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Zorluk: OrtaFluids at Rest, Archimedes' Principle and Viscosity

A solid block of mass 0.60 kg0.60\text{ kg} and density 600 kg/m3600\text{ kg/m}^3 is held fully submerged in water of density 1000 kg/m31000\text{ kg/m}^3 by a light vertical string attached to the bottom of a container. What is the tension in the string? (Take g=10 m/s2g = 10\text{ m/s}^2)

  1. A
    10.0 N10.0\text{ N}
  2. 4.0 N4.0\text{ N}Cevap
  3. C
    6.0 N6.0\text{ N}
  4. D
    16.0 N16.0\text{ N}

Cevap

The tension in the string is 4.0 N4.0\text{ N}.
First, the volume of the block is computed as V=mρ=0.60600=1.0×10−3 m3V = \frac{m}{\rho} = \frac{0.60}{600} = 1.0 \times 10^{-3}\text{ m}^3. According to Archimedes' principle, the upthrust exerted by the displaced water is U=ρwaterVg=1000×1.0×10−3×10=10.0 NU = \rho_{\text{water}} V g = 1000 \times 1.0 \times 10^{-3} \times 10 = 10.0\text{ N}. The weight of the block is W=mg=0.60×10=6.0 NW = mg = 0.60 \times 10 = 6.0\text{ N}. For the block to remain completely submerged in equilibrium, the upward upthrust must balance the downward forces (the weight of the block and the tension TT pulling downward). Thus, T=U−W=10.0 N−6.0 N=4.0 NT = U - W = 10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}.

Adım Adım Çözüm

1
Calculate the volume of the block using its mass and density.
V=mρblock=0.60 kg600 kg/m3=1.0×10−3 m3V = \frac{m}{\rho_{\text{block}}} = \frac{0.60\text{ kg}}{600\text{ kg/m}^3} = 1.0 \times 10^{-3}\text{ m}^3
The volume of fluid displaced equals the total volume of the fully submerged block.
2
Calculate the upward upthrust force exerted by the water.
U=ρwater⋅V⋅g=1000 kg/m3×1.0×10−3 m3×10 m/s2=10.0 NU = \rho_{\text{water}} \cdot V \cdot g = 1000\text{ kg/m}^3 \times 1.0 \times 10^{-3}\text{ m}^3 \times 10\text{ m/s}^2 = 10.0\text{ N}
Archimedes' principle states upthrust equals the weight of the displaced fluid.
3
Calculate the downward gravitational weight of the block.
W=m⋅g=0.60 kg×10 m/s2=6.0 NW = m \cdot g = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}
Weight is the force exerted on the mass of the block by gravity.
4
Apply vertical force equilibrium to solve for string tension.
U=W+T⇒T=U−W=10.0 N−6.0 N=4.0 NU = W + T \Rightarrow T = U - W = 10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}
The string is tied to the bottom, so tension acts downward to hold the buoyant block in equilibrium.

Anahtar Kavram

Archimedes' Principle and Static Equilibrium of Submerged Bodies
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Konular

Fluids at Rest, Archimedes' Principle and Viscosity

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