Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 14, Problem 33AP

(a)

To determine

To determine: The appropriate model to describe the system when balloon is stationary.

(a)

Expert Solution
Check Mark

Answer to Problem 33AP

Answer: The appropriate model to describe the system is particle in equilibrium.

Explanation of Solution

Explanation:

Given information: The mass of the balloon is 0.250kg tied to a uniform length 2.00m and mass 0.050kg . The balloon is spherical with radius 0.4m and the density of the air is 1.20kg/m3 .

If a system remains stationary, the sum of all forces acted on a system in all direction vertical as well as horizontal is equal to zero. This condition is also called is equilibrium condition.

Conclusion:

Therefore, appropriate model to describe the system is particle in equilibrium.

(b)

To determine

To determine: The force equation for the balloon for this model.

(b)

Expert Solution
Check Mark

Answer to Problem 33AP

Answer: The force equation for the balloon for this model is BFbFHeFs=0 .

Explanation of Solution

Explanation:

Given information: The mass of the balloon is 0.250kg tied to a uniform length 2.00m and mass 0.050kg . The balloon is spherical with radius 0.4m and the density of the air is 1.20kg/m3 .

In equilibrium condition, sum of all forces in vertical direction is equal to zero.

Fy=0Fy=BFbFHeFs=0 (I)

  • B is buoyant force.
  • Fb is the weight of the balloon (envelope)
  • FHe is the weight of the helium gas.
  • Fs is the weight of the string.

Conclusion:

Therefore, the force equation for the balloon for this model is Fy=BFbFHeFs=0 .

(c)

To determine

To determine: The mass of the string in terms of mb , r , ρHe and ρair .

(c)

Expert Solution
Check Mark

Answer to Problem 33AP

Answer: The mass of the string in the terms of mb , r , ρHe and ρair is 43πr3(ρairρHe)mb .

Explanation of Solution

Explanation:

Given information: The mass of the balloon is 0.250kg tied to a uniform length 2.00m and mass 0.050kg . The balloon is spherical with radius 0.4m and the density of the air is 1.20kg/m3 .

From equation (I),

BFbFHeFs=0

The buoyant force act on the balloon is equal to the displaced volume of the air by the balloon.

Formula to calculate the buoyant force acting on the balloon is,

B=ρairg×43πr3

  • ρair is the density of the air.
  • r is the radius of the balloon.
  • g is the acceleration due to gravity.

Formula to calculate the weight of the balloon is,

Fb=mbg

  • mb is the mass of the balloon.

Formula to calculate the weight of the helium gas is,

FHe=mHeg

  • mHe is the mass of the helium gas.

Formula to calculate the weight of the string is,

Fs=msg

  • ms is the mass of the string.

Substitute ρairg×43πr3 for B , mHeg for FHe , mbg for Fb and msg for Fs in equation (I).

ρairg×43πr3mbgmHegmsg=0

Formula to calculate the mass of the helium gas is,

mHe=ρHe×43πr3

  • ρHe is the density of the helium gas.

Substitute ρHe×43πr3 for mHe in above expression.

ρairg×43πr3mbgρHe×43πr3gmsg=0

Rearrange the above expression for ms

ms=ρair×43πr3ρHe×43πr3mbms=43πr3(ρairρHe)mb

Conclusion:

Therefore, the mass of the string in terms of mb , r , ρHe and ρair is 43πr3(ρairρHe)mb .

(d)

To determine

To determine: The mass of the string.

(d)

Expert Solution
Check Mark

Answer to Problem 33AP

Answer: The mass of the string is 0.0237kg .

Explanation of Solution

Explanation:

Given information: The mass of the balloon is 0.250kg tied to a uniform length 2.00m and mass 0.050kg . The balloon is spherical with radius 0.4m and the density of the air is 1.20kg/m3 .

From equation (II),

ms=43πr3(ρairρHe)mb

Substitute 1.20kg/m3 for ρair , 0.250kg for mb , 0.4m for r and 0.179kg/m3 for ρHe to find ms .

ms=43π(0.4m)3(1.20kg/m30.179kg/m3)0.250kg=0.0237kg

Conclusion:

Therefore, the mass of the string is 0.0237kg .

(e)

To determine

To determine: The length h of the string if mass of the string is 0.050kg .

(e)

Expert Solution
Check Mark

Answer to Problem 33AP

Answer: The length h of the string is 0.948m if mass of the string is 0.050kg .

Explanation of Solution

Explanation:

Given information: The mass of the balloon is 0.250kg tied to a uniform length 2.00m and mass 0.050kg . The balloon is spherical with radius 0.4m and the density of the air is 1.20kg/m3 .

From equation (II),

ms=43πr3(ρairρHe)mb

The mass of the string of height h is equal to the ms×hl .

Substitute ms×hl for ms in above expression.

ms×hl=43πr3(ρairρHe)mb

Substitute 1.20kg/m3 for ρair , 0.250kg for mb , 0.4m for r , 0.050kg for ms 2.0m for l and 0.179kg/m3 for ρHe to find h .

0.050kg×h2.0m=43π(0.4m)3(1.20kg/m30.179kg/m3)0.250kgh=0.0237kg×2.0m0.050kg=0.948m

Conclusion:

Therefore, the length h of the string is 0.948m if mass of the string is 0.050kg .

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Chapter 14 Solutions

Physics for Scientists and Engineers with Modern Physics

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