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1. As a prank, a student's friends put a harness on him and hook him to a suction cup on the ceiling. If the student has a mass of 70 kg, what must the area of the suction cup be to support him?
   

   The "suction force" is supplied by atmospheric pressure on the area of the cup, F = P_ATMA. This force must equal the weight W = mg of the student. So

   A	= mg / PATM
   	= (70 kg)(9.81 m/s2) / 1.01 × 105  Pa
   	= 0.0068 m2

   This is less than the area of a 9.0 cm × 9.0 cm square!



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2. The Mindanao Trench, off the coast of the Philippines, is the world's deepest at 11,515 m. What is the pressure at this depth? How can sea creatures live at this depth? Why is it so dangerous for submersibles to go to this depth? Sometimes biologists catch fish at this depth and bring them to the surface. If they raise the fish too quickly, they get instant sushi. How come?

   Pressure in a liquid is governed by the formula P = P₀ + ρgh. The pressure at the surface, p₀, is just atmospheric pressure and the density of seawater is ρ = 1024 kg/m³ (according to the table of densities in the text). Thus

   p = 1.01 × 10⁵ Pa + (1024 kg/m³)(9.81 m/s²)(11,515 m) = 1.16 × 10⁸ Pa.

   This is over 1100 times atmospheric pressure.

   Sea animals survive at this depth as long as their internal fluids are at the same pressure as the surrounding seawater. Only a difference of pressure would cause a problem.

   Submersibles have a difficult time getting to this depth because people breath air, a gas. At such enormous pressures air would be difficult to breath, it may even become a liquid. As a result the submersibles innards are at a much lower pressure than the outer seawater and it is this tremendous difference in pressure that is dangerous.

   Fish brought up from the bottom too fast suffer the reverse problem. There is a difference between their inner pressure and the pressure of the seawater near the surface.

   

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3. A tube contains 12.2 cm of water. A second tube contains 15.8 cm of an unknown fluid. The pressure at the bottom of both tubes is the same but unknown. What is the density of the unknown fluid?

   The pressure at any depth of fluid is given by p = p₀ + ρgh. Since we have two fluids we write this equation down twice, one for each case

   pW = p0W + ρWghW	(1)
   pU = p0U + ρUghU	(2)

   Both have the same surface pressure, since both are exposed to air, so p_0W = p_0U. We are also told p_W = p_U. Thus (1) and (2) combine to become

   ρ_Wgh_W = ρ_Ugh_U.

   Eliminating the common factor g and solving for ρ_U, we get

   ρ_U = ρ_Wh_W/h_U = (1000 kg/m³)(12.2)/(15.8) = 772 kg/m³.



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4. An 80-kg person lies on a 1.5-m wide, 2.0-long, and 12-cm thick water mattress. Assuming that the deformation of the top of the mattress is negligible, what is the pressure at the bottom of the water?
   

   The pressure at the bottom of the mattress when the person is not there is given by P =P_ATM + ρgh, where ρ is the density of water and and h is the thickness of the mattress. This yields P = 1.01 × 10⁵  Pa + (1000 kg/m3)(9.81 m/s²)(.12 m) =  1.022  × 10⁵  Pa.

   The extra pressure from the weight of the person, P_extra = mg / A = (80 kg)(9.81 m/s²) / (1.5 m)(2.0 m) = 262 Pa, is transmitted through out the fluid. So the total  pressure at the bottom is 1.025  × 10⁵  Pa.

   

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5. A patient with no arms visits the doctor. The doctor places a blood pressure cuff around the patient's ankle and measures BP to be 235/190. If the patient's ankle is 1.42 m below where his elbow would have been, what would the reading have been at elbow height?
   

   We need to make the assumption that blood in the body can be treated as a static fluid. Blood isn't static so this answer is only approximate. The pressure at the ankle is higher, since in is farther down the liquid, by ΔP = ρ_bloodgh, where h is the change in height bewteen elbow (heart) and ankle. Using the value for the density of whole blood from the text and calculating, we find ΔP = (1050 kg/m³)(9.81 m/s²)(1.42 m) =14626 Pa. We need to convert to mm Hg usuing the relationship that 760 mm Hg = 1.01 × 10⁵  Pa. So ΔP = 14626 Pa ×  (760 mm Hg / 1.01 × 10⁵  Pa) = 110 mm Hg. So the BP at the elbow would be approximately 125/80.

