A container of uniform cross-sectional area A is filled with liquid of uniform d

Question

A container of uniform cross-sectional area A is filled with liquid of uniform density ?. (Figure 1) Consider a thin horizontal layer of liquid (thickness dy) at a height y as measured from the bottom of the container. Let the pressure exerted upward on the bottom of the layer be pand the pressure exerted downward on the top be p+dp. Assume throughout the problem that the system is in equilibrium (the container has not been recently shaken or moved, etc.). What is the weight wlayer of the thin layer of liquid? Express your answer in terms of quantities given in the problem introduction and g, the magnitude of the acceleration due to gravity. Since the liquid is in equilibrium, the net force on the thin layer of liquid is zero. Complete the force equation for the sum of the vertical forces acting on the liquid layer described in the problem introduction. Express your answer in terms of quantities given in the problem introduction and taking upward forces to be positive.

Explanation / Answer

Force F up= P A

F down = (P+dP)A

Weight of the layer W = mg = volume x density x g = (rho) dV g

= (rho)(A dy)g

Sum of the forces = Fup - F down - weight = 0

==> P A - ( P + dP)A - (rho)(A dy)g = 0

so, dP = -(rho)g dy

Integrate both sides of the differential equation you found for dP to obtain an equation

for P . Your equation should then include a constant that depends on initial conditions.

Determine the value of this constant by assuming that the pressure at some reference

height yo is Po

==> Po = -(rho) g yo + C

=> C = Po + (rho) g yo

therefore P = Po + (rho)g(yo -y)

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