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.(adsbygoogle = window.adsbygoogle || []).push({});

Answer (1.01)

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

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# Homework Help: Equilibrium and a log

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