   

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6. If someone your size sat on your chest you would have difficulty breathing. Now imagine you go swimming using a long straw as a snorkel. Estimate how deep underwater you could swim using the straw?
   

   The following answer is approximate because we will use estimates and so the values below may differ from your answer.

   Let us assume a 90-kg person and a rectangular chest of 30 cm by 20 cm. The extra pressure on your chest is then P = mg / A = (90 kg)(9.81 m/s²) / (0.3 m)(0.2 m) = 14715 Pa. This pressure is besides the 1 Atmosphere of pressure on your chest. Now the pressure underwater is P =P_ATM + ρgh, where ρ is the density of water and h is your depth under water. 

   Now P_ATM is common to both so we have ρgh = 14715 Pa or h = (14715 Pa) / (1000 kg/m³)(9.81 m/s²) = 1.5 m! You cannot physically breath at a greater depth.
   

   This is why scuba divers use pressurized air − so that the pressure in your lungs matches the pressure from the surrounding water.

   

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7. A hydraulic lift is used to raise a 2000-kg car 2.0 m. The base under the car has an area of 9.0 m². The other side of the area has an area of 0.50 m². What force must be exerted at that end to lift the car? What distance must the force move through?
   

   We have the relationship F_out/F_in = A_out/A_in. Rearranging F_in = A_in F_out /A_out = ( 0.50 m²)(2000 kg)(9.81 m/s²) / (9.0 m²) = 1090 N (equivalent to a 111 kg block).

   The work in must equal the work out, F_in d_in = F_out d_out.  Solving for d_in yeilds  d_in = F_out d_out / F_in = (2000 kg)(9.81 m/s²) (2.0 m) / (1090 N) = 36 m.

   

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8. A sightseeing balloon can lift a total mass of 500 kg including the balloon material, basket, and passengers but excluding the helium gas. What volume of helium is needed to lift the balloon?

   When doing bouyancy problems, it is best to consider a free-body diagram. Note that the helium-filled ballooon shows two weights, the weight of the 500-kg load and the unknown weight of helium gas. The bouyant force acts upwards.

   

   The equation of the diagram is F_B = w_Load + w_He. We know F_B = ρ_{fluid displaced}V_{fluid displaced} g. The weight of the load is w_Load = (500 kg) × (9.81 m/s²) = 4905 N. Since we don't know the mass of helium gas and we only want the volume anyway, w_He = ρ_HeV_He g. So our force equation is

   ρ_{fluid displaced}V_{fluid displaced}g = 4905 N + ρ_HeV_He g.

   Now the balloon is totally immersed in air so V_{fluid displaced} = V_He. Our force equation simplifies to

   ρ_{fluid displaced}V_He g = 4905 N + ρ_HeV_He g.

   We collect the tems with volume in them to one side. This yields

   (ρ_{fluid displaced} − ρ_He)V_He g = 4905 N.

   Rearranging

   V_He = (4905 N) / (ρ_{fluid displaced} − ρ_He)g.

   The displaced fluid is air which has a density of 1.29 kg/m³ while the density of helium is 0.179 kg/m³ (see table in textbook). Using these values V_He = 450 m³.

   

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9. A large balloon, filled with an unknown gas, exerts a 5.25 N force on a spring. If the volume of the balloon is 1.70 m³, what is the density of the gas. Take the density of air to be 1.21 kg/m³.

   

   The problem mentions a force, the spring force F_S, and we know that the acceleration is zero since the balloon is not moving. This suggests that we apply Newton's second law. The others forces acting are the weight of the balloon, F_W, and the buoyancy force, F_B.

   

   Apply Newton's Second Law, we find

   F_B - F_S - F_W = 0.           (1)

   The buoyant force is equal to the weight of the displaced air, m_airg. In turn the mass of the displaced air is related to the volume of the balloon, F_B = ρ_airVg. The weight of the gas is F_W = ρ_gasVg. Thus equation (1) becomes

   ρ_airVg - F_S - ρ_gasVg = 0.           (2)

   Solving for ρ_gas,

   ρ_gas = ρ_air - F_S/Vg = 1.21 kg/m³ - (5.25 N)/(1.70 m³)(9.81 m/s²) = 0.895 kg/m³ .

