Finding the magnitude necessary to balance two forces

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To balance two forces F1 (10.0N) and F2 (4.0N), the third force F3 must counteract their net effect. The smallest magnitude of F3 required depends on the direction of F1 and F2. When both forces act in the same direction, F3 must equal the sum of F1 and F2, totaling 14.0N. If F1 and F2 are at angles to each other, a graphical approach can help determine the minimum F3 needed by analyzing vector sums. Understanding the equilibrium condition is crucial for solving this problem effectively.
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Homework Statement


Two forces are acting on an object. The magnitudes of them are F1=10.0N, F2=4.0N. If a third force F3 is applied on the object, what is the smallest magnitude of F3 needed to balance the first two forces?


Homework Equations


N/A


The Attempt at a Solution


I've done a lot of internet searching for this question and had no luck. With that being said, I am really starting from ground zero here (hence no equation). I'm guessing that this problem has to do with FNET and is probably easier than it looks? If someone could give me an equation to use I would probably be in good shape. Thank you!
 
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Yes, it's very easy and you guessed the right equation. The word "balanced" can only mean the object is in equilibrium.
 
Try a graphical approach. For example choose arbitrary values, say F1 = 3 and F2 = 4 and draw a few vector sum diagrams to scale with F1 and F2 pointing in different directions such as...

a) F2 and F1 pointing in the same direction
b) F2 at 90 degrees to F1
c) F2 at 45 degrees to F1
d) F2 at 135 degrees to F1
c) F2 in the opposite direction to F1

In which direction must they point to give the smallest Fnet that F3 must oppose?
 
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Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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