Two force vectors, F1 = (-3.0 N) x-hat + (-2.5 N) y-hat and F2 = (-6.0 N) x-hat + (3.5 N) y-hat are applied to a particle. What third force F3 would make the net force on the particle zero?

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Two force vectors, F1 = (-3.0 N) x-hat + (-2.5 N) y-hat and F2 = (-6.0 N) x-hat + (3.5 N) y-hat are applied to a particle. What third force F3 would make the net force on the particle zero?

Calculus1
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Are you familiar with the bra -ket vector notation F1 = = < -3 , -2.5> ?
Nope. :0
ok, it is just that, the x and y components

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This is technically for a physics class but no one in the physics section on here is helping me
You want to find a third force F3, so that the object does not accelerate,
Net force is Zero \[F _{NET} = F _{1}+F _{2}+F _{3}\]
With vector addition, you can simply add the components of each dimension together...
Instead of using the unit vector multiplier x-hat and y hat, it is simpler with where a is multiplied by x-hat and b is by y-hat
\[<0 , 0> = <-3.0 ,~ -2.5> + <-6.0,~ 3.5> + \]
\[F _{3}= a* xroof + b*yroof = \]
So would it be <9,-1>?
F3 =
With that vector equation, you can solve for the a and b components of F3 by equating the x direction and the y direction values... 0 = -3.0 +(-6.0) + a [N]
Total force in the x direction is 0 or -3 - 6 + a, newtons, same for Y, then you have your F3 force
The sum of the components in the X-direction is zero The sum of the components in the Y-direction is zero (all the xroofs have to equal zero) (all the yroofs) have to equal zero) solve for the components of the third force
So is that a yes or a no to being <9,-1> For F3?
That looks good
Did you do a section on vectors before this? if i remember i think there was in physics 1, maybe not

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