   The density of the gas is 0.895 kg.m³.

   

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10. A class conducts a number of experiments on the behaviour of fluids using an instructor, a rope, a force scale, and a very large graduated cylinder. The cylinder is 3.00 m deep and has a diameter of 2.00 m. The 80-kg instructor is completely immersed in the water in the cylinder and tied to a force scale at the bottom of the cylinder as shown in the diagram below. The water in the cylinder rises 2.65 cm after the instructor is immersed.
    (a) How dense is the instructor?
    (b) What is the reading on the scale?
    

    (a) We know that density is given by ρ = M/V. We are given the mass of the instructor so we need to know his volume. We also know that the volume of fluid displaced by a submerged object equals the volume of the object − Archimedes' Principle. For a cylinder

    V = π(d/2)²h = π(1 m)²(0.0265 m) = 0.08325 m³ .

    Hence the density of the instructor is

    ρ = (80 kg) / (0.08325 m³) = 960.9 kg/m³ .

    This is less than the density of water, 1000 kg/m³, which is why the instructor is tied to the bottom to keep him from floating to the top.

    (b) The scale reading is a force, call it F_Scale, and we know that the acceleration is zero since the instructor is not moving (if the students wait long enough anyway). This suggests that we apply Newton's second law. The others forces acting are the weight of the instructor, F_W, and the buoyancy force, F_B.

    

    Apply Newton's Second Law, we find

    F_B - F_Scale - F_W = 0 . (1)

    The buoyant force is equal to the weight of the displaced water, m_waterg. In turn the mass of the displaced water is related to the volume of the instructor, F_B = ρ _waterVg. The weight of the instructor is F_W = ρ Vg. Thus equation (1) becomes

    ρ _waterVg - F_Scale - ρ Vg = 0. (2)

    Solving for F_Scale,

    F_Scale = (ρ _{water− ρ} )Vg = (1000 kg/m³ − 960.9 kg/m³)( 0.08325 m³)(9.81 m/s²) = 31.9 N.



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11. A thin steel slab is 1.0 m × 1.0 m × 2.0 mm. A thicker wooden slab, t = 5.0 mm, of the same area is stuck to the steel slab. What is the density of the combined slab if the density of the wood is 0.5 × 10³ kg/m³? How thick would the wood slab have to be for the combined steel and wood slab to just float on water?

    Density is defined as mass over volume, ρ = M / V. We need to determine the mass of the combined slab using the density of iron given in the textbook, ρ_Fe = 7.8 × 10³ kg/m³, and the volumes described in the question. The total volume of the slab is the combined volumes of the the individual pieces.

    ρ	= [ρFeVFe + ρwoodVwood] / [VFe + Vwood]
    	= [(7.8 × 103 kg/m3)(1.0 m)(1.0 m)(0.002 m) + (0.5 × 103 kg/m3)(1.0 m)(1.0 m)(0.005 m)] / [(1.0 m)(1.0 m)(0.002 m) + (1.0 m)(1.0 m)(0.005 m)]
    	= 2.59 × 103 kg/m3

    A object float when is density is just barely less than that of water, 1000 kg/m³. We can thus use the above equation replacing ρ with the density of water and replacing 5 mm by an unknown thickness t.

    1 × 10³ kg/m³ = [(7.8 × 10³ kg/m³)(1.0 m)(1.0 m)(0.002 m) + (0.5 × 10³ kg/m³)(1.0 m)(1.0 m)t] / [(1.0 m)(1.0 m)(0.002 m) + (1.0 m)(1.0 m)t]

    The power of 10³ kg/m³ can be cancelled from each side. As well, (1.0 m)(1.0 m) is in the top and bottom of the fraction and may be cancelled. We are left with

    1 = [(7.8)(0.002 m) + (0.5)t] /[(0.002 m) + t]

    Cross-multiplying yields

    (0.002 m) + t = (7.8)(0.002 m) + (0.5)t

    Collecting terms with t in them on the right gives

    t − (0.5)t = (7.8)(0.002 m) − (0.002 m)

    Solving, we find the necessary thickness of wood to be t = 0.0272 m = 27.2 mm.



