# Homework Help: Mechanics Part II

1. Apr 20, 2006

### 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 dont 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 dont really understand how you get....

THANKS :)

Last edited: Apr 20, 2006
2. Apr 20, 2006

### Staff: Mentor

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).

3. Apr 20, 2006

### 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 isnt one...

Yeah?

4. Apr 20, 2006

### Staff: Mentor

Of course.

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

5. Apr 20, 2006

### turnstile

last question..

are you god? :D ....

Thank you so much.