A cube with sides of length L is tethered to the bottom of a pool of fluid by a
ID: 1683518 • 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 the density of 3
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?
Explanation / Answer
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 the density of 3 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? Consider the free body diagram of the cube. There is tension pulling it up (T). There is a buoyancy force (B) also acting up and the weight acting down (W). Thus, T+B = W since it is at rest. Archimedes law says the buoyance is equal to the weight of the fluid displaced by the object. The volume displaced is equal to the volune of the cube which is L^3. Thus, mass = V x density = 3L^3. Thus, weight of fluid displaced is 3gL^3. Weight of cube is mass x g = volume x density x g = 1 x L^3 x g = gL^3. thus, Tension T = gL^3 - 3gL^3 = -2gL^3 (we assumed tension is acting upwards so - sing means its downwards. ). (b) If the rope is cut, what will the acceleration of the cube be (both magnitude and direction)? If rope is cut, tension is 0. So only B and W act. B is 3gL^3 upwards upwards and W is gL^3 downwards. Resultant force is 2gL^3 upwards and hence, it will accelerate in this direction with a = F/mass = 2gL^3 / L^3 = 2g. (c) How long will it take after the rope is cut for the cube to reach the surface of the water? distance = 5L (given) aceleration is 2g and initial speed is 0. s = ut + 0.5at^2 but u=initial speed = 0. find t. (d) Once it come to rest on the surface, what fraction of the cube’s volume will still be underwater? when its at rest, weight = buoyancy force. thus, gL^3 = (volume submerged) x density of fluid x g Thus, volume submerged = gL^3 / 3g = L^3/3. this means a 1/3 of it is sumerged.