Solved Problems

Draw a free body diagram as shown in the figure: 

Apply Newton's second law to the cork:\[\Sigma F_y=0\Rightarrow B=mg+T\] Where $T$ is tension in the spring scale and $B$ is buoyancy force. 

Buoyancy force: when a rigid object is submerged in a fluid (completely or partially), there exists an upward force on the object that is equal to the weight of the fluid that is displaced by the object, i.e., $B=\rho_{fluid}V_{submerged}\,g$

The mass of the cork is known (weight measured in the air), so using the buoyancy force formula we can find its volume.
\begin{gather*} B=mg+T=\rho_{water}V_{cork}g \\ \\ 0.285+0.855=1000\times 9.8\times V_{cork} \\ \\ V_{cork}=1.16\times {10}^{-4}\quad {\rm m^3}\end{gather*} 
Given the $V_{cork}$, one can determine the density of the cork by \begin{gather*} \rho_{cork}=\frac{m_{cork}}{V_{cork}}=\frac{W_{cork}}{gV_{cork}}\\ \\ \rho_{cork}=\frac{0.285}{9.8\times 1.16\times {10}^{-4}}=250.7\,{\rm kg/m^3}\end{gather*}
 

Recall from Archimedes's principle examples, that the buoyant force is equal to the weight of the displaced fluid.

In this case, since the object is totally submerged so the volume of the displaced fluid and object is the same.

Remind that the weight of an object is mass times g and mass is density times volume of an object.

Applying buoyant force formula  gives  \begin{align*}F_B&=\underbrace{\rho_{fluid}V_{submerged}}_{M}g\\ \\ &=500\times 0.2\times 9.8=980\quad {\rm N}\end{align*}
 

By applying Archimedes' principle and Newton's second law, we can solve this problem. 

Stage (1): before loading the lead block, Newton's second law states that in the vertical direction there is no motion. 

On the other hand, recall from Archimedes' principle problems that the buoyant force equals to the weight of displaced fluid.

\begin{gather*} \Sigma F_y=0 \\ \\ F_B-m_{block}g=0\\\\ \Rightarrow \rho_{water}V_{sub}g=m_{block}g \\\\ \Rightarrow \quad \rho_{water}\left(0.68\ V_{block}\right)=m_{block} \\\\ \rho_{water}V_{block}=\frac{m_{block}}{0.68}\end{gather*}

Stage (2): after loading \begin{gather*} \Sigma F_y=0\\\\ \left(m_{lead}+m_{block}\right)g=F_B=\rho_{water}V_{block}g\\\\ \Rightarrow m_{lead}=\rho_{water}V_{block}-m_{block}\end{gather*}

Now substituting stage 1 into the stage 2, we obtain
\begin{align*} m_{lead}&=\frac{m_{block}}{0.68}-m_{block}\\\\&=1.5\times 0.47\\\\&=0.71\quad {\rm kg}\end{align*}

The volume of displaced water is equal to the volume of the object that is submerged i.e. $\frac{1}{2}\left(\frac{4}{3}\pi R^3_1\right)$.

Because the object is motionless (in equilibrium) so $\Sigma F_y=0$
\[mg=F_B=\rho_wV_{subm}g\] 
\[\rho_{copper}V_{shell}g=\rho_wV_{subm}g\] 
\[\rho_{copper}\frac{4}{3}\pi\left(R^3_1-R^3_2\right)=\rho_w\frac{1}{2}\left(\frac{4}{3}\pi R^3_1\right)\Rightarrow \ R^3_2=R^3_1\left(1-\frac{1}{2}\frac{\rho_w}{\rho_{copper}}\right)\] 
\[\therefore R_2={\left({\left(0.12\right)}^3\left(1-\frac{1}{2}\frac{1000}{8940}\right)\right)}^{\frac{1}{3}}=0.118\ {\rm m}= 11.8\ {\rm cm}\] 
Where $\frac{4}{3}\pi(R^3_1-R^3_2)$ is the volume of the spherical shell with outer radius $R_1$ and inner radius $R_2$. 
 

