- #1

- 15

- 0

**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!