Physics problem involving sum of torques

[hholzer]

Homework Statement

Given a beam of 5 meters and uniform weight 100N position on a fulcrum
such that it subtends an angle of 20 degrees and the length of the beam
to the left of the fulcrum is 3.3m, how far could an 80N cat walk before
the beam tips?

Here is an image that I drew up in mspaint:
http://i52.tinypic.com/2ntbqqb.png

Homework Equations

Clearly, we want the sum of the torques.

T_1 + T_2 + T_3 = 0

where T_i = r*F

where r is the perpendicular distance from the axis of rotation
to the force F.

The Attempt at a Solution

My thoughts were as follows:

we have three torques(from left to right):

T_1(keeping the beam down),
T_2(from the fulcrum),
T_3(from the cat)

T_2 can be ignored as it is in line
with the axis of rotation, i.e.,
it contributes nothing.

T_1(from the weight of the beam,
keeping it down) would be
the center of gravity of the beam of
3.3m, which for a uniform beam that
is 100N over 5m, would be 66N.

This center of gravity would intersect
with the perpendicular distance from the
axis of rotation, which would be (3.3/2)*cos(20).

Hence, T_1 = 66N*(3.3/2)*cos(20)

and it is positive as it works in the CCW direction.

T_3 would operate in the CW direction and
hence be negative; also, the distance from the
point to where the beam tips from the cat walking
on to it would be x*cos(20) and given that
the cat is 80N, we end up with:
- T_3 = - x*cos(20)*80N

Thus,
T_1 - T_3 = 0
<=>
T_1 = T_3

and we solve for x.

But, this is not right. I am missing
something conceptually. However,
I'm not seeing what that is.

Maybe I need to account for the weight
of the beam on the right side
of the fulcrum?

 

[PhanthomJay]

But, this is not right. I am missing
something Maybe I need to account for the weight
of the beam on the right side
of the fulcrum?

That's correct. Or you could consider the weight of the entire beam located at its cg half way (2.5 m) up the beam.

 

[hholzer]

That is much easier, actually. Thanks for the reply.