- #1

EnricoHendro

- 80

- 11

- Homework Statement
- The weight of a rectangular block of low-density material is 15.0 N. With a thin string, the center of the horizontal bottom face of the block is tied to the bottom of a beaker partly filled with water. When 25.0% of the block’s volume is submerged, the tension in the string is 10.0 N. (a) Find the buoyant force on the block. (b) Oil of density 800 kg/m3 is now steadily added to the beaker, forming a layer above the water and surrounding the block. The oil exerts forces on each of the four sidewalls of the block that the oil touches. What are the directions of these forces? (c) What happens to the string tension as the oil is added? Explain how the oil has this effect on the string tension. (d) The string breaks when its tension reaches 60.0 N. At this moment, 25.0% of the block’s volume is still below the water line. What additional fraction of the block’s volume is below the top surface of the oil?

- Relevant Equations
- B=T+Mg

B-Mg-T=0

Hello there, I have a quite different approach on answering the part c and d of this problem. Can you guys help me to see if my method is acceptable or logical? (I got the same answers as the back of the textbook, however my methods differ from the solutions manual). Here is my answer/attempt at a solution :

Part A :

B = T+Mg

B = 25 N

Part B :

Perpendicular to each sides of the block (towards the centre of the block)

Part C :

if the density of the object is less than the density of the fluid, then the object would accelerate upward until it floats which at that point, the object reaches static equilibrium (assuming the liquid does not generate any waves). We see that the density of the object is less than the density of oil. The proof is as follows :

thus, vol = 0.010204081 Meter Cubic. Since :
, and M = 1.530612245 Kg, therefore, the density of the object is 150 Kg/Meter Cubic.

Therefore, the additional presence of oil would exert a buoyant force towards the remaining part of the cube that is not submerged in water. This increases the tension.

Part D :

B-Mg-T=0

with x being the part of the volume that is under the top surface of oil. Therefore :

Thus, 25% of the cube is submerged in water, while 62.5% of the volume of the cube is submerged in oil, and the remaining 12.5% of the cube is not submerged in anything.

Part A :

B = T+Mg

B = 25 N

Part B :

Perpendicular to each sides of the block (towards the centre of the block)

Part C :

if the density of the object is less than the density of the fluid, then the object would accelerate upward until it floats which at that point, the object reaches static equilibrium (assuming the liquid does not generate any waves). We see that the density of the object is less than the density of oil. The proof is as follows :

thus, vol = 0.010204081 Meter Cubic. Since :

Therefore, the additional presence of oil would exert a buoyant force towards the remaining part of the cube that is not submerged in water. This increases the tension.

Part D :

B-Mg-T=0

with x being the part of the volume that is under the top surface of oil. Therefore :

Thus, 25% of the cube is submerged in water, while 62.5% of the volume of the cube is submerged in oil, and the remaining 12.5% of the cube is not submerged in anything.