turnstile
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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.
------
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 :)
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.
------
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 :)
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