Solve Force Equilibrium Problem: Angle Between 1N & 2N

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The discussion revolves around solving a force equilibrium problem involving three forces: 1N, 2N, and √3N. The goal is to determine the angle between the 1N and 2N forces. Participants suggest visualizing the forces as vectors and constructing a right triangle, with the hypotenuse representing the resultant force. The equilibrium condition is expressed as the sum of the forces equaling zero. Understanding vector addition is crucial for solving this problem effectively.
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there are three forces (1N, 2N, 3^(1/2)N) working on an object and have reached an equilibrium. what is the angle formed between 1N and 2N? can someone please help me out? I've been working on this problem for a while not, but can't figure it out.
 
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Draw a triangle starting from the origin and since its in equilibrium it should end at the origin as well. Think of it in terms of vector addition.

\vec{F1} + \vec{F2} + \vec{F3} = 0

Define F1 as the hypotenuse and simply use the other two to construct a right angle triangle. Be sure that the hypotenuse you've defined contains the correct properties.
 
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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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