1. The figure below shows a tank open to the atmosphere and filled to depth D with a liquid of density ρL. Suspended from a string is a block of density ρB (which is greater than ρL), whose dimensions are x, y, and z (meters). The top of the block is at depth h meters below the surface of the liquid.
In each of the following, write your answer in simplest form in terms of ρL, ρB, x, y, z, h, D, and g.
(a) Find the force due to the pressure on the top surface of the block and on the bottom surface. Sketch these forces in the diagram below:
(b) What are the average forces due to the pressure on the other four sides of the block? Sketch these forces in the diagram above.
(c) What is the total force on the block due to the pressure?
(d) Find an expression for the buoyant force on the block. How does your answer here compare to your answer to part (c)?
(e) What is the tension in the string?
Correct Answer:
(a)
The pressure at the top surface of the block is Ptop = Patm + 
. Since the area of the top of the block is A = xy, the force on the top of the block has magnitude 
The pressure at the bottom of the block is Pbottom = Patm + 
. Since the area of the bottom face of the block is also A = xy, the force on the bottom surface of the block has magnitude 
These forces are sketched below:
(b)
Each of the other four faces of the block (left and right, front and back) is at an average depth of h + 
z, so the average pressure on each of these four sides is
The left and right faces each have area A = xz, so the magnitude of the average force on this pair of faces is
The front and back faces each have area A = yz, so the magnitude of the average force on this pair of faces is
These forces are sketched below:
(c)
The four forces sketched in part (b) add up to zero, so the total force on the block due to the pressure is the sum of Fon top and Fon bottom; because Fon bottom > Fon top, this total force points upward and its magnitude is
(d)
By Archimedes' Principle, the buoyant force on the block is upward and has magnitude
This is the same as the result you found in part (c).
(e)
The weight of the block is
If FT is the tension in the string, then the total upward force on the block, FT + Fbuoy, must balance the downward force, Fg; that is, FT + Fbuoy = Fg, so
2. The figure below shows a large, cylindrical tank of water, open to the atmosphere, filled with water to depth D. The radius of the tank is R. At a depth h below the surface, a small circular hole of radius r is punctured in the side of the tank, and the point where the emerging stream strikes the level ground is labeled X.
In parts (a) through (c), assume that the speed with which the water level in the tank drops is negligible.
(a) At what speed does the water emerge from the hole?
(b) How far is point X from the edge of the tank?
(c) Assume that a second small hole is punctured in the side of the tank, a distance of h/2 directly above the hole shown in the figure. If the stream of water emerging from this second hole also lands at Point X, find h in terms of D.
(d) For this part, do not assume that the speed with which the water level in the tank drops is negligible, and derive an expression for the speed of efflux from the hole punctured at depth h below the surface of the water. Write your answer in terms of r, R, h, and g.
Correct Answer:
(a)
See the section on Torricelli's Theorem for the derivation of the efflux speed from the hole; applying Bernoulli's Equation to a point on the surface of the water in the tank (Point 1) and a point at the hole (Point 2), the assumption that v1 ≈ 0 leads to the result
(b)
The initial velocity of the water, as it emerges from the hole, is horizontal. Since there's no initial vertical velocity, the time t required to drop the distance y = D - h to the ground is found as follows:
Therefore, the horizontal distance the water travels is
(c)
The second hole would be at a depth of h/2 below the surface of the water, so the horizontal distance it travels-from the edge of the tank to the point where it hits the ground-is given by the same formula you found in part (b) except it will have h/2 in place of h; that is,
If both streams land at the same point, then the value of x from part (b) is the same as x2 :
(d)
Once again, apply Bernoulli's Equation to a point on the surface of the water in the tank (Point 1) and to a point at the hole (Point 2). Choose the ground level as the horizontal reference level; then y1 = D and y2 = D - h. If v1 is the flow speed of Point 1-that is, the speed with which the water level in the tank drops-and v2 is the efflux speed from the hole, then, by the Continuity Equation, A1v1 = A2v2, where A1 and A2 are the cross-sectional areas at Points 1 and 2, respectively. Therefore, v1 = (A2/A1)v2. Bernoulli's Equation then becomes
Since P1 = P2 = Patm, these terms cancel out; and substituting v1 = (A2/A1)v2, you have
Now, since A1 = πR2 and A2 = πr2, this final equation can be written as
[Note that if r << R, then (r/R)4 ≈ 0, and the equation above reduces to v2 = >
, as in part (a).]
3. The figure below shows a pipe fitted with a Venturi U-tube. Fluid of density ρF flows at a constant flow rate and with negligible viscosity through the pipe, which constricts from a cross-sectional area A1 at Point 1 to a smaller cross-sectional area A2 at Point 2. The upper portion of both sides of the Venturi U-tube contain the same fluid that's flowing through the pipe, while the lower portion is filled with a fluid of density ρV (which is greater than ρF). At Point 1 in the pipe, the pressure is P1 and the flow speed is v1; at Point 2 in the pipe, the pressure is P2 and the flow speed is v2. All the fluid within the Venturi U-tube is stationary.
(a) What is PX, the hydrostatic pressure at Point X? Write your answer in terms of P1, ρF, h1, and g.
(b) What is PY, the hydrostatic pressure at Point Y? Write your answer in terms of P2, ρF, ρV, h2, d, and g.
(c) Write down the result of Bernoulli's Equation applied to Points 1 and 2 in the pipe, and solve for P1 - P2.
(d) Since PX = PY, set the expressions you derived in parts (a) and (b) equal to each other, and use this equation to find P1 - P2.
(e) Derive an expression for the flow speed, v2, and the flow rate, f, in terms of A1, A2, d, ρF, ρV, and g. Show that v2 and f are proportional to 
.
Correct Answer:
(a)
Point X is at a depth of h1 below Point 1, where the pressure is P1. Therefore, the hydrostatic pressure at X is PX = P1 + 
.
(b)
Point Y is at a depth of h2 + d below Point 2, where the pressure is P2. The column of static fluid above Point Y contains fluid of density of depth h2 and fluid of density 
of depth d. Therefore, the hydrostatic pressure at Y is PY = P2 + 
.
(c)
First, notice that Points 1 and 2 are at the same horizontal level; therefore, the heights y1 and y2 are the same, and the terms 
and 
will cancel out of the equation. Bernoulli's Equation then becomes
By the Continuity Equation, you have A1v1 = A2v2, so v1 = (A2/A1)v2. Therefore,
(d)
In parts (a) and (b) above, you found that PX = P1 + ρFgh1 and PY = P2 + ρF gh2+ρv gd. Since PX = PY, you have
so
(e)
In parts (c) and (d), you found two expressions for P1 - P2. Setting them equal to each other gives
The flow rate in the pipe is
Since
you see that f is proportional to >, as desired.
