Problem 12.74 Block A in the figure (Figure 1) hangs by a cord from spring balan

Question

Problem 12.74

Block A in the figure (Figure 1) hangs by a cord from spring balance D and is submerged in a liquid Ccontained in beaker B. The mass of the beaker is 1.35 kg ; the mass of the liquid is 2.30 kg . Balance D reads 3.20 kg and balance E reads 7.50 kg. The volume of block A is 4.30×10?3 m3 .

Part A

What is the density of the liquid? (in  kg/m3 )

Part B

What will the balance D read if block A is pulled up out of the liquid? (in kg)

Part C

What will the balance E read if block A is pulled up out of the liquid? (in kg)

Explanation / Answer

A)

for the spring balance

W1 + Fb - W = 0

buoyancy force = Fb = rho*V*g

W = m*g

W1 = 3.2*g

3.2*g + rho*V*g = m*g

3.2 + rho*V = m .........(1)

+++++++++

for balance E

E measures the mass of the beaker, liquid and down ward buoyancy force

W2 = (1.35+2.3)*g + (rho*V*g)

W2 = 7.5*g

7.5*g = (1.35+2.3)*g + (rho*V*g)

3.65 + rho*V = 7.5 .....(2)

from 1 & 2

mass of the block A = m = 7.05 kg

A)

from 1

3.2+ (rho*4.3*10^-3) = 7.05

density = rho = 895.35 kg/m^3

B)

now D will read only mass of the block = 7.05 kg

C)

E will read the mass of fluid , beaker = 1.35+2.3 = 3.65

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