A cube with sides of length L is tethered to the bottom of a pool of fluid by a
ID: 1683677 • Letter: A
Question
A cube with sides of length L is tethered to the bottom of a pool of fluid by a rope whose length is also L.The top of the cube is a distance 5L below the water's surface. The density of the cube is 1/3. the density of the fluid in the pool. Express your answers to the following questions in terms of the density of the fluid
(f ), the length of a side of the cube (L), and g.
(a) What is the tension in the rope?
(b) If the rope is cut, what will the acceleration of the cube be (both magnitude and direction)?
(c) How long will it take after the rope is cut for the cube to reach the surface of the water?
(d) Once it come to rest on the surface, what fraction of the cube's volume will still be underwater?
1
Explanation / Answer
(a) What is the tension in the rope? Let f' be density of the cube. f'=f/3. so the cube weight P=f'*g*L^3=fgL^3/3. upward force of the fluid acted on the cube. F=f*gL^3. so tension of the rope T+P=F. so T=F-P=2fgL^3/3 (b) If the rope is cut, what will the acceleration of the cube be (both magnitude and direction)? upward. magnitude a=T/m=(2fgL^3/3) / (fL^3/3)=2g. (c) How long will it take after the rope is cut for the cube to reach the surface of the water? we have at^2/2=s. so 2g*t^2/2=5L. so t=sqrt(5L/g) (d) Once it come to rest on the surface, what fraction of the cube's volume will still be underwater? now upward force will equal to weight of the cube. so let that fraction be s. f'gL^3=s*L^3*g*f. so f'=s*f=f/3. so s=1/3