Question 9.1: Pressure and Weight of Water GOAL Relate density, pressure, ...

(a) Calculate the weight of a cylindrical column of water with height 40.0 m and radius 1.00 m.
Calculate the volume of the cylinder:

V=\pi r^2 h=\pi(1.00  m )^2(40.0  m )=126  m ^3

Multiply the volume by the density of water to obtain the mass of water in the cylinder:

m=\rho V=\left(1.00 \times 10^3  kg / m ^3\right)\left(126  m ^3\right)=1.26 \times 10^5  kg

Multiply the mass by the acceleration of gravity g to obtain the weight w:

w=m g=\left(1.26 \times 10^5  kg \right)\left(9.80  m / s ^2\right)=1.23 \times 10^6  N

(b) Calculate the force exerted by air on a disk of radius 1.00 m at the surface of the lake.
Write the equation for pressure:

P=\frac{F}{A}

Solve the pressure equation for the force and substitute A = πr² :

F = PA = Pπr²

Substitute values:

F=\left(1.01 \times 10^5  Pa \right) \pi(1.00  m )^2=3.17 \times 10^5  N

(c) What pressure at a depth of 40.0 m supports the water column?
Write Newton’s second law for the water column:

-F_{\text {down }}-w+F_{\text {up }}=0

Solve for the upward force:

F_{\text {up }}=F_{\text {down }}+w=\left(3.17 \times 10^5  N \right)+\left(1.23 \times 10^6  N \right)=1.55 \times 10^6  N

Divide the force by the area to obtain the required pressure:

P=\frac{F_{ up }}{A}=\frac{1.55  \times  10^6  N }{\pi(1.00  m )^2}=4.93 \times 10^5  Pa

REMARKS Notice that the pressure at a given depth is related to the sum of the weight of the water and the force exerted by the air pressure at the water’s surface. Water at a depth of 40.0 m must push upward to maintain the column in equilibrium. Notice also the important role of density in determining the pressure at a given depth.