All Solutions
Page 434: Lesson Check
begin{enumerate}[-]
item
The pressure in a gas acts equally in all directions.
item
The pressure exerted on a given surface is always at right angles to that surface.
end{enumerate}
In the great salt lake at water, the density of the water $(rho)$ is very high, owing to the high salt content. so the buoyant force acting on our body balances the weight of our body, and we are able to float without effort.
When the ice melts its weight remains the same, and it will fully fill the volume $V_1$ (just as it did in the easier problem with only the ice block in the cup), but the pebble with volume $V_3$ will now drop into the cup.
The whole question boils down to which is bigger, the volume of the pebble $V_3$, or the volume of the pushed out water $V_2$ owing to the weight of the rock.
Equating the weights of the pebble and the water pushed out by its weight we get:
$$
begin{align*}
V_2 cdot rho_{text{water}} cdot g = V_3 cdot rho_{text{pebble}} cdot g
end{align*}
$$
Since we know that the density of the pebble is higher that that of water it follows that
$$
begin{align*}
V_3 < V_2
end{align*}
$$
We have found that the volume $V_2$ pushed out because of the pebble is greater than its own volume, so we conclude that the final level of water will fall when the ice melts.
Force of buoyancy $(F_B)$ is equal in magnitude to the weight of the fluid displaced by the object.
The lead and aluminium spheres have the same diameter , which means that they displace the same volume of water $(V)$ .
Now , buoyant force acting on both spheres is : $F_B=Vrho_wg$ , where $(rho_w )$ is the density of water.
Hence, the buoyant force acting on both the spheres is the same.
$$
m=rho V
$$
$$
begin{equation*}
F = P_{text{atmospheric}} cdot A
end{equation*}
$$
Where $A$ is the area of the football field:
$$
begin{align*}
A = a cdot b = 110text{ m} cdot 49text{m}
end{align*}
$$
Finally:
$$
begin{align*}
F &= 101.3text{ kPa} cdot 110text{ m} cdot 49text{ m} \
F &= 546 cdot 10^{6}text{ N}
end{align*}
$$
$$
F = rho Vg
$$
$$
F = (1000kg/m^3)[(4.2m)(6.5m)(0.027m)](9.81m/s^2)
$$
$$
F = 7231.0N
$$
To find the weight of the pushed off water we first find its mass:
$$
begin{equation*}
m = rho , V
end{equation*}
$$
To find the volume we use the dimensions of the raft $a, b$ and the heights it sinks into the water $h$:
$$
begin{align*}
V = A cdot h = a , b , h
end{align*}
$$
Finally the weight is:
$$
begin{equation*}
F = m , g
end{equation*}
$$
Putting this all together we have:
$$
begin{align*}
F = m , g = rho , V , g = rho , a , b , h , g
end{align*}
$$
Plugging in the values:
$$
begin{align*}
F &= 1000 ; frac{text{kg}}{text{m}^3} cdot 4.2text{ m} cdot 6.5text{ m} cdot 2.7 cdot 10^{-2}text{ m} cdot 9.81 ; frac{text{m}}{text{s}^2} \
F &= 7230.95text{ N}
end{align*}
$$
$$
begin{align*}
P_1 &= P_2 \
frac{F_1}{A_1} &= frac{F_2}{A_2}
end{align*}
$$
We know the force on the small piston $F_1 = 750text{ N}$, and the areas of the small and large piston $A_1 = 0.0075text{ m}^2$, $A_2 = 0.13text{ m}^2$
Rearranging for $F_2$ we get:
$$
begin{align*}
F_2 = F_1 cdot frac{A_2}{A_1}
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
F_2 = 750text{ N} cdot frac{0.13}{0.0075} = 13000text{ N}
end{align*}
$$
