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A 1.00-kg beaker containing 2.50 kg of oil (density = 916 kg/m3) rests on a scale. A 1.90 kg block of iron is suspended from a spring scale and completely submerged in the oil (Fig. P9.30). Find the equilibrium readings of both scales.

Respuesta :

mass of iron block given as

[tex]m_1 = 1.90 kg[/tex]

density of iron block is

[tex]\rho = 7860 kg/m^3[/tex]

now the volume of the iron piece is given as

[tex]V = \frac{m}{\rho}[/tex]

[tex]V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3[/tex]

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

[tex]F_b = \rho_L V g[/tex]

here we know that

[tex]\rho_L[/tex] = density of liquid = 916 kg/m^3

[tex]F_b = 916* 2.42 * 10^{-4} * 9.8[/tex]

[tex]F_b = 2.17 N[/tex]

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

[tex]F_s + F_b = mg[/tex]

[tex]F_s + 2.17 = 1.90* 9.8[/tex]

[tex]F_s = 16.45 N[/tex]

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

[tex]F_n = F_g + F_b[/tex]

[tex]F_n = (1 + 2.50)*9.8 + 2.17[/tex]

[tex]F_n = 34.3 + 2.17 = 36.47 N[/tex]

So the other scale will read 36.47 N

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