Equilibrium and a log

[Supernerd2004]

A log weighs 800N and has its center of gravity at the midpoint. Find the lowest coefficient of static friction necessary to keep the log from slipping.

Answer (1.01)

http://img115.imageshack.us/img115/8407/physicsproblem3ox.th.png

Ok, so, i started out by finding my torque and force equations.
Torque = (800)*(3.5)*(cos(25)) = (Tv)*(6.5)*(cos(25)) + (Th)*(6.5)*(sin(25))

Normal Force = (800) - (Tv)

Force of Friction = (Normal Force) (Coefficient of Friction)

Force of Friction = (Th)

tan(25) = (Th)/(Tv)

So, solving for (Th) I got...
(Th) = tan(25)/Tv

So, Torque = (800)*(3.5)*cos(25) = (Tv)*(6.5)*cos(25) + (tan(25)/(Tv))*(sin(25))

Where (Tv) is the vertical component of the tension and (Th) is the horizontal component of tension.

Solving for (Tv) I get (Tv) = 353.83
Solving for (Th) (Th) = tan(25)(353.83) = 164.99

If Normal Force = Weight - (Tv) then N = 800 - 353.83 = 446.17

so (Normal Force)(Coefficient of Friction) = (Th)
Then (446.17)(Coe. of Friction) = 164.17
The coe. of friction is not equal to 1.01

Any help as to where I went wrong would be much appreciated! Thanks so much in advance for any help!

Dan

 

[Delzac]

hmm, i don't claim to be a physics pro but the coeff i calculated is the same as yours. Maybe someone want to come in and help?

 

[kyle8921]

I can see that you have used the correct equations and have approached the problem correctly. However, there seems to be a small error in your calculations. When solving for (Th), you have multiplied tan(25) by 353.83 instead of dividing by it. This would give you a value of 164.99, which is slightly different from the correct answer of 164.17. This small error could account for the difference in the coefficient of friction you have calculated. Therefore, I would suggest double checking your calculations and correcting this error to get a more accurate result. Other than that, your approach and equations seem to be correct.