All Solutions
Page 353: Practice Problems
So ;
$F_{2} = dfrac{F_{1}A_{2}}{A_{1}}$ = $dfrac{(1600N)(72cm^{2})}{1440 cm^{2}} = 80 N$
textit{color{#c34632}80N}
$$
Weight of the chair: $F_g=1600 mathrm{N}$
Cross-sectional area of a piston: $A_c=1440 mathrm{cm^2}$
Cross-sectional area of a smaller piston: $A_p=72 mathrm{cm^2}$.
We need to calculate the force $F_p$, acting on the smaller piston, required to lift the chair.
$$begin{align*}
P_p&=P_g\
frac{F_p}{A_p}&=frac{F_g}{A_c}\
end{align*}$$
We solve this equation for $F_p$ to get the required force
$$begin{align*}
F_p&=frac{F_gA_p}{A_c}\
&=frac{1600cdot72}{1440}\
&=boxed{80 mathrm{N}}
end{align*}$$
So;
$F_{2} = dfrac{F_{1}A_{2}}{A_{1}} = dfrac{(55N)(2.4m^{2})}{(0.015m^{2})} = 8.8 times 10^{3} N$
Force exerted on the hydraulic piston: $F_p=55 mathrm{N}$
Cross-sectional area of the hydraulic piston: $A_p=0.015 mathrm{m^2}$
Cross-sectional area of the piston that the car sits on: $A_c=2.4 mathrm{m^2}$
We need to calculate the weight of the car $F_g$.
$$begin{align*}
P_g&=P_c\
frac{F_g}{A_c}&=frac{F_p}{A_p}
end{align*}$$
we solve this equation for $F_g$ to find the weight of the car
$$begin{align*}
F_g&=frac{F_p A_c}{A_p}\
&=frac{55cdot 2.4}{0.015}\
&=boxed{8800 mathrm{N}}
end{align*}$$
textbf{Given:} \
$F_1 = 400 text{N}$ \
$F_2 = 1100 text{N}$ \
textbf{Find:} \
$frac{A_1}{A_2} = ?$ \
textbf{Calculation:}\
To get get an expression for the ratios of the areas of the pistons, we start with the equation below
$$
F_2 = frac{F_1A_2}{A_1}
$$
Isolating $A_1$ and $A_2$ on the left side of the equation, we have
$$
frac{A_1}{A_2} = frac{F_1}{F_2}
$$
Plugging in the given values, we have
$$
frac{A_1}{A_2} = frac{400}{1100}
$$
$$
boxed{frac{A_1}{A_2} = 0.36}
$$
frac{A_1}{A_2} = 0.36
$$
So ;
$dfrac{A_{2}}{A_{1}} = dfrac{F_{2}}{F_{1}} = dfrac{400 N}{1100N} = 0.4$
textbf{Given:} \
$A_1 = 7.0 times 10^{-2} text{m}^2$ \
$A_2 = 2.1 times 10^{-1} text{m}^2$ \
$F_2 = 2.7 times 10^3 text{N}$ \
$h_2 = 0.20 text{m}$ \
textbf{Find:} \
(a) $F_1 = ?$ \
(b) $h_1 = ?$
textbf{Calculation:}\
textbf{(a)} \
To get an exression for $F_1$, we start with the equation below
$$
F_2 = frac{F_1A_2}{A_1}
$$
Isolating $F_1$ on the left side of the equation, we have
$$
F_1 = frac{F_2A_1}{A_2}
$$
Plugging in the given values
$$
F_1 = frac{(2.7 times 10^3) cdot (7.0 times 10^{-2})}{(2.1 times 10^{-1})}
$$
$$
boxed{F_1 = 900 text{N}}
$$
textbf{(b)} \
To get an expression relating $A$ and $h$, we assume that the volumes traced by the movements of the two pistons are equal thus we have
$$
V_1 = V_2
$$
Since $V = Ah$, we can express the equation above as follows
$$
A_1h_1 = A_2h_2
$$
Isolating $h_1$ on the left side of the equation, we have
$$
h_1 = frac{A_2h_2}{A_1}
$$
Plugging the given values
$$
h_1 = frac{(2.1 times 10^{-1}) cdot (0.20)}{(7.0 times 10^{-2})}
$$
$$
boxed{h_1 = 0.60 text{m}}
$$
(b) $h_1 = 0.60 text{m}$
$F_1 = dfrac{F_2 A_1}{A_2} = dfrac{(2.7e3)*(7.0e-2)}{(2.1e-1)} = 900 N$
b)
$$
h_1 = dfrac{h_2 A_2}{A_1} = dfrac{(0.20)*(2.1e-1)}{(7.0e-2)} = 0.6 m
$$
