Is Fg the Only Force to Consider for Ft4 in a Pulley System?

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In a pulley system, when analyzing the forces acting on the piano, Ft1 and Ft2 are equal, and Fg (the weight of the piano) is a crucial factor for Ft4. The force required to lift the piano, Ft3, must also consider the applied force, especially if the piano is accelerating. If the piano is moving at a constant speed, Ft4 equals the weight of the piano, simplifying calculations to just m*g. The number of strands in the pulley system and the tension in each strand are important for understanding the overall force dynamics. Understanding these principles is essential for solving pulley system problems effectively.
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I have some questions about the following problem:

I know that Ft1 and Ft2 are equal. Is Fg the only force that needs to be taken into account for Ft4? Does the force being applied to lift the piano need to be taken into account for Ft3? If not, is it just m*g?

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Assuming that the pulleys are frictionless, the tension is the same throughout the rope (frictionless pulleys can be found only in textbook problems).

Also assuming that the piano is moving at a constant speed, F_T4 is equal to the weight of the piano. More force here is necessary to accelerate the piano upward.

Look at the top pully: how many strands are pulling it down? How much tension in each strand? ("strand" = "rope segment")
 
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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