Here there is two buoyant force acting on the cylinder that must be balanced by the weight of the cylinder (since the cylinder is in equilibrium) i.e. $\Sigma F_y=0$
\begin{gather*}F_{Bo}+F_{Bw}=W_{cyl} \\ \\ \rho_oV_{sub,o}g+\rho_wV_{sub,w}g=m_{cyl}g \end{gather*} Where the above volumes are submerged volumes of the cylinder in the oil and water.  
Because the cylinder totally submerged in the fluids ,$V_{sub,o}+V_{sub,w}=V_{cyl}$
Therefore from the above equations, we obtain
\[\rho_o\left(V_{cyl}-V_{sub,w}\right)+\rho_wV_{sub,w}=m_{cyl}\] 
\begin{align*} \Rightarrow V_{sub,w}&=\frac{m_{cyl}-\rho_oV_{cyl}}{\rho_w-\rho_o}\\ &=\frac{6.5-725\times \left(\pi{\left(0.15\right)}^2\times 0.12\right)}{1000-725}\\ &=1.27\times {10}^{-3}\,\rm m^3 \end{align*} Use the definition of the volume of the cylinder to find its height \begin{align*} h_w&=\frac{V_{sub,w}}{\pi r^2}\\ \\ &=\frac{1.27\times {10}^{-3}}{p\times {\left(0.15\right)}^2}\\ \\ &=0.018\quad{\rm m}\end{align*} The sum of the submerged heights in the oil and water is the height of the cylinder.  \[h_w+h_o=H_{cyl}\Rightarrow h_o=0.12-0.018=0.102\ {\rm m}\] 
$h_o$ is the height of the cylinder floating in the oil. 
 

Because the flux passing through the artery is constant, so if we called $I_V$ as the volume flow rate then the Poiseuille's law states 
\[\Delta P=\frac{\eta L}{\pi r^4}I_V\] 
\[{\left(I_V\right)}_{obst}={\left(I_V\right)}_{free}\to \frac{\pi R^4_{obst}\Delta P_{obst}}{8\eta L}=\frac{\pi R^4_{free}\Delta P_{free}}{8\eta L}\] 
\[R^4_{obst}\Delta P_{obst}=R^4_{free}\Delta P_{free}\] 
\[\Delta P_{obst}=\Delta P_{free}{\left(\frac{R_{free}}{R_{obst}}\right)}^4=450{\left(\frac{1}{0.8}\right)}^4=1048\ {\rm Pa}\] 
\[\therefore \Delta P_{obst}-\Delta P_{free}=\left(1048-450\right)=648\ {\rm Pa}\] 
 

Because all the blood flowing through the aorta flows through the capillaries, then the volume flux is conserved. By definition, volume flux ($I_V$) is equal to the area of the tube times the speed of fluid through it.

Thus,in this case we have
\[{\left(I_V\right)}_{aorta}={\left(I_V\right)}_{capillaries}\] 
\[{\left(A\times v\right)}_{aorta}=N{\left(A\times v\right)}_{capillary}\] 
\[N_{capillary}=\frac{A_{aorta}\times v_{aorta}}{A_{capillary}\times v_{capillary}}=\frac{\left(\pi r^2_{aorta}\times v_{aorta}\right)}{\pi r^2_{capillary}\times v_{capillary}}\] 
\[=\frac{{\left(0.25\times {10}^{-6}\right)}^2\times 1}{{\left(5\times {10}^{-6}\right)}^2\times {10}^{-2}}=2.5\times {10}^7\ \] 
 

Use the Poiseuille's law to determine how the pressure drop relates to the flux $I_V$
\[I_V=Av=\frac{\pi R^4\Delta P}{8\eta L}\] 
\[v=\frac{\pi R^4\Delta P}{8\eta L}\frac{1}{A}=\frac{\eta R^4\Delta P}{8\eta L}\frac{1}{\eta R^2}=\frac{R^2\Delta P}{8\eta L}\] 
\[=\frac{{\left(2.4\times {10}^{-3}\right)}^2\times 450}{8\times 4\times {10}^{-3}\times 8.5\times {10}^{-2}}=0.45\frac{{\rm m}}{{\rm s}}\] 
 
 

