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A uniform slender beam of mass 10kg and length 1.2m is resting on a corner which is knife edged. The vertical wall on the left is smooth. What is the angle for equilibrium? a = .5m

I can't figure out this problem for the life of me. Any help in setting up formulas would be great.

I thought of using the knife-edge as the pivot point. This would give me a torque for one side of the beam and an opposite and equal torque for the other side. I gave the left side a mass m1 and the right a mass m2 for a total mass of the 10kg. if the length for the portion of the beam with mass m1 was x then the length for portion m2 is 1.2-x. so with it being in equilibrium i get a torque formula: 10(m1)(x/2)sin(theta)-10(m2)((1.2-x)/2)sin(theta)=0. I also know that cos(theta)=.5/x and sin(theta)=(sqrt(x^2-.5^2) but none of this really helps me come up with a system of equations i can solve. I am not sure if i am going in the right direction or not, so any heklp would be absolutely amazing!

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## Homework Statement

A uniform slender beam of mass 10kg and length 1.2m is resting on a corner which is knife edged. The vertical wall on the left is smooth. What is the angle for equilibrium? a = .5m

## Homework Equations

I can't figure out this problem for the life of me. Any help in setting up formulas would be great.

## The Attempt at a Solution

I thought of using the knife-edge as the pivot point. This would give me a torque for one side of the beam and an opposite and equal torque for the other side. I gave the left side a mass m1 and the right a mass m2 for a total mass of the 10kg. if the length for the portion of the beam with mass m1 was x then the length for portion m2 is 1.2-x. so with it being in equilibrium i get a torque formula: 10(m1)(x/2)sin(theta)-10(m2)((1.2-x)/2)sin(theta)=0. I also know that cos(theta)=.5/x and sin(theta)=(sqrt(x^2-.5^2) but none of this really helps me come up with a system of equations i can solve. I am not sure if i am going in the right direction or not, so any heklp would be absolutely amazing!

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