Question

A sphere of plastic of radius = .100 m is completely immersed in water. The density of the plastic is 600 kg/m3. Since the plastic would otherwise float it is restrained from moving upward by a thread attached to the bottom of the container. What is the tension in the thread?

Answer 1

To solve this problem we need to use the proportional relationships between density, mass and volume, together with Newton's second law.

The force can be described as

`F = ma \rightarrow mg`

Where,

m = Mass

g = Gravitational acceleration

At the same time the Density can be defined as

`\rho = (m)/(V) \rightarrow m = \rho V`

Where,

m = mass

V = Volume

Replacing the value of the mass at the equation of Force we have,

`F = \rho V g`

Since the difference between the two forces gives us the total Force then we have to

`F_T = F_w - F_p`

Where

`F_w =` Force of the water

`F_p`= Force of plastic

Therefore with the values for this force we have,

`F_T = \rho_w Vg - \rho_p Vg`

`F_T = Vg(\rho_w - \rho_p)`

`F_T = ((4)/(3) \pi r^3) g(\rho_w - \rho_p)`

`F_T = ((4)/(3) \pi (0.1)^3) (9.8)(1000 - 600)`

`F_T = 16.412 N`

Therefore the tension in the thread is 16.412N