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?

Respuesta :

Answer:

Tension = 227.6 N

Explanation:

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

[tex]Tension + Weight\ of\ Cylinder - Bouyant\ Force = 0\\Tension = \rho Vg - Weight\ of\ Cylinder[/tex]

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

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

Therefore,

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

Tension = 227.6 N