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12. An aluminum box is 1.00 m on each side with only air inside. The aluminum is 4.0 mm thick. What is the density of the box? How deep will it sink if placed in water?

    A box has six sides, so the volume of aluminum is V = 6 × (1.00 m)(1.00 m)(0.0040 m) = 0.0240 m³. Using the density given in the text book, the mass of aluminum is m = ρV = (2700 kg/m³)(0.0240 m³) =  64.8 kg. The cubic metre of air in the box has a mass of 1.29 kg. The total mass of the box is therefore 66.09 kg. The density of the box is thus ρ_Al = m / V = ( 66.09 kg) / ( 1 m³)  = 66.09 kg/m³. To find the depth of the box in water, recall first that the density of the box is less than that of water so it will float. Next recall Archimedes' Principle that the to float, the box will displace a volume of water with the equivalent weight (or mass). Mathematically this is ρ_waterV_water = m_box. As a result the volume of water displaced is V_water = m_box  / ρ_water  = ( 66.09 kg) / ( 1000 kg/m³) = 0.06609 m³. The shape of the water is a 1.00 m by 1.00 m bottom with a height of h. We can solve for the height, so the box sinks 0.066 m or 6.6 cm.

    

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13. Water leaves the exterior faucet of your house at 30 ℓ/min though a 5/8" (1.59 cm) diameter faucet. A garden hose of the same radius but with a tapered nozzle is attached. The nozzle has a exit diameter of 3 mm. What is the speed of the water leaving the nozzle?

    We are given a volume flow rate, Q = 30 ℓ/min = ΔV/ Δt  = AΔL/ Δt  =  Av, where A is the cross-section al area of the hose and  v is the initial speed of the water. Note that one litre is  a volume  of  a cube of side length 10 cm, so it is  a volume of (0.1 m)³ =  0.001 m³.  The initial speed of the water leaving the house is thus 

    v = Q/A = (30 ℓ/min)×(0.001 m³ / ℓ) × (1 min / 60 s) / (π [(0.0159 m)/2]²) = 2.52 m/s.

    We can now use the Equation of Continuity to find the speed of the water out of the nozzle, 

    v_out = A_in v_in / A_out  = (π [(0.0159 m)/2]²) × 2.52 m/s / (π [(0.003 m)/2]²) = 70.8 m/s.
    



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14. A barrel 1.75 m high is filled to the brim with water. How far from the top would a hole have to be punched so that the stream of water would land 0.85 m from the barrel? Hint you need to know the speed at which the water leaves the barrel and then treat the water as a projectile.

    

    To find the speed that the water leaves the barrel, we use Bernoulli's Equation,

    p₀ + ½ρ(v₀)² + ρgy₀ = p₁ + ½ρ(v₁)² + ρgy₁.           (1)

    The top of the barrel is exposed to the atmosphere, as is the hole, so both p₀ and p₁ are equal to atmospheric pressure. The water at the top of the barrel is barely moving compared to the water coming out of the hole, so v₀ = 0. Thus equation (1) becomes

    ρgh₀ = ½ρ(v₁)² + ρgh₁.           (2)

    Dropping the common factor ρ, setting h = y₀ - y₁ and rearranging, yields

    v₁ = [2gh]^½.

    Surprisingly, this is the same result as for dropping a stone the same distance.

    Next we treat the water as a projectile leaving a surface horizontally. As always, we separate the x and y components.

    i	j
    Δx = 0.85 m	Δy = h-H
    v0x = v1 = [2gh]½	v0y = 0
    ax = 0	ay = -g
    t = ?

    We get a kinematic equation from each column,

    Δx = [2gh]^½t and,

    h - H = -½gt² .

    Eliminating t from the above yields,

    h - H = -(Δx)²/4gh.

    Cross-multiplying yields an equation quadratic in h,

    h² - hH + ¼(Δx)² = 0 .

    The given values make this

    h² - 1.75h + 0.180625 = 0 .

    The roots are both positive, h = 1.64 m and h = 0.11 m.

    Another surprise, there are two possible answers! This occurs because water coming out near the top, h = 0.11 m, is travelling more slowly that water from a hole near the bottom, but being higher up the slow water has more time in the air.

