Buoyancy Force and Block Equilibrium

[lafalfa]

Hi,

A cylindrical log with volume 2.0m^3, length 4.0m and density 700 kg/m^3 is anchored to the sea floor by a light cable attached to one end, and with the long axis vertical. Density of sea water is 1030 kg/m^3. g=9.80 m/s^2.

a) What is the buoyancy force acting on the wood?

I did this: rho_f * V_o * g = 1030*2.0*9.8 = 20188 N

However, the answer is 20.2 N and I can't figure out why it appears I am off by a factor of 1000!

b) What is the weight of the log?

I did this: rho_o * V_o * g = 700*2.0*9.8 = 13720 N

However, the answer is 13.7 N and again, I can't figure out why I am off by a factor of 1000!

c) What is the tension in the cable?

I did this: Since rho_f * V_o * g = F_tension + rho_o * V_o * g, 20188-13720 = 6468 N

The correct answer is 6470, which I'm assuming is the same as my answer, but rounded differently (correct me if I'm wrong!).

d) If the cable is released and the log remains vertical, what length of the log will be out of the seawater, after it reaches equilibrium?

I did this: rho_f * V_f * g = rho_o * V_o * g, so V_f=700*2/1030 = 1.35922, thus 0.5*h=1.35922 and h=2.718 m
(0.5 is the area of the base of the cylinder.)
The correct answer is 1.28m and I can't figure out where I went wrong!

What should be the correct answers for parts A and B? Am I doing the question wrong? I have a final exam on this stuff in a couple of days! Please help me!

Thank you very much!

 

[diazona]

a) What is the buoyancy force acting on the wood?

I did this: rho_f * V_o * g = 1030*2.0*9.8 = 20188 N

However, the answer is 20.2 N and I can't figure out why it appears I am off by a factor of 1000!

I think you might actually be right. 20.2N seems much too small for the buoyant force on a 2-cubic-meter volume. Perhaps whoever prepared the answers mistakenly used 1.03 kg/m^3 as the density of water. (I've done that)

b) What is the weight of the log?

I did this: rho_o * V_o * g = 700*2.0*9.8 = 13720 N

However, the answer is 13.7 N and again, I can't figure out why I am off by a factor of 1000!

Again, I can't see anything you've done wrong.

c) What is the tension in the cable?

I did this: Since rho_f * V_o * g = F_tension + rho_o * V_o * g, 20188-13720 = 6468 N

The correct answer is 6470, which I'm assuming is the same as my answer, but rounded differently (correct me if I'm wrong!).

The correct answer that you've found appears to be rounded to 3 significant figures, and if you round your answer to 3 significant figures, it agrees with the correct one. I would consider your answer correct by that standard.

d) If the cable is released and the log remains vertical, what length of the log will be out of the seawater, after it reaches equilibrium?

I did this: rho_f * V_f * g = rho_o * V_o * g, so V_f=700*2/1030 = 1.35922, thus 0.5*h=1.35922 and h=2.718 m
(0.5 is the area of the base of the cylinder.)
The correct answer is 1.28m and I can't figure out where I went wrong!

You appear to have calculated the length of the log that remains in the water.

 

[lafalfa]

Thank you! I hope the answer key is the one that's wrong.