Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 15, Problem 58P

(a)

To determine

The appropriate analysis model for the system when the balloon remains stationary.

(a)

Expert Solution
Check Mark

Answer to Problem 58P

A particle in equilibrium model is the appropriate analysis model for the system when the balloon remains stationary.

Explanation of Solution

An analysis model is a simplified version of any physical system that strips away the unnecessary aspects of the situation. There are different kinds of analysis models such the particle under constant velocity, particle under constant acceleration and particle in equilibrium etc.

In the given situation, the balloon remains stationary which implies the balloon has zero acceleration. The net force on the balloon in any direction is zero. The balloon is in equilibrium. The particle in equilibrium is the appropriate analysis model for the given physical situation.

Conclusion:

Thus, a particle in equilibrium model is the appropriate analysis model for the system when the balloon remains stationary.

(b)

To determine

The force equation for the balloon.

(b)

Expert Solution
Check Mark

Answer to Problem 58P

The force equation for the balloon is BFbFHeFs=0.

Explanation of Solution

Write the equation for equilibrium.

  ΣFy=0        (I)

Here, ΣFy is the net force acting on the balloon in the vertical direction.

Take the upward direction to be +y direction.

Write the equation for ΣFy .

  ΣFy=BFbFHeFs

Here, B is the buoyant force, Fb is the weight of the balloon, FHe is the weight of the helium and Fs is the weight of the segment of the string above the ground.

Put the above equation in equation (I).

  BFbFHeFs=0        (II)

Conclusion:

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

(c)

To determine

The expression for the mass of the segment of string.

(c)

Expert Solution
Check Mark

Answer to Problem 58P

The expression for the mass of the segment of string is ms=(ρairρHe)43πr3mb.

Explanation of Solution

Write the equation for the buoyant force.

  B=ρairgV        (III)

Here, ρair is the density of air, g is the acceleration due to gravity and V is the volume of the balloon.

Write the equation for the weight of the helium.

  FHe=mHeg        (IV)

Here, mHe is the mass of the helium filled.

Write the equation for density of the helium.

  ρHe=mHeV

Here, ρHe is the density of helium.

Rewrite the above equation for mHe.

  mHe=ρHeV

Put the above equation in equation (IV).

  FHe=ρHeVg        (V)

Write the equation for the weight of the balloon.

  Fb=mbg        (VI)

Here, mb is the mass of the balloon.

Write the equation for the weight of the segment of the string above the ground.

  Fs=msg        (VII)

Here, ms is the mass of the string.

Put equations (III), (V), (VI) and (VII) in equation (II) and rewrite it for ms.

  ρairgVmbgρHeVgmsg=0ρairVmbρHeV=msms=(ρairρHe)Vmb        (VIII)

Write the equation for the volume of the balloon.

  V=43πr3        (IX)

Here, r is the radius of the balloon.

Conclusion:

Put equation (IX) in equation (VIII).

  ms=(ρairρHe)43πr3mb        (X)

Therefore, the expression for the mass of the segment of string is ms=(ρairρHe)43πr3mb.

(d)

To determine

The numerical value of the mass ms.

(d)

Expert Solution
Check Mark

Answer to Problem 58P

The numerical value of the mass ms is 0.0237 kg.

Explanation of Solution

Equation (X) can be used to determine the numerical value of the mass ms.

Conclusion:

The density of air is 1.20 kg/m3 and the density of helium is 0.179 kg/m3.

Substitute 1.20 kg/m3 for ρair , 0.179 kg/m3 for ρHe , 0.400 m for r and 0.250 kg for mb in equation (X) to find ms.

  ms=(1.20 kg/m30.179 kg/m3)43π(0.400 m)30.250 kg=0.0237 kg

Therefore, the numerical value of the mass ms is 0.0237 kg.

(e)

To determine

The numerical value of the length h.

(e)

Expert Solution
Check Mark

Answer to Problem 58P

The numerical value of the length h is 0.948 m.

Explanation of Solution

Write the equation for the mass ms in terms of the total mass of the string.

  ms=mhl

Here, m is the total mass of the string, h is the length of the string above the ground and l is the total length of the string.

Rewrite the above equation for h .

  h=lmsm        (XI)

Conclusion:

Substitute 2.00 m for l , 0.0237 kg for ms and 0.050 kg for m in equation (XI) to find h.

  h=(2.00 m)0.0237 kg0.050 kg=0.948 m

Therefore, the numerical value of the length h is 0.948 m.

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Chapter 15 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

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