    
    

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15. A tube of fluid is shown in the diagram below. There is a hole on the side through which the fluid is pouring out. The hole has an area A. The mouth of the tube has a sealed gasket of area 3A. A gasket is a lid which is free to move up and down. The seal around the gasket keeps water from leaking out. A force F is applied to the gasket. The fluid is currently at a height H while the hole is at height h. Find an expression for the velocity of the fluid leaving the hole. Take atmospheric pressure to be P_atm.

    Since we are looking for the speed that the water leaves the barrel, we use Bernoulli's Equation,

    p_top + ½ρ(v_top)² + ρgy_top = p_side + ½ρ(v_side)² + ρgy_side. (1)

    The top and hole of the barrel are exposed to the atmosphere, so both p_top and p_side depend on P_atm. However the top also has an extra force F pushing down on it. The total pressure at the top is therefore p_top = P_atm + F/3A, while the pressure at the hole is just p_side = P_atm.

    The speed of the water at the top of the barrel is related to the speed of the water coming out of the hole by

    v_topA_top = v_sideA_hole ,

    Since we are given these areas, this reduces to 3v_top = v_side.

    Equation (1) becomes

    P_atm + F/3A + ½ρ(¹/₃v_side)² + ρ gH = P_atm + ½ρ(v_side)² + ρgh. (2)

    Cancelling the common term P_atm, and collecting terms with v_side on the right hand side yields,

    F/3A + ρg(H− h) = ½ρ(1 − ¹/₉)(v_side)² .

    The equation for the speed of the water leaving the hole is thus

    v_side = (³/₂)[F/3ρA + g(H − h)]^½ .


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16. The 80-kg owner of a 12-cm thick 1.5 m × 2.0 m water mattress wants to drain the water. He attaches a 5/8" (1.59 cm) diameter garden hose to the mattress and hangs the other end out the window so that it lies on the ground 5.0 m below.
    a) How fast does water exit the hose? If the flow remains constant how long does it take to drain the full water mattress?
    b) To speed things up the owner lies on the mattress. Assuming that the deformation of the top of the mattress is negligible, how fast does water now exit the hose? If the flow remains constant how long does it take to drain the full water mattress?

    The change in height and speed of water suggests that we should use Bernoulli's Equation.

    P_mattress + ½ρ(v_mattress)² + ρgy_mattress = P_ground + ½ρ(v_ground)² + ρgy_ground.       [1]

    Both the mattress and hose opening at ground level are exposed to air, so P_mattress = P_ground = P_Atm. These will just cancel leaving

    ½ρ(v_mattress)² + ρgy_mattress = ½ρ(v_ground)² + ρgy_ground.

    Or, eliminating the common constant ρ,

    ½(v_mattress)² + gy_mattress = ½(v_ground)² + gy_ground.

    Now the surface area of the mattress is much greater than the surface area of the hose opening, so we can safely assume v_mattress = 0. Let's take the ground as zero height, y_ground = 0. That makes y_mattress = 5.0 m (ignoring the small thickness of the mattress).

    g(5.0 m) = ½(v_ground)².

    We thus find v_ground = 9.90 m/s.

    To find out how quickly the mattress empties remember that the volume flow rate, Q = V/t = Av. Here A is the surface area of the hose opening. Thus Q = (π[0.0159 m)/2]²)(9.90 m/s) = 0.00197 m³/s. We started with a volume of water in the mattress of V = (1.5 m) × (2.0 m) × (0.12 m) = 0.36 m³. The time required is t = V / Q = (0.36 m³) / (0.00197 m³/s) = 183 s or just over three minutes.

    When the person is on the bed, there is an extra pressure of P_person = W / A = (80 kg)(9.81 m/s²) / (1.5 m)(2.0 m) = 261.6 Pa. Our equation for this case is

    261.6 Pa + ½ρ(v_mattress)² + ρgy_mattress = ½ρ(v_ground)² + ρgy_ground.

    Our assumption that v_mattress = 0 is still valid. Note that ρ = 1000 kg/m³. The equation becomes

    261.6 Pa + (1000 kg/m³)g(5.0 m) = ½(1000 kg/m³)(v_ground)².

    Isolating velocity

    (2)(261.6 Pa)/(1000 kg/m³) + (2)g(5.0 m) = (v_ground)².

    We solve and find v_ground = 9.93 m/s.

