Resultant Force #2 - Calculating ##\vec{F}_R##

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SUMMARY

The resultant force ##\vec{F}_R## is calculated to be 15.65N at an angle of 39.81° north of due west. The calculation involves three vectors: the normal force ##F_N## at 8N plus 17Ncos(45°), the southward force ##F_S## at 10N, and the westward force ##F_W## at 17Nsin(45°). The vertical component is derived by adjusting the direction of the southward force to north, resulting in a final vertical force of 10.02N. The angle is determined using the tangent function, confirming the direction of the resultant force.

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Homework Statement



I need to calculate the resultant force ##\vec{F}_R## for the following :

http://gyazo.com/5f2d7aad6aaed178135909480016e04a

I had another thread where I clarified a few things I hope will be good now.

Homework Equations





The Attempt at a Solution



So first I should get my components. There are three vectors to consider by the looks of the diagram.

##F_N = 8N + 17Ncos(45°)##
##F_S = 10N##
##F_W = 17Nsin(45°)##

Let ##F_V## denote the final vertical force. Note that I switch the direction of south to north as to add the vectors together :

##F_V = F_N - F_S = -2N + 17Ncos(45°) = 10.02N##

##∴\vec{F}_V = 10.02N [N]##

Thus ##F_R = \sqrt{(10.02N)^2 + (17Nsin(45°))^2} = 15.65N##.

To find the direction of ##F_R## consider :

##tan(θ) = \frac{10.02N}{17Nsin(45°)}##
##∴ θ = 39.81°##

##∴ \vec{F}_R = 15.65N [W 39.81° N]##

I hope all the arrows and the contexts are okay now.
 
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Looks good -- if "W 39.81° N" means 39.81° north of due west.
 
Redbelly98 said:
Looks good -- if "W 39.81° N" means 39.81° north of due west.

Yup, I opt to read from right to left when using that notation.

Thanks for looking over it :)
 

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