# Equilibrium and a log

1. Jun 13, 2006

### 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.

http://img115.imageshack.us/img115/8407/physicsproblem3ox.th.png [Broken]

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

Last edited by a moderator: May 2, 2017
2. Jun 14, 2006

### Delzac

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