ME106
Homework Set 3 - Solutions
1) Using the ME106 lemma to Gauss’s theorem, RS P n d ^ (area) = - RV (rP) d(volume). However, if the pressure is constant, then rP = 0, and the integrand is zero everywhere, so the integral,
or buoyancy is zero.
2) BArhimedes = -gρair(2=3)πR3 z^. α ≡ jBj=[(πR2P(z = ceiling)] = 3P(2zgρ =air ceiling R ). Use ρair =
1:225kg=m3; P(z = ceiling) = 101; 325kg=(ms2); and g = 9:8m=s2. With R = 0:1m, α =
0:000008, so this a 0.0008% error. For a 1% error, we need R = (1=:0008) 0:1m or 120 m. For a
10% error, we need R = (10=:0008) 0:1m or 1200 m.
3a) The pressure P(z) in the fluid can be obtained by integrating the hydrostatic equation dP=dz =
-gρ , or by finding the weight of the column fluid above a certain point at height z, or by “inspection”. At this point we can try the last of these methods, knowing that because ρ is constant in each
of the two sections of the fluid, that the pressure in each section will be a linear function of z. We
“guess” the following solution:
P(z) =  | Pair - gρ1z  | for - H ≤ z ≤ 0
P(z) = Pair + gHρ1 - gρ2(z + H) for - h ≤ z ≤ -H:  | (1)Note that the by differentiating the solution above with respect to z that we obtain the correct
hydrostatic equation in each of the two sections. Note that we have found the integration constants
1
correctly because at z = 0, we obtain Pair. Also note that at z = -H, both the top and bottom
expressions for P(z) in equation (1) give the same value.
All four problems 3–6 in this problem set 3 (corresponding to figure 1 -4) have the same P(z),
regardless of what is inside the suction cup or the condition of the left wall.
3b) F2 can be calculated directly by knowing the normal vector of an ellipse as a function of
position and being able to do a messy integral quickly. We avoid that approach, and rather rely on
finding F2 with Archimedes’ help. Figure 1a (at the end of these solutions) shows the lower half
of the suction cup as it appears in the Problem Set 3 (Figures 1-4). Figure 1b (at the end of these
solutions) shows the lower half of the suction cup if it were detached from the wall and if the upper
part of the suction cup did not exist. The fictional control volumes in Figs 1b and 1c (at the end
of these solutions) (made of the curved black line, the green straight line, and the red straight line)
are surrounded on all sides by a fluid with pressure P(z) given by eq.(1).
To make the lower half of the suction cup part of a closed surface that surrounds a control
volume, we added a fictitious left-side (in red) and fictitious top-side (in green). The red and green
additions, the lower part of the suction cup, and the original sides of the suction cup (that are not
shown because they are in x-z planes parallel to the paper at y = ±L=2) surround the control
volume. This control volume has a length L in the y direction, like the original suction cup. The
volume of this control volume is half that of the volume enclosed by the original suction cup, or
πDdL=16. Because this control volume is surrounded by fluids with the same pressure P(z) as
the suction cup in the homework problem, the pressure force F2 exerted by the fluid of density ρ2
on the lower, outer, convex portions of the suction in Figure 1b (at the end of these Solutions) is
the same as it is in the homework problem. Because the control volume in Figure 1b is completely
surrounded by fluid, the sum of all the pressure forces on all the sides is equal to the buoyancy B
2
force. From Archimedes, B = ρ2gπDdL=16 z^ because ρ2πDdL=16 is the mass of the displaced
fluid. From Figure 1c (at the end of these solutions),
B = F2 + Ftop + Fleft; (2)
where Ftop is the downward pressure force at z = -H acting on the green lid of the lower part of
the suction cup (note that the lid has area Ld=2), and where Fleft is the pressure force acting on
the red side. Because pressure force is the pressure times the unit inward normal vector integrated
over the area:
F
top = -[P(z = -H)Ld=2] z^ = -(Pair + gHρ1)Ld=2 z^;  | (3)
and
Fleft = L x^ Z | P(z) dz:  | (4)--HH-D=2 Using equation (2),
F2 = B - Ftop - Fleft; (5)
or
F2 = (ρ2gπDdL=16) z^ + [(Pair + gHρ1)Ld=2)] z^ - x^ L Z--HH-D=2 P(z) dz (6)
Because P(z) in eq. (1) is the same in problems 3–6 and because the shape and position of the
suction cup is the same in all four, F2 is the same in all of problems 3 – 6, as is F1.
3c) We are interested in finding Fwall, which is defined in the problem set to be the force from
the left wall acting on the circular edge (and only on the circular edge) of the suction cup where
the cup touches the wall. This is very hard to do directly, so we will find Fwall indirectly by first
finding a force F on a “fictitious rectangle” shown in Fig. 2 (at the end of these solutions).
The figure shows a blue “picture frame” in the y - z plane at x = 0, which represents the area
or glue between the suction cup and the right side of the left wall. Interior to this area is a red
3
region in the y-z plane that separates the contents inside the suction cup from the outside world.
That outside world might be the left wall of the container or, as in Figure 3 of the Problem, part
of it might be the air in the hole in the wall. The union of the red and blue regions in Figure 2 is
the “fictitious rectangle” with area L × D referred to in the problem set. This fictitious area does
not consist of atoms or molecules, but rather is a mental construction only. This fictitious rectangle
along with the suction cup’s outer convex boundary and the two sidewalls in the x-z plane define
the surface that surrounds a control volume V .
The control volume V is made of the fictitious rectangle, the suction cup, and its contents are
in equilibrium, which means that all of the forces acting on it must sum to zero. What are those
forces? They are the pressure forces from the fluids with densities ρ1 and ρ2 acting on the suction
cup, the unknown force F acting on the fictitious rectangle, and the weight W of the suction cup
and its contents. Thus,
0 = F1 + F2 + F + W: (7)
We already calculated or were given the values of F1 and F2. For Problem 3, the suction cup