Mechanics Part II: P, Q, and F Forces

[turnstile]

Mechanics Part II :)

Hi.
Sorry, i got stuck again..

well...

Two Forces, P and Q, act on a particle. The force P has a magnitude 5N and the force Q has magnitude 3N. The angle between the directions of P and Q is 40 degrees. The RESULTANT of P and Q is F

(a) Find, the magnitude of F

(b) Find, in degrees to one decimal place the angle between the directions of F and P.

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Now, my stab at part (a) was in the form
F^2 = Fx^2 + Fy^2
- = (5cos 40)^2 + (3sin 40)^2
However, it gives me a different answer to the mark scheme...
who have... (5+ 3cos40)^2 + (3 in 40)^2 ...

Why the 5 +3 cos40 ... i don't really get it...

Any help would be greatly appreciated.

(b) similarly for this.. my version would be ; tan(theta)= Fy / Fx ..but then for Fx... there would be 5+3cos 40... which i don't really understand how you get...

THANKS :)

 

[Doc Al]

Take the direction of P to be along the x-direction. Thus the x-component of F will be Px + Qx. But Px = 5 N and Qx = 3 cos(40) N, so Fx = 5 + 3 cos(40).

 

[turnstile]

ah... so you have to take the x component for force Q as well... correct?

and there will be no y component for force P.. since well.. there isn't one...

Yeah?

 

[Doc Al]

ah... so you have to take the x component for force Q as well... correct?

Of course.

and there will be no y component for force P.. since well.. there isn't one...

Right. Because we cleverly chose our x-axis to be parallel to P.

 

[turnstile]

last question..

are you god? :D ...

Thank you so much.