Max Mass for Equilibrium Problem Homework

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The discussion revolves around solving a homework problem involving a mass hanging from a string attached to a wall, requiring the determination of the maximum mass before the rod slips. Key equations include the net forces in both the x and y directions, with tension in the string equated to the weight of the hanging mass. The angle of 50 degrees is crucial, affecting both the frictional force and the torque calculations. Participants emphasize the importance of considering torques, which must sum to zero, and suggest incorporating the rod's length into the torque equation. Understanding these concepts is essential for correctly solving the equilibrium problem.
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Homework Statement


Basically, the string is attached a to a wall at the top right, with a mass m hanging on the other end. It asks to find the max mass m that can hang before bottom of the rod slips.

Homework Equations


Fnetx=0
Fnety=0

The Attempt at a Solution


I know that Ft=mg for the hanging mass m.
Then for the x forces on the rod: Ft=Ffs
so usFn=mg
But I don't know what to do with the angle 50 degrees. You have to include it somewhere right?
I was thinking us(50)(9.8)(sin50)=m(9.8) but I don't know if that's right.

Thanks
Adam
 

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Welcome to PF.

There are also torques to consider, which must sum to zero. (The angle will play a role here.)

Also, while the tension in the vertical section of string is mg, in the horizontal section it can be something different.
 
I was thinking about the torque, but don't you need to know the length of the rod to find them?
 
Let the rod length be L, and see how the torque equation works out.
 
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