A large plastic cylinder with mass 30.0 kg and density 395 kg/m3 is in the water of a lake. A light vertical cable runs between the bottom o

Question

A large plastic cylinder with mass 30.0 kg and density 395 kg/m3 is in the water of a lake. A light vertical cable runs between the bottom of the cylinder and the bottom of the lake and holds the cylinder so that 30.0% of its volume is above the surface of the water. What is the tension in the cable?

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Sapo 4 years 2021-08-23T07:44:11+00:00 1 Answers 61 views 0

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  1. King
    0
    2021-08-23T07:45:52+00:00
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    Answer:

    Tension = 227.6 N

    Explanation:

    Applying the equilibrium condition on the cylinder with the downward direction taken as positive:

    Tension + Weight\ of\ Cylinder - Bouyant\ Force = 0\\Tension = \rho Vg - Weight\ of\ Cylinder

    where,

    ρ = density of water = 1000 kg/m³

    g = acceleration due to gravity = 9.81 m/s²

    V = Volume of water displaced = 70% of Volume of Cylinder

    V = 0.7(\frac{Mass\ of\ Cylinder}{Density\ of\ Cylinder} ) = 0.7(\frac{30\ kg}{395\ kg/m^3}) = 0.0532 m³

    Weight of Cylinder = (mass)(g) = (30 kg)(9.81 m/s²) = 294.3 N

    Therefore,

    Tension = (1000\ kg/m^3)(0.0532\ m^3)(9.81\ m/s^2) - 294.3\ N\\

    Tension = 227.6 N

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