can someone solve this ( cube floats question)

Question

An open tank with square horizontal cross-section, of constant side length L, contains liquid of uniform density ρ and depth H. A solid cube ,of mass m and side length l (where l<L), is placed into the liquid in the tank. The cube floats in the liquid with one pair of opposite faces horizontal, the lower of these faces being submerged to a depth h below the surface of the liquid.

 

(a)Deduce that once the block has been placed in the liquid, the depth K of liquid in the tank is given by

K = H +m ρL2

 

(b) Write down a condition ,in terms of m, ρ and l, that ensures that the cube will indeed float (if there is a su?cient depth of liquid in the tank) and not sink.

(c) Assuming that the condition of part (a)(v) is satis?ed, show that there is a su?cient depth of liquid in the tank for the cube to float only if

H>m(L^2−l^2) /ρL^2l^2 .

please help me with this question

thanks

 

Answer 1

This is a classic buoyancy problem.   The mass m, must equal the mass of water displaced.  So

 

 m = ρ l² h    which implies l² h  = m/ρ

 

The volume of water before the cube is placed is  V =  L² H  ,

 the volume of water after the cube is placed is   V = L² H - l² h       These two volumes must be equal so,

 

  L² K =  L² H + l² h      which implies K =  H + l² h /L²     substituting for l² h gives

 

   K = H + m/(ρ L²)      This is the answer to part (a)

 

 

The condition for floatation is simply  m < ρ l³    This is the answer to (b)

 

To answer (c) ,  just note that the condition is  K > h

 

   Substitutions from the equations above to eliminate K and h  yield

 

  H >  m/(ρ l² ) - m/ (ρ L²)    after rearrangement this gives

 

  H  >  (m/ρ) (L² - l² ) /(L² l² )   This is the answer for (c)