Question

A perfectly spherical balloon of air is tethered to the floor of a stagnant body of water by cable AB, and sits just below the surface. The density of water is 1000kg/m3 and the density of air is 1.225kg/m3, Given the following: r = 0.53 m A B = 7.95 m What is the tension in cable AB? Matlab input: r = 0.53; AB =7.95;

Answer 1

To solve this problem it is necessary to apply the concepts related to Newton's second Law and its definition of density.

By Newton's second law we understand that,

F = ma

Where,

m = mass

a = acceleration (at this case the gravity acceleration)

In the case of density, we know that it is described as the proportion of mass versus volume, that is,

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

Where,

m = mass

V = Volume

The total tension of the AB cable would be given by the tension exerted upwards by the water and the tension exerted by the weight, therefore,

$F_t = F_u - F_w$

$F_t =m_wg - m_a g$

Mass can be expressed as,

$F_t = \rho_w(\frac{4}{3}\pi r^3)*g -\rho_a(\frac{4}{3}\pi r^3)*g$

$F_t = (1000))\frac{4}{3}\pi 0.53^3)*9.8 -(1.225)(\frac{4}{3}\pi 0.53^3)*9.8$

$F_t = 6103.94N$

Therefore the tension in cable AB is 6103.94N