(a) In an incompressible fluid, the volume flow rate is defined as $I_V=Av$, where $v$ is the velocity of the fluid perpendicular to the cross-sectional area $A$. Therefore, 
\begin{align*} I_V&=A_{disk}v\\ \\ &=\left(\pi r^2\right)v\\ \\ &=\pi{\left(0.002\right)}^2\left(1\right)\\ \\&=1.256\times {10}^{-3}\quad {\rm m^3/s}\end{align*}

(b) Use the Bernoulli equation, $P+\rho gh+\frac{1}{2}\rho v^2={\rm Constant}$ at the two points to find the velocity of the water at $H$, then given that the volume flow rate is constant determines the desired radius! At $h=H/2$, we have \[P_h+\rho gh+\frac{1}{2}\rho v^2=Constant\] and at $h=0$, \[P_0+\frac{1}{2}\rho v^2_0=Constant\]
\begin{gather*} \Rightarrow \ quad P_h+\rho gh+\frac{1}{2}\rho v^2=P_0+\frac{1}{2}\rho v^2_0 \\ \\  \Rightarrow\quad  v={\left(v^2_0+2gh\right)}^{\frac{1}{2}}={\left(1^2+9.8\times 0.5\right)}^{\frac{1}{2}}=\ 3.286\ \frac{{\rm m}}{{\rm s}}\end{gather*} Since the initial and final entry are open to air so the air pressure at these points is the same that is $P_h=P_0$. \begin{align*} {\left(I_V\right)}_{ini}&={\left(I_V\right)}_{fin} \\ \\ \Rightarrow \quad A_iv_i&=A_fv_f \\ \\ \Rightarrow \quad \left(\pi r^2_i\right)v_i&=\left(\pi r^2_f\right)v_f \end{align*} Thus, the radius of the column at that distance is \begin{align*} r_f&={\left(\frac{v_i}{v_f}\right)}^{\frac{1}{2}}r_i\\ &=(0.002)\sqrt{\frac{1}{3.286}}\\ \\&=1.1{\rm mm}\end{align*}

(a) The pressure at a point $h$ below the free surface of a fluid is given by $P=P_0+\rho gh$, where $P_0$ is the air pressure. Therefore \begin{align*} P&=1.013\times {10}^5+800\left(9.8\right)\left(2\right)\\ \\ &=1.17\times {10}^5\ {\rm Pa}\end{align*}

(b) Since the fluids are at equilibrium (motionless) so the pressure at the bottom of the tubes must be the same so \begin{gather*} P_1=P_2 \\ \\ \Rightarrow P_0+\rho_1gh_A=P_0+\rho_2gH+\rho_1gh \\ \\ \rho_1h_A=\rho_2H+\rho_1h \\ \\ \Rightarrow h_A=\frac{\rho_2}{\rho_1}H+h\end{gather*} Since stated that there is equal volume thus \begin{gather*} V_1=V_2 \\ \\ \Rightarrow HA=\left(h_A+h\right)A \\ \\ \Rightarrow H=h_A+h \\ \\  \Rightarrow\quad h_A=\frac{\rho_2}{\rho_1}H+h \\ \\ \xrightarrow{H=h_A+h} h_A=\frac{\rho_2}{\rho_1}H+\left(H-h_A\right) \\ \\ \Rightarrow h_A=\frac{1}{2}H\left(\frac{\rho_2}{\rho_1}+1\right) \end{gather*} Therefore, we get \[h_A=\frac{1}{2}\left(2\right)\left(\frac{800}{4800}+1\right)=1.167\ {\rm m}\] 
(c) The three forces acting on the ball are, buoyancy force $F_b$, Drag force $F_D$, and gravity. Terminal velocity means that the acceleration of the ball must be zero so the net force on the ball is zero i.e. $\Sigma F_{net}=0$
\begin{gather*} \Sigma F_y=0 \\ \\ F_D+F_b-mg=0 \\ \\ 6\pi\eta Rv+\rho_2V_{sub}g-mg=0 \\ \\ \Rightarrow v_T=\frac{m-\rho_2V_{sub}}{6\pi\eta R}g\end{gather*} Using the definitions of mass and density $m=\rho V$, we have \begin{align*} v_T&=\frac{\left(\rho_b-\rho_2\right)\left(\frac{4}{3}\pi R^3\right)}{6\pi\eta R}g\\ \\&=\frac{2}{9}\left(\rho_b-\rho_2\right)\frac{R^2g}{\eta}\\ \\ &=\frac{2}{9}\left(1200-800\right)\frac{{\left(0.01\right)}^2\left(9.8\right)}{0.03}\\ \\&=2.90\quad {\rm m/s}\end{align*} 

