Question

When a body is accelerated under water, some of the surrounding water is also accelerated. This makes the body appear to have a larger mass than it actually has. For a sphere at rest this added mass is equal to the mass of one half of the displaced water. Calculate the force necessary to accelerate a 10 kg, 300-mm-diameter sphere which is at rest under water at the acceleration rate of 10 m/s2 in the horizontal direction. Use H2O

Answer 1

Solution :

Mass of the sphere, m = 10 kg

Diameter, D = 300 mm = 0.3 m

Volume of the sphere is $V=\frac{4}{3} \pi \left(\frac{0.3}{2}\right)^3$

                                       $V=0.01414 \ m^3$

So the volume of displaced water, $V_w=V = 0.01414 \ m^3$

Additional mass, $m_w = \frac{1}{2} \ \rho_w\times v_w$

                                  $=\frac{1}{2} \times 1000 \times 0.01414$

                                  $=7.0686 \ kg$

So the total mass, $M = m_w+m$

                                   = 7.0686 + 10

                                   = 17.0686

Force required, F = Ma

$F=17.0686 \times 10$

   = 170.686 N