A cube of wood having an edge dimension of 19.4 cm and a density of 652 kg/m³ floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface?

Principles of Physics: A Calculus-Based Text
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Chapter15: Fluid Mechanics
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A cube of wood having an edge dimension of 19.4 cm and a density of 652 kg/m³ floats on water.
(a) What is the distance from the horizontal top surface of the cube to the water level?
(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the
water surface?
Transcribed Image Text:A cube of wood having an edge dimension of 19.4 cm and a density of 652 kg/m³ floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface?
The buoyant force supports the weight of both blocks, so we have
B = F₁ + Mg,
Fg
where M is the mass of the lead. The buoyant force of the displaced volume of water is equal to the buoyant
force of the submerged wood plus the weight of the lead. Using the expression for buoyant force pVg we have
the following equation.
Pwater³g = PwoodL³g + Mg
Solving for the mass of the lead, we have
M = (Pwater - Pwood)
3
0
X
= The correct answer is not zero. kg/m³)
The correct answer is not zero. m
m) ³
=
0
X
kg.
Transcribed Image Text:The buoyant force supports the weight of both blocks, so we have B = F₁ + Mg, Fg where M is the mass of the lead. The buoyant force of the displaced volume of water is equal to the buoyant force of the submerged wood plus the weight of the lead. Using the expression for buoyant force pVg we have the following equation. Pwater³g = PwoodL³g + Mg Solving for the mass of the lead, we have M = (Pwater - Pwood) 3 0 X = The correct answer is not zero. kg/m³) The correct answer is not zero. m m) ³ = 0 X kg.
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