Chapter 14, Problem 1Q

We fully submerge an irregular 3 kg lump of material in a certain fluid. The fluid that would have been in the space now occupied by the lump has a mass of 2 kg. (a) When we release the lump, does it move upward, move downward, or remain in place? (b) If we next fully submerge the lump in a less dense fluid and again release it, what does it do?

To determine

To find:

a) Movement of block when we release the irregular lump in a certain fluid.

b) Movement of block when we release the lump in a less dense fluid.

Explanation of Solution

1) Concept:

To float the object in fluid, buoyant force must be greater than or equal to the weight of object.

2) Formulae:

Buoyant force $\left(F_B\right)=\rho Vg$

3) Given:

Mass of irregular submerged lump $=3 kg$

Mass of fluid $\left(m_f\right)=2kg$

4) Calculations:

We have

Buoyant force $F_B=m_{f}g$

Weight of lump $\left(W\right)=mg$

We know that to float an object in a fluid, the buoyant force must be greater than or equal to the weight of the object.

i.e. $F_B\ge W$

a) As the mass of the submerged lump is greater than the mass of fluid, when the lump is released, it moves downwards.

b) If another fluid has less density, buoyant force will be less. Hence, the lump will move downwards due to greater weight than buoyant force.

Conclusion:

By comparing the buoyant force and the weight of object for each case, we can conclude whether an irregular lump of material floats or moves downwards.