Equation for gravitational torque

  • Thread starter alem
  • Start date
  • #1
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Homework Statement


A thin rod (uniform density and thickness)
with mass M and length L attached to floor at a fixed location by a frictionless hinge.

while balanced vertically the gravitational torque acting on the rod is
a.) zero
b.) 1/2 MgL
c.)1/3 MgL
d.)1/4 MgL
e.)1/6 MgL

Homework Equations



I know that torque is F x D(or L)
F=Ma
and a is g in this case,
so T= MgL

The Attempt at a Solution



as stated above I got to T=MgL
but I don't understand where the fractions are coming from? can someone please explain this?
 

Answers and Replies

  • #2
390
1
You forgot something when computing the cross product, there's a trig function in there somewhere ;)
 

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