What is the third force needed for equilibrium?

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To achieve equilibrium with a force of 6N in the negative i direction and 3N in the positive j direction, a third force must be calculated. The magnitude of this force is determined using the Pythagorean theorem, yielding a result of the square root of 6^2 + 3^2. The direction of the third force can be found using trigonometric functions, specifically tangent, to determine the angle relative to the horizontal. The force must counteract the existing forces, meaning it should point opposite to the resultant direction of the other two forces. Properly visualizing the forces through a free body diagram can clarify the necessary direction for equilibrium.
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Homework Statement



If I have a force of 6N in the negative i hat direction and a force of 3N in the positive j hat direction , what third force will cause the object to be in equilibrium?

Homework Equations





The Attempt at a Solution



I got the answer to be the square root of 6^2+3^2 to make it like a triangle. But I am not sure if this is right and in what direction it should be pointed.
 
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for the direction you can use trig .. let the angle of the force required from horizontal be x..then how can you find tan x in terms of known data?

Your magnitude of force is correct so you've got to worry only about direction; :smile:
 
The possible answers says it can be clockwise or counter clock wise I can't figure out which one. I used tan and got my degrees. But idk if the vector is pointing toward the y-axis force or away from it.
 
try drawing the FBD and see how the forces act.
 
Ok I did, so the object is moving to the left and up. So the force I need will be opposite of that? And it will be the degrees below the positive x-axis instead of to the right of the negative y axis, because its moving faster to the left than up?
 
yes ... just like 2 guys push a car from the front and other two from back! :biggrin:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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