b) 0.60 m
textbf{Known:} \
$rho_text{brick} = 1.8rho_text{water}$ \
$rho_text{water} = 1.00 times 10^3 frac{text{kg}}{text{m}^3}$ \
$V = 0.20 text{m}^3$ \
$g = 9.8 frac{text{m}}{text{s}^2}$ \
textbf{Unknown:} \
$F_text{apparent} = ?$ \
textbf{Calculation:}\
To solve for the apparent weight of the brick in water, we use the equation below
begin{equation}
F_text{apparent} = F_g – F_text{buoyant}
end{equation}
To get an expression for $F_g$, we use the equation
$$
F_g = mg
$$
We know that the definition of density is $rho_text{brick} = frac{m_text{brick}}{V}$. Plugging this into the equation above, we have
begin{equation} tag{2}
F_g = rho_text{brick}Vg
end{equation}
To get an expression for the buoyant force, we use the equation
begin{equation} tag{3}
F_text{buoyant} = rho_text{water}Vg
end{equation}
Plugging in the Equation (2) and (3) into Equation (1), we have
$$
F_text{apparent} = rho_text{brick}Vg – rho_text{water}Vg
$$
$$
F_text{apparent} = (rho_text{brick} – rho_text{water})Vg
$$
Since $rho_text{brick} = 1.8rho_text{water}$, we can express the equation above as follows
$$
F_text{apparent} = (1.8rho_text{water} – rho_text{water})Vg
$$
$$
F_text{apparent} = 0.8rho_text{water}Vg
$$
Plugging in the given values, we have
$$
F_text{apparent} = 0.8(1.00 times 10^3)(0.20 text{m}^3)(9.8)
$$
$$
boxed{F_text{apparent} = 1600 text{N}}
$$
F_text{apparent} = 1600 text{N}
$$
$F_{apparent} = rho_{brick} V g – rho_{water} V g$
$F_{apparent} = (rho_{brick} – rho_{water}) V g$
$F_{apparent} = ((1.8)*(1000) – (1000)) * (0.20) * (9.80)$
$$
F_{apparent} = 1600 N
$$
1600 N
$$
textbf{Known:} \
$F = 610 text{N}$ \
$rho_text{water} = 1.0 times 10^3 frac{text{kg}}{text{m}^3}$ \
$g = 9.8 frac{text{m}}{text{s}^2}$ \
textbf{Unknown:} \
$V = ?$ \
textbf{Calculation:}\
To get an expression for the volume of the submerged part of the her body, we use the equation
$$
F = rho Vg
$$
Isolating $V$ on one side of the equation
$$
V = frac{F}{rho g}
$$
Plugging in the given values, we have
$$
V = frac{(610)}{(1.0 times 10^3) cdot (9.8)}
$$
$$
boxed{V = 0.062 text{m}^3}
$$
V = 0.062 text{m}^3
$$
Solve for V:
$$
V = dfrac{F}{rho g} = dfrac{610}{(1000)*(9.80)} = 0.0622 m^3
$$
0.0622 m^3
$$
textbf{Known:} \
$F_g = 1250 text{N}$ \
$V = 16.5 times 10^{-3} text{m}^3$ \
$g = 9.8 frac{text{m}}{text{s}^2}$ \
$rho_text{water} = 1.0 times 10^3 frac{text{kg}}{text{m}^3}$ \
textbf{Unknown:} \
$F_text{apparent} = ?$ \
textbf{Calculation:}\
To solve for the tension in the wire supporting the camera, we determine the apparent weight of the camera using the equation below
$$
F_text{apparent} = F_g – F_text{buoyant}
$$
$$
F_text{apparent} = F_g – rho_text{water}Vg
$$
Plugging in the given values, we have
$$
F_text{apparent} = 1250 – (1.0 times 10^3) cdot (16.5 times 10^{-3}) cdot (9.8)
$$
$$
boxed{F_text{apparent} = 1090 text{N}}
$$
F_text{apparent} = 1090 text{N}
$$
$F = F_g – rho_{bouyant} V g$
$F = 1250 – (1000)*(16.5e-3)*(9.80)$
$$
F = 1090 N
$$
1090 N
$$
textbf{Known:} \
$rho_text{foam} = 0.10rho_text{water}$ \