(a) The weight of an object that is suspended from a string (scale) equals the tension in the string i.e. $\Sigma F_y=0\Rightarrow T-W=0$
\[\Rightarrow \ \ T=mg=2.26\times 9.8=26.1\ {\rm N}\] 
(b) The volume of the sphere is defined as \begin{align*} V&=\frac{4}{3}\pi r^3\\ \\ &=\frac{4}{3}\pi{\left(3.34\times {10}^{-2}{\rm m}\right)}^2\\ \\&=1.56\times {10}^{-4}\quad {\rm m^3}\end{align*}

(c) There is a buoyancy force acting on the objects in the fluids. This force equals the weight of the displaced fluid.

Therefore, \begin{align*} \Sigma F_y=0 &\Rightarrow \ T+F_b-mg=0\\ &\Rightarrow T=mg-\rho V_{sub}g\\ &=26.1-\left({10}^3\times 1.56\times {10}^{-4}\right)\left(9.8\right)\\ &=24.57\ {\rm N} \end{align*} Where $V_{sub}$ is the amount of volume that submerged in the water.
 

(a) The pressure at depth $h$ below the surface of a fluid is given by $P=P_0+\rho gh$ where $P_0$ is the air pressure. So 
\begin{align*} P_2&=P+\rho gh\\ \\&=\left(2\times {10}^5\right)+1000\times 9.8\times 2.2\\ \\&=2.21\times {10}^5\ {\rm Pa}\end{align*} 
(b) The force acting on the open end of the tube due to the tank ($P_2A$) and air ($P_0A$) must be balanced with the weight of $W$. So using the definition of pressure we have \begin{align*} F_{net} &=A\Delta P=A\left(P_2-P_0\right)=\pi{\left(\frac{d}{2}\right)}^2\left(P_2-P_0\right)\\ &\Rightarrow F_{net}=\pi{\left(\frac{0.03}{2}\right)}^2\left(2.21-1\right)\times {10}^5=86.2\ {\rm N} \end{align*} Thus,\[W=F_{net}=86.2\, {\rm N}\to {\rm \ \ W=mg}\Rightarrow {\rm m=8.8\ kg}\] 
(c) Use the Bernoulli's equation at points A and B 
\[P_A+\rho gh_A+\frac{1}{2}\rho v^2_A=P_B+\rho gh_B+\frac{1}{2}\rho v^2_B\] 
Let the base to be as shown in the figure, so $h_A=h_B=0=v_B$

\begin{align*} &\Rightarrow \frac{1}{2}\rho v^2_A=P_B-P_A\\ &\Rightarrow z_A=\sqrt{\frac{2\left(P_B-P_A\right)}{\rho}}=\sqrt{\frac{2\left(2.21-1\right)\times {10}^5}{1000}}\\ &=15.55\quad {\rm m/s} \end{align*}
(d) The flow rate is defined as $Q=Av$ so 
\begin{align*} Q &=\pi r^2v=\pi{\left(\frac{d}{2}\right)}^2v\\ &=\pi{\left(\frac{0.03}{2}\right)}^2\times 15.6\\ &=0.099\quad {\rm m^3/s} \end{align*}

