# Thread: Magnitude of the reultant force Help

1. ## Magnitude of the reultant force Help

300N is on the x-axis (-300,0)
and 200N (100,173) is in a 60 degree angle from the x-axis
How do I find the magnitude of the resultant force and the angle it makes w/ the positive x-axis?

2. Originally Posted by khuezy
300N is on the x-axis (-300,0)
and 200N (100,173) is in a 60 degree angle from the x-axis
How do I find the magnitude of the resultant force and the angle it makes w/ the positive x-axis?

Let $\displaystyle \bold F_1=\left<-300,~0\right>~N$ and $\displaystyle \bold F_2=\left<100,~173\right>~N$

The resultant force vector $\displaystyle \bold R=\bold F_1+\bold F_2\implies \bold R=\left<-300+100,~0+173\right>~N\implies \color{red}\boxed{\bold R=\left<-200,~173\right>~N}$

The magnitude of the resultant force would be $\displaystyle ||\bold R||=\sqrt{(-200)^2+(173)^2}\implies \color{red}\boxed{||\bold R||\approx 264.44~N}$

I leave it for you to find the angle, given that $\displaystyle \vartheta=\tan^{-1}\left(\frac{R_y}{R_x}\right)$

Does this make sense?

--Chris

3. ## ic

thanks a lot
the angle is arctan(173/-200) + 180 = 139.1degree?

4. Originally Posted by khuezy
thanks a lot
the angle is arctan(173/-200) + 180 = 139.1degree?
Yes!

--Chris

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# the angle the resultant makes with the positive x axis

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