    The new flow rate is Q' = (π[0.0159 m)/2]²)(9.90 m/s) = 0.001972 m³/s. It takes time t = V / Q = (0.36 m³) / (0.001972 m³/s) = 182.6 s or just under three minutes.



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17. In storms, it is sometimes said that the winds blew the roof off a house. Explain what really happens. 

    The fast winds travelling over the roof of the house create a low pressure according to Bernoulli's Principle. The air in the house is at higher pressure, so the air in the house blows the roof off.
    



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18. If blood flow to the stomach goes from 1.0 L/min to 5.0 L/min to aid digestion, by how much must the arteries dilate? Assume all other quantities stay the same.

    We are given volume rates of flow. Recall that one litre is the same as 0.001 m³. The fact that the question gives flow rate and is interested in the radius of the arteries, suggest that we use Poiseuille's Law (which was in fact developed to describe blood flow). We can use the value of viscosity of blood given in the textbook, if necessary, η = 1.5 × 10^-3 Pa·s. Now Q₁ = πR₁⁴ΔP / 8ηL, where ΔP, η, and L will stay the same. So Q₂ = πR₂⁴ΔP / 8ηL. The ratio of these two equations is Q₂ / Q₁ = R₂⁴ / R₁⁴. We have the values of Q_{1 and} Q₂, so R₂ / R₁ = 5^¼ = 1.50. So the arteries have to be 50% bigger. 



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19. A hypodermic syringe contains 15 cm³ of a solution with a viscosity of 2 × 10^-3 Pa·s. The syringe has a diameter of 1.0 cm and the needle had an inner radius of 5.0 × 10^-4 m. The needle is 3.0 cm long. The solution is injected into a patient's vein (vein pressure is 3 mm Hg). How hard must the plunger be pressed to inject the solution in 5.0 seconds?

    Poiseuille's Equation would apply to the needle, Q = πR⁴ΔP / 8ηL, where L is the length of the needle and R is it's radius. The volume flow rate is Q =  (15 cm³)×(1 m / 100 cm)³ / (5.0 s) = 3.0  × 10^-6 m³/s. The pressure difference, ΔP, is between the pressure exerted on the syringe plunger given by F / A snd the pressure at the vein P_vein = 3 mm Hg = (3 / 760) × (101 000 Pa) =  398.7 Pa. Note that the plunger pressure and the vein pressure are both guage pressures. To get the actual pressure we would need to add P_ATM to both, but that is necessary since a difference in pressures is the same as the difference in gauge pressures. Rearranging Poiseuille's Equation yields ΔP = 8ηLQ / πR⁴ or (F/A − P_vein ) = 8ηLQ / πR⁴. We can further rearrange to get F by itself
    

    F	= A( Pvein + 8ηLQ / πR4)
    	= ¼π(0.01 m)2 × [398.7 Pa + 8(2 × 10-3 Pa·s)(0.03 m)(3.0  × 10-6 m3/s) / π(5.0 × 10-4 m)4]
    	= ¼π(0.01 m)2 × [398.7 Pa + 7333.9 Pa]
    	= 0.61 N



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20. Another problem with attempting to use a long straw to swim deeply under water is that you must blow out the stale air to get fresh air in. What is the smallest radius of snorkel you can use if you need 6.0 ℓ/min of fresh air? Assume your breathing pressure is 100 Pa.

    We are given the volume flow rate Q which we must convert to m³/s, Q = (6.0 ℓ/min)×(0.001  m³/ ℓ) / (60 s) = 1.0  × 10^-4 m³/s. The coefficient of the viscosity of air is given in the textbook, η = 0.018 × 10^-3 Pa·s. We can take Poiseuille's Equation, and rearrange for R, R = [8ηLQ / πΔP]^¼, where L is the snorkel length of 1.5 m we found in the last question and ΔP = 100 Pa, the breathing guage pressure.Evaluating  R = [8(0.018 × 10^-3 Pa·s)(1.5 m)(1.0 × 10^-4 m³/s) / π(100 Pa)]^¼ = [6.88 × 10^-11 m⁴]^¼ = 4.84 × 10^-3 m = 4.84 mm. The diameter of the straw needs to be at least 9.7 mm which is much bigger than most straws. 

Questions?mikec@kwantlen.bc.ca