Tension in an Accelerated System

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In an accelerated system, objects attached to a rope share the same acceleration, but the tensions in the rope vary based on the mass being accelerated. The discussion highlights that without specific mass values, assuming equal masses leads to the conclusion that the tensions are not equal. Tension FT3 only needs to accelerate the third mass, while FT2 must accelerate both the second and third masses, and FT1 has to account for all attached masses. This results in different tension values throughout the system. Understanding these dynamics is crucial for solving problems related to tension in accelerated systems.
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Homework Statement



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The Attempt at a Solution



"Any objects attached to the rope will have the same magnitude of acceleration as the rope." So since no mass is given, I assumed all the masses are the same. So wouldn't A be the answer?
 
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The tension FT3 only has to accelerate the third mass upwards. FT2 has to accelerate both the second and third mass. FT1 has to... you get the picture. So the tensions are not equal.
 
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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