Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 14, Problem 72AP
To determine

The equilibrium readings of both top scale and bottom scale.

Expert Solution & Answer
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Answer to Problem 72AP

The equilibrium reading of the top scale is T=(1ρ0ρFe)mFeg_ and the reading on the bottom scale is n=[mb+mo+(ρoρFe)]g_.

Explanation of Solution

The weight of the iron block is balanced by the sum of tension on spring and the buoyant force exerted on iron block by the oil when viewed from the upper part of the scale.

    T+B=Fg,iron                                                                                                        (I)

Here, T is the tension force on the spring scale, B is the buoyant force on iron block, and Fg,iron is the force of gravity on the iron block.

Write the expression for density of iron block.

    ρiron=mironViron                                                                                                         (II)

Here, ρiron is the density of iron block, miron is the mass of iron block, and Viron is the volume of block.

Rearrange equation (II) to find Viron.

    Viron=mironρiron                                                                                                        (III)

By Archimedes law, volume of iron block dipped in oil is equal to the volume of oil displaced from the jar.

    Vdisplaced oil=mironρiron                                                                                                (IV)

Here, Vdisplaced oil is the volume of the displaced oil.

Write the expression for the buoyant force exerted by the oil on the iron block.

    B=ρoilVirong                                                                                                       (V)

Here, ρoil is the density of oil, Viron is the volume of iron block, and g is the acceleration due to gravity.

Rearrange equation (I) to find T.

    T=Fg,ironB                                                                                                    (VI)

Use expression (V) in (VI) to find T.

    T=Fg,ironρoilVirong                                                                                        (VII)

Write the expression for force of gravity on iron block.

    Fg,iron=mirong                                                                                                 (VIII)

Here, miron is the mass of the iron block.

Use expression (VIII) in (VII).

    T=mirongρoilVirong                                                                                           (IX)

Use expression (III) in (IX) to find T.

    T=mirongρoil(mironρiron)g=(1ρoilρiron)mirong                                                                                   (X)

Now observe the system from the bottom side of scale. Let n be the upward normal force acting on the system.

Write the sum of all the vertical forces acting on the system.

    Fy=T+nFg,beakerFg,oilFg,iron                                                                (XI)

Here, Fy is the sum of vertical forces acting on the system, Fg,beaker is the force of gravity on beaker.

At equilibrium the sum of all vertical forces is equal to zero.

    T+nFg,beakerFg,oilFg,iron=0                                                                     (XII)

Write the expression for Fg,beaker.

    Fg,beaker=mbeakerg                                                                                            (XIII)

Here, mbeaker is the mass of the beaker.

Write the expression for Fg,oil.

    Fg,oil=moilg                                                                                                    (XIV)

Here, moil is the mass of the oil.

Use expressions (XIV), (XIII), and (VIII) in expression (XII) and solve for n.

    T+nmbeakergmoilgmirong=0n=(mbeaker+moil+miron)gT                            (XV)

Substitute expression (X) in (XV) to find n.

    n=(mbeaker+moil+miron)g(1ρoilρiron)mirong=[mbeaker+moil+(ρoilρiron)mFe]g

Conclusion:

Therefore, the equilibrium reading of the top scale is T=(1ρoilρiron)mirong_ and the reading on the bottom scale is n=[mbeaker+moil+(ρoilρiron)]g_.

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Chapter 14 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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