how did you know to draw it like that?? that is the most confusing part sorry any good tricks in understanding to draw those confusing drawings?

anonymous · Answer

I draw the vector first, than the angle.
When you have these 2 you need to find another angle that is 90 degrees so that you can use trigonometry.
$$\cos 40=\frac{ y }{ 400 }$$

adunb8 · Answer

oh i see but on the answer it has 200sin40 that we found and - 400 cos 40 . is 400cos40 x component or is it y component because it has $$\sum_{}^{}Fx$$