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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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Section
BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A 1.00-kg beaker containing 2.00 kg of oil (density = 916 kg/m3) rests on a scale. A 2.00-kg block of iron is suspended from a spring scale and is completely submerged in the oil (Fig. P9.31). Find the equilibrium readings of both scales.

images

Figure P9.31

To determine
The equilibrium readings of both scales.

Explanation

Section1:

To determine: The equilibrium reading of the upper scale.

Answer: The equilibrium reading of the upper scale is 17.3N .

Explanation: The volume of the iron block is V=miron/ρiron and the buoyancy force on the block is B=ρoilVg=ρoil(miron/ρiron)g . Using Newton’s equations of motion at equilibrium condition, the upper scale reading would be Fy=0=Fupper+BmirongFupper=miron[1(ρoil/ρiron)]g .

Given info: The mass of iron block is 2.00kg , acceleration due to gravity is 9.80m/s2 , density of iron is 7.86×103kg/m3 , and the density of oil is 916kg/m3 .

The formula for the upper scale reading is,

Fupper=miron[1ρoilρiron]g

  • miron is mass of iron.
  • g is acceleration due to gravity.
  • ρiron is density of iron.
  • ρoil is density of oil.

Substitute 2.00kg for miron , 9.80m/s2 for g , density of oil is 7.86×103kg/m3 , 7.86×103kg/m3 for ρiron , and 916kg/m3 for ρiron to find Fupper .

Fupper=(2.00kg)(1916kg/m37.86×103kg/m3)(9.80m/s2)=17.3N

Thus, the equilibrium reading of the upper scale is 17.3N .

Section2:

To determine: The equilibrium reading of the lower scale.

Answer: The equilibrium reading of the lower scale is 31.7N .

Explanation: Use Newton’s equation of motion as FlowerB(moil+mbeaker)g=0 and it is rearranged for the lower scale reading is Flower=[ρoil(miron/ρiron)+(moil+mbeaker)]g

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