How Does Tension in a Frictionless Pulley System Behave with Balanced Forces?

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In a frictionless pulley system with a 2.0 kg mass on a horizontal surface and a 1.5 kg mass hanging, the tension in the strings can be determined by analyzing the forces. The gravitational force acting on the 1.5 kg mass equals the tension in the string connected to it, leading to the conclusion that the tension in both strings is equal to the weight of the suspended mass. Since there is no acceleration in the system, the tension remains constant across both strings. The additional horizontal force applied to the 2.0 kg mass does not affect the tension calculation as long as it maintains equilibrium. This confirms that the tension in the strings is indeed equal to the weight of the hanging mass.
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1. A 2.0 kg mass sits on a frictionless horizontal surface. The mass is attached by a weightless string running over a frictionless pulley attached to a 1.5 kg mass. An additional string exerts a force on the block on the mass on the surface so that no motion results. Find the tension in each of the strings.



2.



3. I just wonder if this problem is as easy as it seems. Fg on mass 2 is equal to the tension force on mass 2 therefore the tension in string 2 is equal to the weight of the object. If that is the case and there is no acceleration to either mass in the problem the tension force for both strings is equal to the weight of the suspended mass.
 
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Sounds right as long as the additional string is pulling horizontally.
 
yea diagram looks like

---------block---------
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|
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block
 
Good. Then I think your reasoning is correct! :smile:
 
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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