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

$\rho = \frac{m}{V} \rightarrow m = \rho V$

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,

$F_T = \rho_w Vg - \rho_p Vg$

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

$F_T = (\frac{4}{3} \pi r^3) g(\rho_w - \rho_p)$

$F_T = (\frac{4}{3} \pi (0.1)^3) (9.8)(1000 - 600)$

$F_T = 16.412 N$

Therefore the tension in the thread is 16.412N