1. Two cylinders with the same mass density r C = 713 kg / m 3 are floating in a

Question

1. Two cylinders with the same mass density rC = 713 kg / m3 are floating in a container of water (with mass density rW = 1025 kg / m3). Cylinder #1 has a length of L1 = 20 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 10 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water (h2 / h1)?

2. A rectangular block of ice 15 m on each side and 0.7 m thick floats in seawater. The density of the seawater is 1025 kg/m3. The density of ice is 917 kg/m3.

a) How high does the top of the ice block float above the water level?

= m   

b) How many penguins of mass 16 kg each can stand on the ice block before they get their feet wet?

A tube of radius 5 cm is connected to tube of radius 1 cm as shown above. Water is forced through the tube at a rate of 10 liters/min. The pressure in the 5 cm tube is 1×105 Pa. The density of water is 1000 kg/m3. Assume that the water is nonviscous and uncompressible.

(a) What is the velocity of the water in the 5 cm radius tube?

v1 = m/s     

(b) What is the velocity of the water in the 1 cm radius tube?

v2 = m/s     

(c) What is the water pressure in the 1 cm radius tube?

Explanation / Answer

If Vc the cylinder's volume, hc is its height, and A is its cross sectional area, and if Vw is the volume of the displaced water, which is also the volume of the submerged part of the cylinder, hw is the height of the submerged part of the cylinder, then
Vc = hc * A and

Vw = hw * A

If c is the cylinder's mass density, w is the water's mass density, Mc is the mass of the cylinder, and Mw is the mass of the displaced water, then
Mc = c * Vc

= c * hc * A
and the mass of the displaced water is
Mw = w * Vw

= w * hw * A

But, because the cylinders are floating, Mc = Mw and
c * hc = w * hw

Note at this point that the cross sectional area of the cylinder A has dropped out of the equation. The cylinders' masses and volumes have also dropped out. The radii also dropped out with the area. In fact, this derivation does not depend on the cylinders' cross section being circular or even of constant shape; only that the cross sectional area be be independent of the height at which the cross section is taken.

If hx is the height of the cylinder x sticking out of the water, then
Lx = hc and
hx = hc hw

= Lx hw.
It turns out that
hw / hc = c / w
Subtracting both sides of the above equation from 1, taking the least common denominator on each side, and substituting hx and Lx gives
hx / Lx = 1 c / w
By substituting 1 and 2 for x, we get one equation each for cylinder 1 and 2.

Dividing equation #2 by equation #1 also causes the 1 c / w factor to drop out, and with it, c and w. After doing a bit more algebra, we find that all we really needed was L1 and L2 and the fact that c < w:

h2 / h1 = L2 / L1 = 11 / 24

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