Calculate the net force on the object

Homework Statement

Two forces are acting on an object and make a 130 degree angle between the two. One force is horizontal to the x axis and is 13N. The other is 9 N.
Calculate the net force on the object

Homework Equations

Not really equations, but I broke the one force into its x and y components. Since the 13 Newton force is just along the x axis with no y, hats the only force in that direction. So I guess trig functions are my equations

The Attempt at a Solution

Sin40= -x/9
X= -5.79
Cos40= y/9
Y= 6.89

Adding the components gives the net force for that portion:
1.1 N

And since there is no y component in the other direction, the force is just 13 N.

Here's where I'm stuck, the stronger force has no y component and is going directly right (free body diagram). The other component is headed up and left in the somewhat opposite direction, but is weaker.
Here's my thinking, to get the net force I have to add them together, which gives 14.1 N. But if the stronger force is to the right, would I subtract the 1.1 N? To get 11.9 N?
[/B]

SammyS
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Homework Helper
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Homework Statement

Two forces are acting on an object and make a 130 degree angle between the two. One force is horizontal to the x axis and is 13N. The other is 9 N.
Calculate the net force on the object

Homework Equations

Not really equations, but I broke the one force into its x and y components. Since the 13 Newton force is just along the x axis with no y, hats the only force in that direction. So I guess trig functions are my equations

The Attempt at a Solution

Sin40= -x/9
X= -5.79[/B] ##\ \ \ \ \ ## This tells you the x component of the 9 N force .
Cos40= y/9
Y= 6.89 ##\ \ \ \ \ ## This tells you the y component of the 9 N force .

Adding the components gives the net force for that portion:
1.1 N

And since there is no y component in the other direction, the force is just 13 N.

Here's where I'm stuck, the stronger force has no y component and is going directly right (free body diagram). The other component is headed up and left in the somewhat opposite direction, but is weaker.
Here's my thinking, to get the net force I have to add them together, which gives 14.1 N. But if the stronger force is to the right, would I subtract the 1.1 N? To get 11.9 N?
Adding the ##x##-component and ##y##-component as numbers usually makes no sense.