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Difficulty: HardFluids at Rest, Archimedes' Principle and Viscosity

A beaker containing water of density 1000 kg/m31000\text{ kg/m}^3 rests on a digital weighing scale, giving an initial reading of 1.50 kg1.50\text{ kg}. A solid aluminum block of mass 0.80 kg0.80\text{ kg} and density 2500 kg/m32500\text{ kg/m}^3 is suspended from a string and completely immersed in the water without touching the bottom or sides of the beaker. What is the new reading on the digital weighing scale, in kilograms? (Take g=10 m/s2g = 10\text{ m/s}^2)

Answer: 1.82 kg

Answer

The new reading on the digital weighing scale is 1.82 kg1.82\text{ kg}.
When the aluminum block is fully submerged in the water, it displaces a volume of water equal to its own volume (V=0.802500=3.2×10−4 m3V = \frac{0.80}{2500} = 3.2 \times 10^{-4}\text{ m}^3). The mass of this displaced water is mwater=1000×3.2×10−4=0.32 kgm_{\text{water}} = 1000 \times 3.2 \times 10^{-4} = 0.32\text{ kg}. The upthrust exerted by the water upward on the block is equal to the weight of the displaced water (3.2 N3.2\text{ N}). By Newton's Third Law, the block exerts an equal and opposite downward reaction force (3.2 N3.2\text{ N}) on the water. This extra downward force adds an equivalent mass of 0.32 kg0.32\text{ kg} to the digital scale reading, making the new reading 1.50 kg+0.32 kg=1.82 kg1.50\text{ kg} + 0.32\text{ kg} = 1.82\text{ kg}.

Step-by-Step Solution

1
Calculate the volume of the submerged block
Volume V=3.2×10−4 m3V = 3.2 \times 10^{-4}\text{ m}^3
The volume of fluid displaced by a completely submerged body equals the volume of the body itself.
2
Find the mass of the displaced water
Mass of displaced water mwater=0.32 kgm_{\text{water}} = 0.32\text{ kg}
According to Archimedes' principle, the upthrust equals the weight of the displaced fluid, which corresponds to a displaced mass of ρwaterV\rho_{\text{water}} V.
3
Apply Newton's Third Law to determine the change in scale reading
Scale reading increase Δm=0.32 kg\Delta m = 0.32\text{ kg}
The fluid exerts an upward buoyant force on the block, so by Newton's Third Law, the block exerts an equal downward reaction force on the fluid, transferring an effective weight equal to the upthrust onto the scale.
4
Compute the total new scale reading
New scale reading =1.82 kg= 1.82\text{ kg}
Sum the initial mass reading of the beaker system (1.50 kg1.50\text{ kg}) and the mass of the displaced water (0.32 kg0.32\text{ kg}).

Key Concept

Apparent weight transfer, Archimedes' principle, and Newton's Third Law
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Topics

Fluids at Rest, Archimedes' Principle and Viscosity

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