How Can I Solve Viscosity and Bouyant Force Problems?

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SUMMARY

This discussion addresses two physics problems related to buoyancy and viscosity. The first problem involves calculating the radius of a hot-air balloon with a total mass of 500 kg, where the density of hot air is 0.95 kg/m³ and cold air is 1.3 kg/m³. The second problem focuses on determining the viscosity of oil through which a steel sphere of mass 1.9 g and diameter 8 mm falls, with the oil density at 900 kg/m³. Key concepts include the application of buoyancy principles and Newton's law of viscosity, which relates stress to the velocity gradient in a fluid.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Familiarity with Newton's law of viscosity
  • Basic knowledge of fluid dynamics and density calculations
  • Ability to manipulate equations involving mass, volume, and density
NEXT STEPS
  • Study the derivation of Archimedes' principle for buoyancy calculations
  • Learn how to apply Newton's law of viscosity in practical scenarios
  • Explore the relationship between drag force and viscosity in fluid mechanics
  • Investigate the effects of temperature on the density of gases and liquids
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Students preparing for physics exams, educators teaching fluid mechanics, and professionals in engineering fields dealing with buoyancy and viscosity calculations.

Struggling
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hi all just doing some questions and come along these two which have me stumped. There multiple choice questions but id rather know the working out to them as i have exams in a couple of weeks. thanks.

1. The mass of a hot-air balloon and its fully loaded basket is 500kg excluding the air in the balloon. if we assume the balloon is spherical, what radius must the balloon be for it to float if the density of the hot air inside the balloon is 0.95kg/m^3 and the density of the cold air outside is 1.3kg/m^3.

--- with this question i believe you use the P = Po + pgh but i cannot set it out to find the radiusi have tried to use 4/3 pi R^3 where m = pV.

2. A sphere of mass 1.9g & diameter 8mm made of steel is dropped into a tank of liquid with density = 900kg/m^3 and falls a distance of 30cm in 27 seconds. What is the Viscosity of the oil?

--- i looked through textbooks but i could only find Newtons law for viscosity and i had no idea how to work it out.

and just one quick check i believe this is right.

Newtons law of viscosity says that: - the stress depends on the acceleration( i believe this is it because the equation is stress is the viscosity of the liquid x the change in velocity i believe?)

the other options for the answer were
1. the stress is proportianal to the rate of strain
and
2. The velocity gradient is linear.

but i see no relation with these other 2 answers.

any help is much appreciated thanks!
 
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Pressure inside the hot-air balloon is the same as the
Pressure outside the balloon, at the bottom of the bag.
The cold air is "Buoyant enough" to hold up a Volume of cold air.
The basket mass + hot air mass should be the same amount
as the cold air mass that the same Volume encloses.

the gradient of velocity is linear ...
at the falling ball surface, the oil has same velocity as the ball.
the oil is dragged slightly slower, at a ring slightly larger than ball.
CHANGE in velocity (time-wise) is totally distinct from
Difference in velocity (at different places) .
z-component of v_ball - v_oil is proportional to r-R .
 
This might help on viscosity

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri.html#c3

In the first part, one is concerned with bouyancy.

As lightgrav indicated, the effective density of the balloon (total mass including hot air, balloon and basket divided by volume of hot air) must be equal to the density of cold air. Or mass mass of balloon and hot air must equal mass of cold air displaced by balloon.

In the second problem, a metal sphere falling through oil is subject to buoyant force and drag (resistant force). The drag is determined by the viscosity of the oil.

One can calculate how fast the steel ball would fall (or accelerate) in air (neglecting air resistance) and compare to how fast it falls in oil.
 

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