Maximum mass hanging from a rod at 50 degrees before slipping

[bluejay1]

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

 

[Redbelly98]

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.

 

[bluejay1]

I was thinking about the torque, but don't you need to know the length of the rod to find them?

 

[Redbelly98]

Let the rod length be L, and see how the torque equation works out.