All Solutions
Page 358: Section Review
$rho_text{water} = 1.0 times 10^3 frac{text{kg}}{text{m}^3}$
$frac{V_text{submerged}}{V_text{cork}} = 0.10$
**Unknown:**
$rho_text{cork} = ?$
Since the buoyant force $F_text{buoyant}$ equalizes the weight of the cork $F_text{g}$, we can express their relationship as follows
$$ tag{1}
F_text{g} = F_text{buoyant}
$$
$$
F_text{g} = m_text{cork}g
$$
The buoyant force on the submerged part of the cork is given by the equation
$$
F_text{buoyant} = rho_text{water}V_text{submerged}g
$$
Thus, we can express Equation (1) as follows
$$ tag{2}
m_text{cork}g = rho_text{water}V_text{submerged}g
$$
$$
rho_text{cork} = frac{m_text{cork}}{V_text{cork}}
$$
Isolating $m_text{cork}$ one the left side of the equation
$$
m_text{cork} = rho_text{cork}V_text{cork}
$$
We can now express Equation (2) as follows
$$
rho_text{cork}V_text{cork}g = rho_text{water}V_text{submerged}g
$$
$$
rho_text{cork} = rho_text{water}frac{V_text{submerged}}{V_text{cork}}
$$
Plugging in the given values, we have
$$
rho_text{cork} = (1.0 times 10^3) cdot (0.10)
$$
$$
boxed{rho_{cork} = 100 frac{text{kg}}{text{m}^3}}
$$
rho_text{cork} = 100 frac{text{kg}}{text{m}^3}
$$
$F_{buoyant} = F_{g}$
$rho_{water} V_{floated} g = rho_{cork} V_{total} g$
cancel g:
$rho_{water} V_{floated} = rho_{cork} V_{total}$
$rho_{cork} = rho_{water} (dfrac{V_{floated}}{V_{total}})$
$rho_{cork} = (1000) * (dfrac{1}{10})$
$$
rho_{cork} = 100 kg/m^3
$$
100 kg/m^3
$$
$rho_text{helium} = 0.18 frac{text{kg}}{text{m}^3}$
$rho_text{air} = 1.3 frac{text{kg}}{text{m}^3}$
$F_text{brick} = 10 text{N}$
**Unknown:**
$V = ?$
Since the buoyant force of the helium balloon must be greater than or equal to the weight of the brick in order to lift it, we can express their relationship as follows
$$ tag{1}
F_text{brick} = F_text{buoyant}
$$
$$
F_text{buoyant} = (rho_text{air}-rho_text{helium})Vg
$$
Plugging this into Equation (1), we have
$$ tag{2}
F_text{brick} = (rho_text{air}-rho_text{helium})Vg
$$
$$
V = frac{F_text{brick}}{(rho_text{air}-rho_text{helium})g}
$$
Plugging in the given values, we have
$$
V = frac{(10)}{(1.3 – 0.18) cdot (9.8)}
$$
$$
boxed{V = 0.91 text{m}^3}
$$
V = 0.91 text{m}^3
$$
Solve for V:
$V = dfrac{F_{brick}}{(rho_{air} – rho_{helium}) g}$
$V = dfrac{10}{(1.3 – 0.18) * (9.80)}$
$$
V = 0.91 m^3
$$
0.91 m^3
$$
$F_1 = 150 text{N}$
$A_1 = 2.5 times 10^{-3} text{m}^2$
$A_2 = 4.0 times 10^{-4} text{m}^2$
**Unknown:**
$F_2 = ?$
To solve for the force exerted by the hydraulic lift, we use the equation below
$$
F_2 = frac{F_1A_2}{A_1}
$$
$$
F_2 = frac{(150) cdot (4.0 times 10^{-4})}{(2.5 times 10^{-3})}
$$
$$
boxed{F_2 = 24 text{N}}
$$
F_2 = 24 text{N}
$$
So;
$$
F_{2}=dfrac{F_{1}A_{2}}{A_{1}} = dfrac{(150N)(4.0 times 10^{-4} m^{2})}{2.5 times 10^{-3}m^{2}} = 24 N
$$
textit{color{#c34632} 24N }
$$
$F_1 = 2.3 times 10^4 text{N}$
$A_1 = 0.15 text{m}^2$
$A_2 = 0.0082 text{m}^2$
**Unknown:**
a. $P = ?$
b. $F_2 = ?$
(a)
To calculate the pressure in the hydraulic cylinder, we use the equation below
$$
P = frac{F_1}{A_1}
$$
$$
P = frac{2.3 times 10^4}{0.15}
$$
$$
boxed{P = 1.5 times 10^5 text{Pa}}
$$
To calculate the force exerted on the small cylinder to lift the automobile, we use the equation below
$$
F_2 = frac{F_1A_2}{A_1}
$$
$$
F_2 = frac{(2.3 times 10^4) cdot (0.0082)}{(0.15)}
$$
$$
boxed{F_2 = 1300 text{N}}
$$
**b.** $F_2 = 1300 text{N}$
$P = dfrac{F}{A} = dfrac{2.3 times 10^4}{0.15} = 1.5 times 10^5 Pa$
b)
The pressure is the same:
$P_1 = P_2$
$dfrac{F_1}{A_1} = dfrac{F_2}{A_2}$
$$
F_1 = dfrac{F_2 A_1}{A_2} = dfrac{(2.3e4)*(0.0082)}{0.15} = 1300 N
$$
b) $1300 N$
The density of the block of lead is greater than aluminum, therefore the volume of the aluminum is greater and displaces more water.
b)
The volume of the blocks is the same and therefore they displace equal amount of water when they are placed in water.
b) same
1. Reinforce the doors and the windows so that it can withstand the force exerted on them
2. From the equation $F = PA$, we can reduce the force by making the surface area of the doors and windows smaller.
