AP Physics B – Answer to Free Response Practice Question on Fluid Mechanics

He who joyfully marches in rank and file has already earned my contempt. He has been given a large brain by mistake, since for him the spinal cord would suffice.

–Albert Einstein

In the post dated 30^th September 2011 a free response practice question on fluid mechanics was given to you. As promised, I give below a model answer along with the question:

The adjoining figure shows an empty thin walled cubical vessel of side 0.4 m and mass 6.4 kg floating on kerosene contained in a large tank (not shown). Density of kerosene is 800 kgm^–3 where as the density of water is 1000 kgm^–3. You may take the acceleration due to gravity as 10 ms^–2. Now, answer the following questions:

(a) Calculate the magnitudes of the force of buoyancy and the force of gravity acting on the empty vessel and state their directions.

(b) Calculate the height x of the portion of the empty vessel that is submerged in kerosene.

(c) Water is slowly poured into the vessel so that an additional 0.2 m of the height of the vessel is submerged in kerosene. Calculate the volume of water added to obtain this condition.

(a) The force of buoyancy acts vertically upwards while the force of gravity (which is the weight of the vessel) acts vertically downwards. They have the same magnitude since the vessel is in equilibriumn. Since the mass (m) of the empty vessel is 6.4 kg, its weight (mg) is 6.4×10 = 64 newton.

Therefore the force of buoyancy and the force of gravitaty have the same magnitude of 64 N.

(b) The weight of a floating body is equal to the weight of the displaced liquid (in accordance with the law of floatation). Therefore we have

Axρg = mg where A is the area of the base of the vessel, x is the height of the portion of the empty vessel that is submerged in kerosene, ρ is the density of kerosene and g is the gravitational acceleration.

This gives x = m/Aρ = 6.4/(0.4²×800) = 0.05 m

(c) When an additional 0.2 m of the vessel is submerged in kerosene, the additional volume of kerosene displaced by the vessel is 0.2 A and the weight of this volume of kerosene is 0.2 Aρg. This must be equal to the weight of the water added so that we have

0.2 Aρg = Vdg where V is the volume of water added.and d is the density of water.

This gives V = 0.2 Aρ/d = (0.2×0.4²×800)/1000 = 0.0256 m³

[Suppose you were asked to determine the pressure exerted by water at the bottom of the vessel when the condition mentioned in part (c) in the above question is attained. You will then divide the weight of the water by the area of the bottom of the vessel, remembering that the thrust at the bottom is produced by the weight of the water column in the vessel and pressure is thrust per unit area.

Thus presuure at the bottom = 0.2 Aρg/A =0.2 ρg = 0.2×800×10 = 1600 pascal

(This is the same as Vdg/A)].

Now, go over here to find some more refreshing questions in this section.