(a) Two forces acting on the ice block, buoyancy $F_B$, and gravity (which is the weight force). These forces must be in balance (the ice block does not accelerate). Recall from buoyant force formula, that $F_B=\rho_f V_{dis}g$. Thus, balancing these two forces $F_B=W$, get
\begin{align*} \rho_wV_{sub}g=\underbrace{\rho_{ice}V_{ice}g}_{W}\Rightarrow \rho_wx_0A=\rho_{ice}As\\ \Rightarrow x_0=\frac{\rho_{ice}}{\rho_w}s=\frac{917}{1000}\left(1\right)=0.917\ {\rm m}\\ \therefore s-x_0=1-0.917=8.3\ {\rm cm} \end{align*}
In the above, $V_{sub}$ is the submerged volume of the ice in the water and $A=s^2$ is the cross-sectional area of the ice block.
(b) In this case, since the ice block is moved away from the equilibrium so there is a net force on the block as
\begin{align*}F_{net}&=F_B-W\\ \\ &=\rho_wA\left(x_0+x\right)g-\rho_{ice}Asg\end{align*} From part (a) we know that $\rho_wx_0A=\rho_{ice}As$ so 
\[F_{net}=-\rho_ws^2gx\]  We have inserted a minus to show that the force and displacement are in opposite directions. In the other words $F_{net}$ is a restoring force like the spring force in Hook's law. 
(c) As mentioned in part (b), the $F_{net}$ is similar to the spring force $F_S=-kx$ so
\begin{align*}F_{net}&=F_S\\ \\ -\rho_ws^2gx&=-kx\\ \\ \Rightarrow \quad k&=\rho_ws^2g\end{align*} In the springs the angular frequency is defined as 
\begin{align*} \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{\rho_ws^2g}{\rho_{ice}s^3}}=\sqrt{\frac{\rho_wg}{\rho_{ice}s}}\\ \\ \therefore \quad \omega=\sqrt{\frac{1000\times 9.8}{917\times 1}}=3.26\quad {\rm rad/s} \end{align*} Where $m=\rho_{ice}V=\rho_{ice}s^3$ is the mass of the ice block.
 

(a) The pressure at depth $h$ in a fluid with density $\rho$ is given by $P=P_0+\rho gh$. In this problem, the distance between the bottom of the pipe and the free surface of water is \begin{align*}h_{tot}&=h_{tank}+l_{pipe}\,{\sin  60^\circ}\\ &=5+100\,{\sin 60^\circ}\\ &=91.6\quad {\rm m} \end{align*} Therefore, \begin{align*} P_{Bottom} &=P_0+\rho gh_{tot}\\ &=1.013\times {10}^5+\left({10}^3\right)\left(9.8\right)\left(91.6\right)\\ &=9.99\times {10}^5\quad {\rm N/m^2} \end{align*} (b) Use the Bernoulli's equation between the 6${}^{th}$ floor and end of the pipe.

Since the faucet is open to atmosphere so $P_6=P_0=1\,{\rm am}$ 
\begin{gather*} P_0+\frac{1}{2}\rho v^2_0+\rho gh_0=P_6+\frac{1}{2}\rho v^2_f+\rho gh_6 \\ \\ g(\underbrace{h_0-h_6}_{\Delta h})=\frac{1}{2}v^2_f \\ \\ \Rightarrow v_f=\sqrt{2g(h_0-h_6)}=\sqrt{2\times 9.8\times (91.6-41)}=31.49\quad {\rm m/s}\end{gather*} Where $\Delta h$ is the distance shown in the figure. (Since, as mentioned, the water in the pipe is in steady state, $v_0=0$)
 

Notes: a floating object displaces its own weight of water and a submerged object displaces its own volume of water. Therefore, in the case that the rock is in the boat the volume of the displaced water is 
\[\Delta V_{in}=\frac{m_{rock}}{\rho_w}=\frac{100}{1000}=0.1\ {{\rm m}}^{{\rm 3}}\] 
And when the rock is thrown into the water 
\[\Delta V_{out}=\frac{m_{rock}}{\rho_{rock}}=\frac{100}{5\times {10}^3}=0.02\ {{\rm m}}^{{\rm 3}}\] 
Thus the total change in the water level is $\Delta V_{tot}=\Delta V_{in}-\Delta V_{out}$.
\[\Delta {{\rm V}}_{{\rm tot}}{\rm =0.1-0.02=0.08\ }{{\rm m}}^{{\rm 3}}\] 
Using the definition of the volume of a rectangular shape, we obtain the change in the water level of the pool
\[\Delta V_{tot}=A\Delta h\Rightarrow \Delta h=\frac{\Delta V_{tot}}{A}=\frac{0.08}{50}=1.6\times {10}^{-3}\ {\rm m}\] 
 