$V_text{foam} = 1.0 text{m} times 1.0 text{m} times 0.10 text{m}$ \
$g = 9.8 frac{text{m}}{text{s}^2}$ \
$rho_text{water} = 1.0 times 10^3 frac{text{kg}}{text{m}^3}$ \
textbf{Unknown:} \
$F_text{brick} = ?$ \
textbf{Calculation:}\
The total weight $F_g$ of the brick and the plastic foam is given by the equation
$$
F_g = F_text{brick} + F_text{foam}
$$
To get an expression for $F_text{brick}$, we use the equation below
$$
F_text{apparent} = F_g – F_text{buoyant}
$$
$$
F_text{apparent} = (F_text{brick} + rho_text{foam}V_text{foam}g) – rho_text{water}V_text{foam}g
$$
According to the problem, the buoyant force must equalize the weight of the brick, thus we must set the apparent weight to 0
$$
0 = (F_text{brick} + rho_text{foam}V_text{foam}g) – rho_text{water}V_text{foam}g
$$
Isolating $F_text{brick}$ on the left side of the equation, we have
$$
F_text{brick} = rho_text{water}V_text{foam}g – rho_text{foam}V_text{foam}g
$$
$$
F_text{brick} = (rho_text{water} – rho_text{foam})V_text{foam}g
$$
Since $rho_text{foam} = 0.10rho_text{water}$, we have
$$
F_text{brick} = (rho_text{water} – 0.10rho_text{water})V_text{foam}g
$$
$$
F_text{brick} = 0.90rho_text{water}V_text{foam}g
$$
Plugging in the given values
$$
F_text{brick} = 0.90 cdot (1.0 times 10^3) cdot (1.0 times 1.0 times 0.1) cdot (9.8)
$$
$$
boxed{F_text{brick} = 880 text{N}}
$$
F_text{brick} = 880 text{N}
$$
$F_{bricks} + rho_{foam} V g = rho_{water} V g$
$F_{bricks} = rho_{water} V g – rho_{foam} V g$
$F_{bricks} = (rho_{water} – rho_{foam}) V g$
$F_{bricks} = (1000 – (0.10)*(1000)) * (1.0*1.0*0.10) * (9.80)$
$$
F_{bricks} = 882 N
$$
882 N
$$
textbf{Known:} \
$F_text{canoe} = 480 text{N}$ \
$rho_text{foam} = 0.10rho_text{water}$ \
$rho_text{water} = 1.0 times 10^3 frac{text{kg}}{text{m}^3}$ \
$g = 9.8 frac{text{m}}{text{s}^2}$ \
textbf{Unknown:} \
$V= ?$ \
textbf{Calculation:}\
According to the problem, the buoyant force of the foam must equalize the combined weight of the canoe and the foam. We can express this using the equation below
$$
F_text{buoyant} = F_text{foam} + F_text{canoe}
$$
$$
rho_text{water}Vg = rho_text{foam}Vg + F_text{canoe}
$$
Isolating $V$ on the left side of the equation
$$
rho_text{water}Vg – rho_text{foam}Vg = F_text{canoe}
$$
$$
(rho_text{water} – rho_text{foam})Vg = F_text{canoe}
$$
$$
V = frac{F_text{canoe}}{(rho_text{water} – rho_text{foam})g}
$$
Since $rho_text{foam} = 0.10rho_text{water}$, we can express the equation above as follows
$$
V = frac{F_text{canoe}}{(rho_text{water} – 0.10rho_text{water})g}
$$
$$
V = frac{F_text{canoe}}{0.90rho_text{water}g}
$$
Plugging in the given values, we have
$$
V = frac{(480)}{(0.90) cdot (1.0 times 10^3) cdot (9.8)}
$$
$$
boxed{V = 0.054 text{m}^3}
$$
V = 0.054 text{m}^3
$$
$F_{canoe} + rho_{foam} V g = rho_{water} V g$
$F_{canoe} = rho_{water} V g – rho_{foam} V g$
$F_{canoe} = (rho_{water} – rho_{foam}) V g$
Solve for V:
$V = dfrac{F_{canoe}}{(rho_{water} – rho_{foam}) g}$
$V = dfrac{480}{(1000 – (0.10)*(1000)) * (9.80)}$
$$
V = 0.0544 m^3
$$
0.0544 m^3
$$
