Balancing torques, static equalibrium

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SUMMARY

The discussion focuses on calculating the tension in a supporting string for a uniform rod in static equilibrium, which is attached to a vertical wall at an angle theta. The downward torque is expressed as (L/2)Mg sin(theta), where M is the mass of the rod and L is its length. To achieve equilibrium, the counterclockwise (CCW) torque from the tension must balance the clockwise (CW) torque from the weight of the rod, leading to the conclusion that T = Mg/2, where T is the tension in the string.

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Homework Statement



A uniform rod of mass M sticks out from a vertical wall and points toward the floor. If the smaller angle it makes with the wall is theta, and its far end is attached to the ceiling by a string parallel to the wall, find the tension in the supporting string.

Homework Equations


ok I know that this involves balancing torques and the object is in static equalibrium.
this is a MCAt style question so I'm giving multiple choice answers, I know the answer to the question, but I wasn't sure how it was derived.



The Attempt at a Solution



I know that my downward torque based on the rod is (L/2)Mg sin theta. since net torque is 0, the CCW torque must balance the CW torque, but how does this make T = mg/2?
 
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Bu2009 said:

Homework Statement



A uniform rod of mass M sticks out from a vertical wall and points toward the floor. If the smaller angle it makes with the wall is theta, and its far end is attached to the ceiling by a string parallel to the wall, find the tension in the supporting string.

Homework Equations


ok I know that this involves balancing torques and the object is in static equalibrium.
this is a MCAt style question so I'm giving multiple choice answers, I know the answer to the question, but I wasn't sure how it was derived.



The Attempt at a Solution



I know that my downward torque based on the rod is (L/2)Mg sin theta. since net torque is 0, the CCW torque must balance the CW torque, but how does this make T = mg/2?
Hi BU2009 and welcome to PF,

Your downward torque is correct. Now consider your upward torque, provided by the tension. At what [perpendicular] distance does this tension act from the pivot (wall)?
 

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