When I tried this, the forces in the y I got were Tension - Pivot - weight = 0 (

Question

When I tried this, the forces in the y I got were Tension - Pivot - weight = 0 (doing static first since need to get the tension force to figure out pivot).

When I looked in my notes, a similar problem only have Tension - weight = 0, so I'm not sure there.

The next thing that I was having trouble with was getting the theta for summing up the torques.

Torque Tension = (L-1/4L)*Tension*sin(??)

Torque Bar = (1/2L-1/4L)*mg*Sin(??)

Moment of Inertia = (1/12*M*L^2+(1/4*L)^2*M)

alpha = 0 (static right now)

Torque = (L-1/4L)*Tension*sin(??)-(1/2L-1/4L)*mg*Sin(??) = (1/12*M*L^2+(1/4*L)^2*M)*alpha

From here, this would allow solving for Tension followed by angular acceleration when the bar is cut. Angular acceleration allows the solving of acceleration at the pivot, which allows solving of the force of the pivot on to the bar right when the cable is cut.

Anyways, if someone could walk me through this problem step by step, including where the thetas came from, to finally get the force of the pivot, that would be awesome : ).

Explanation / Answer

similar problem here http://www.cramster.com/answers-dec-11/physics/bar-diagram-shows-thin-bar-fixed-pivot-located-1-3l_1884045.aspx?rec=0

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