Buoyancy and Acceleration: Solving for Tension in a Vessel

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The discussion focuses on deriving the tension in a string holding a solid block submerged in a liquid when the vessel accelerates upward. Initially, the tension is denoted as To when the vessel is at rest. During upward acceleration, the tension T is shown to equal To(1 + a/g) by analyzing the forces acting on the block. The buoyant force before and during acceleration is expressed in terms of the block's volume and the densities involved. The final result is derived by comparing the ratios of buoyant forces in both scenarios, confirming the relationship between tension and acceleration.
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1.The tension in a string holding a solid block below the surface of a liquid (of density greater than the solid) is To when the containing vessel is at rest. Show that when the vessel has an upward vertical acceleration of magnitude a, the tension T is equal to To(1+ a/g)?
 
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The sum of the forces acting on the block equals its mass time its acceleration.
 
Tide said:
The sum of the forces acting on the block equals its mass time its acceleration.

F = To +mg
Fb - T - mg = ma

then?
 
Before acceleration, the buoyant force is

F_b = T_0 + mg

and during acceleration it is

F_b' = T + m(g+a)

Now the buoyant force is F_b = \rho V g before acceleration and F_b' = \rho V (g+a) during acceleration where \rho is the density of the water and V is the volume of the block. Now just write the ratio of the buoyant force in the two cases and arrive at

\frac {g}{g+a} T = T_0

from which the desired result follows.
 
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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