You are invited to a pool party with Alice (54kg), Bob (75kg), and Charlie (82kg

Question

You are invited to a pool party with Alice (54kg), Bob (75kg), and Charlie (82kg), but you can't stop thinking about physics. You estimate that the swimming pool has a surface area of 60m^2, and is filled with chlorinated-water (mass density = 1015kg/m^3) such that the depth is 1.8m. a) What are the gauge pressures at the surface of the pool and at the bottom of the pool? b) What is the magnitude of the total force on the pool floor? c) Bob and Charlie are arguing over who gets to use an inflatable tube. It has a mass of 1.64 kg when deflated, and when inflated with pressurized air of density 2.0 kg/m^3, it has a volume of 0.08 m^3. Calculate the mass of the heaviest person that this inflatable tube can support without sinking completely in the pool, and see if you can settle Bob and Charlie's argument. d) Alice and Bob jump in the pool - Alice jumps feet first and takes 600 ms (milliseconds) to stop, while Bob belly-flops and Is stopped in 10 ms. If they have the same speed of 4m/s just before they hit the water, what are the average forces exerted by the water on Alice, and Bob? e) Charlie is showing off with an acrobatic dive. Initially he has an angular velocity of 7 rad/s and no net external torque is acting on him. He then pulls in his arms and legs into a tuck position, and his angular velocity increases to 14 rad/s -- If Charlie's initial moment of inertia is 1.4 kg m^2, what is his final moment of inertia (i.e. in the tuck position)?

Explanation / Answer

The gauge pressure is the amount pressure above the atmospheric pressure.

a. At the surface of pool, the pressure is the atmospheric pressure. hence gauge pressure at the surface =0.

At the bottom of the pool, the pressure = atmospheric pressure + density of liquid x g x height of liquid ( depth of pool)

so gauge pressure at the bottom of the pool =  density of liquid x g x height of liquid ( depth of pool)

= 1015 x 9.8 x 1.8

= 17904.6 N/m2

b. Total force on the pool floor = surface area x pressure at the bottom of the pool ( not gauge pressure)

= surface are x ( atmospheric pessure + gauge pressure)

= 60 m2 x (17904.6 + 101325) N

= 7153776 N

c. Let mass of the heaviest person that tube can support =m

In order to support the heaviest person,

mass of heaviest person + mass of tube + mass of air should be less than mass of displaced liquid

so, m + 1.64 + 2x 0.08 < 0.08 x 1015

so, m should be less than 79.4 kg

so it can support a heaviest mass of 79.4 kg

it can only support Bob (75 kg)

it can not support Charlie (82 kg)

so argument can be settled

d. Force = mass x acceleration

magnitude of acceleration = initial velocity/time taken to stop

so acceleration of Alice =4/600 x 10-3 = 6.66 m/s2

force on Alice = mass of Alice x acceleration = 54 x 6.66 = 360 N

so acceleration of Bob =4/10 x 10-3 = 400 m/s2

force on Bob = mass of Bob x acceleration = 75 x 400 = 30000 N

ALL THE BEST IN THE COURSE WORK

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.