Worked examples · Sheet 1

Pressures and forces in liquids at rest

Theory used: Stevin's law Absolute pressure and gauge pressure Pressure Force on a vertical wall

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A diver is at a depth of $25.0 m$ in the sea. The density of seawater is $1025 kg/ m^{3}$ and atmospheric pressure is $p_{0} = 1.013 \cdot 1 0^{5} Pa$ . Compute the gauge pressure and the absolute pressure at the diver’s depth.

p - p_{0} = ρ g h, p = p_{0} + ρ g h

p - p_{0} = 1025 \cdot 9.81 \cdot 25.0 Pa = 2.51 \cdot 1 0^{5} Pa

p = 1.013 \cdot 1 0^{5} Pa + 2.51 \cdot 1 0^{5} Pa = 3.53 \cdot 1 0^{5} Pa

The seawater is at rest and is treated as an incompressible liquid, with uniform density; moreover its free surface is in contact with air at pressure $p_{0}$ . The hypotheses of Stevin’s law are therefore satisfied, and at depth $h$ the pressure exceeds atmospheric pressure by $ρ g h$ . This difference is the gauge pressure, and it equals $2.51 \cdot 1 0^{5} Pa$ . The absolute pressure is obtained by adding $p_{0}$ , therefore the diver is subject to an absolute pressure of $3.53 \cdot 1 0^{5} Pa$ , about three and a half times atmospheric pressure.

As an order-of-magnitude check, every $10 m$ of water adds about $1 atm$ ; at $25 m$ the absolute pressure is therefore about $3.5 atm$ , and indeed $\frac{3.53 \cdot 1 0 ^{5}}{1.013 \cdot 1 0 ^{5}} \approx 3.48$ .

Exercise 2

The vertical wall of a tank contains a circular porthole of radius $r = 0.20 m$ , whose center is $3.0 m$ below the free surface of the water. Outside the tank there is air at atmospheric pressure. Compute the resultant force acting on the porthole.

p - p_{0} = ρ g h, F = \overset{p}{ˉ}_{rel} A

\overset{p}{ˉ}_{rel} = ρ g h_{c}, A = π r^{2} ⟹ F = ρ g h_{c} π r^{2}

F = 1000 \cdot 9.81 \cdot 3.0 \cdot π \cdot (0.20)^{2} N = 3.7 \cdot 1 0^{3} N

On the inner face of the porthole the pressure of the water acts, on the outer face atmospheric pressure. Since the absolute pressure of the water is $p_{0} + ρ g h$ , the contribution of $p_{0}$ cancels with that of the outside air, so the resultant force is computed with the gauge pressure $ρ g h$ alone. All the pressure forces are perpendicular to the porthole, which is flat, therefore they add as numbers, and the resultant force is the average gauge pressure multiplied by the area.

The gauge pressure grows linearly with depth and the porthole is symmetric with respect to the horizontal line through its center. To every horizontal strip at depth $h_{c} + z$ there therefore corresponds a strip of the same area at depth $h_{c} - z$ , and the average of the pressures on the two strips is $ρ g h_{c}$ . Therefore the average pressure on the porthole is that at the center, and the resultant force is $3.7 \cdot 1 0^{3} N$ , directed toward the outside of the tank.

The radius is given to two significant figures, so the result is written with two figures.

Exercise 3

A cylindrical container, with a bottom of area $0.50 m^{2}$ , contains $0.60 m$ of water and above it a layer of oil $0.40 m$ thick, of density $800 kg/ m^{3}$ ; the two liquids do not mix. Compute the gauge pressure at the interface between oil and water and at the bottom, and the force that the liquids exert on the bottom in addition to that due to the atmosphere.

Gauge pressure

p

(in

kPa

) as a function of depth

h

(in centimeters): in the water, which is denser, the line has a greater slope than in the oil

p_{1} = p_{0} + ρ_{o} g h_{o}, p_{2} = p_{1} + ρ_{a} g h_{a}

p_{1} - p_{0} = ρ_{o} g h_{o}, p_{2} - p_{0} = ρ_{o} g h_{o} + ρ_{a} g h_{a}, F = (p_{2} - p_{0}) A

p_{1} - p_{0} = 800 \cdot 9.81 \cdot 0.40 Pa = 3.14 \cdot 1 0^{3} Pa

p_{2} - p_{0} = 3.14 \cdot 1 0^{3} Pa + 1000 \cdot 9.81 \cdot 0.60 Pa = 3.14 \cdot 1 0^{3} Pa + 5.89 \cdot 1 0^{3} Pa = 9.03 \cdot 1 0^{3} Pa

F = 9.03 \cdot 1 0^{3} Pa \cdot 0.50 m^{2} = 4.5 \cdot 1 0^{3} N

Stevin’s law holds inside each liquid, each with its own density. In the oil we start from the free surface, where the pressure is $p_{0}$ , and go down by $h_{o} = 0.40 m$ , so at the interface the gauge pressure is $ρ_{o} g h_{o}$ . The pressure is continuous across the interface, therefore in the water we start again from $p_{1}$ and go down by another $h_{a} = 0.60 m$ with the density of water. The graph shows two straight segments: the second has slope $ρ_{a} g$ , greater than the slope $ρ_{o} g$ of the first.

The bottom is horizontal, so on it the pressure is uniform and the force is the product of the pressure and the area. Atmospheric pressure also acts under the bottom of the container, therefore the force due to the liquids is computed with the gauge pressure and equals $4.5 \cdot 1 0^{3} N$ . As a check, the result coincides with the weight of the two liquids, $(800 \cdot 0.40 + 1000 \cdot 0.60) \cdot 0.50 \cdot 9.81 \approx 4.5 \cdot 1 0^{3} N$ , as it must for a container with vertical walls.

Exercise 4

A dam has a rectangular vertical wall, $20.0 m$ wide, and the reservoir is full of water up to a height of $12.0 m$ above the base of the wall. Compute the gauge pressure at the base of the wall and the magnitude of the resultant force that the water exerts on the dam.

p (z) - p_{0} = ρ g z, d F = ρ g z w d z

F = \int_{0}^{H} ρ g z w d z = \frac{1}{2} ρ g w H^{2}

p (H) - p_{0} = 1000 \cdot 9.81 \cdot 12.0 Pa = 1.18 \cdot 1 0^{5} Pa

F = \frac{1}{2} \cdot 1000 \cdot 9.81 \cdot 20.0 \cdot (12.0)^{2} N = 1.41 \cdot 1 0^{7} N

We measure the depth $z$ downward from the free surface. By Stevin’s law the gauge pressure at depth $z$ is $ρ g z$ , so at the base of the wall, where $z = H = 12.0 m$ , it equals $1.18 \cdot 1 0^{5} Pa$ . Atmospheric pressure acts both on the water and on the dry side of the dam, therefore it does not contribute to the resultant force.

The pressure changes with depth, so the wall is divided into horizontal strips of height $d z$ and area $w d z$ , on which the pressure is considered constant. The forces on the strips are all perpendicular to the wall and have the same direction, therefore the magnitude of the resultant is the integral of their magnitudes. The calculation gives the force on a vertical wall $\frac{1}{2} ρ g w H^{2}$ , that is, $1.41 \cdot 1 0^{7} N$ , a value equal to the weight of about one thousand four hundred metric tons.

As a check, the average gauge pressure is that at half height, $\frac{1.18 \cdot 1 0 ^{5}}{2} Pa = 5.89 \cdot 1 0^{4} Pa$ , and multiplied by the wetted area $240 m^{2}$ it gives $1.41 \cdot 1 0^{7} N$ again.