(a) Static equilibrium reaches when the buoyant and gravity force acting on the object are balanced by each other. By definition, the buoyant force is equal to the weight of the fluids displaced by a submerged object in that fluid. Let the amount of the stick meter above the water be $\Delta L$ and the total length of it be $L$. Therefore,
\begin{align*} F_g=F_B\to \underbrace{\rho V}_{M}g=\underbrace{\rho_wV_{sub}}_{M_{sub}}g\\ \rho ALg=\rho_wA\left(L-\Delta L\right)g\Rightarrow \Delta L=\frac{\rho_w-\rho}{\rho_w}L\\ \Delta L=\frac{1000-700}{1000}\left(1\right)=0.3\ {\rm m=30\ cm}\\ \end{align*} Where $A$ is the cross-sectional area of the meter stick and $M_{sub}$ is the amount of the meter stick submerged in the water. 

(b) Since the meter stick moved away from the equilibrium position so there is a net force on it.

If the submerged height is greater than the equilibrium height ($L-\Delta L$) then $F_B>mg$ and the object accelerates upward or vice versa.

Let us assume the $y$ axis is in the upward direction. Since at the instant that the meter stick is pulled up by $y(t)$, the buoyant force is smaller than the weight of the meter stick so there is net force toward the $-y$ axis. Using Newton's second law and applying Archimedes principle formula, we obtain \begin{align*} \Sigma F=Ma\to F_B\hat{j}+W\left(-\hat{j}\right)=Ma\left(-\hat{j}\right)\\ \rho_wA\left(L-\Delta L+y(t)\right)g-\rho ALg=-\rho AL\frac{d^2y(t)}{dt^2}\\ \end{align*} From part (a), we can rearrange the relation above as follows \[\Rightarrow \rho AL\frac{d^2y(t)}{dt^2}=-\rho_wAy(t)g\Rightarrow \frac{d^2y(t)}{dt^2}=-\frac{\rho_wg}{\rho L}y(t)\]  This is similar to the equation of simple harmonic motion $\frac{d^2x}{dt^2}+\omega^2x=0$ where $\omega$ is the angular velocity and is related to the period of the motion by $\omega=2\pi/T$ so 
\[T=2\pi\sqrt{\frac{\rho L}{\rho_wg}}=2\pi\sqrt{\frac{700\times 1}{1000\times 9.8}}=1.68\ {\rm s}\] 
(c) In above the equation of motion of the meter stick is as the harmonic motion that is $y\left(t\right)=A\,{\sin \omega t\ }$, where $A$ is the maximum displacement of the body from equilibrium position that in this case is $10\, {\rm cm}$. By taking time derivative of $y(t)$, we can find the equation of the velocity of the object as $v\left(t\right)=A\omega\,{\cos \omega t\ }$ where $A\omega$ is the maximum speed. So in this case \begin{align*} v_{max}&=A\omega=0.1\times \sqrt{\frac{\rho_wg}{\rho L}}\\ &=0.1\times \sqrt{\frac{1000\times 9.8}{700\times 1}}\\ &=0.374\quad {\rm m/s} \end{align*}

(a) Since there is a vacuum between container and shield, so the heat must be transferred through the valve via radiation. Recall that Stefan-Boltzmann law states that the power radiated by an object with absolute temperature $T$ is given by $P_r=e\sigma AT^4$, where $\sigma$ is the Stefan's constant and $e$ is emissivity of the radiating surface. The net heat absorbed by the helium is 
\begin{align*} P_{net}=\frac{Q_{abs}}{t}=e\sigma A\left(T^4_{shield}-T^4_{He}\right)\\ \Rightarrow Q_{abs}=4\pi r^2e\,t\sigma\left(T^4_{shield}-T^4_{He}\right) \end{align*} \begin{align*} Q_{abs}&=4\pi{\left(0.1\right)}^2(1)(60\times 60)(5.67\times {10}^{-8})({77}^4-{\left(4.2\right)}^4)\\ &=901.68\ {\rm J} \end{align*} This absorbed heat causes $m\, {\rm g}$ helium boils away through valve so  \begin{align*}Q_{abs}&=mL_V\\ \\ \Rightarrow m&=\frac{901.68}{21000}\\ \\&=0.042\ {\rm kg}\end{align*} Use the formulae of density to determine the desired volume. \[V=\frac{m}{\rho}=3.43\times {10}^{-4}\, {\rm m^3}\]