Max Force F on Rod Before Slipping: $\mu_s Mg$

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A rod of mass M rests vertically on the floor, held in place by static friction. IF the coefficient of static friction is [tex]\mu_{s}[/tex], find the maximum force F that can be applied to the rod at its midpoint before it slips.

I'm not exactly sure what i am suppose to be looking for.. Obviously there will be a FBD of the rod and friction will oppose the force, i need to find the maximum force of static friction and than anything greater than that will push the rod over. Where do i start with this?
 
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The rod is going to rotate but not necessarily slip if F is small enough. You must look at the torque and resulting acceleration of the rod, then apply Newton 2 for the translational motion of the rod's center of mass. You need to find I of the rod also before you begin.
 
so are you saying i would use:

[tex]\alpha = LF/I = LF/(1/3)ML^{2}[/tex]
than
[tex]a= \alpha L[/tex]
than
[tex]F = m[ \alpha L + \mu_{s}g][/tex]
 
skateza said:
so are you saying i would use:

[tex]\alpha = LF/I = LF/(1/3)ML^{2}[/tex]
than
[tex]a= \alpha L[/tex]
than
[tex]F = m[ \alpha L + \mu_{s}g][/tex]
good, you are almost there; however, the load is applied at midpoint...adjust your torque and the tangential aceleration of the center of mass accordingly.
 
Ok i did that and i got as an answer

[tex]F = (3F/2) + \mu_{s}Mg[/tex]

is that right?

(I'm trying to think through this verbally here:)
I can't see how its possible for a force to depend on the force itself?... or can i re-arrange the formula and than i would get:

[tex]F = (-2 \mu_{s}Mg)[/tex]

But why the negative?
 
Last edited:
skateza said:
Ok i did that and i got as an answer

[tex]F = (3F/2) + \mu_{s}Mg[/tex]

is that right?

(I'm trying to think through this verbally here:)
I can't see how its possible for a force to depend on the force itself?... or can i re-arrange the formula and than i would get:

[tex]F = (-2 \mu_{s}Mg)[/tex]

But why the negative?
I don't know what you did here...your original approach was good, its just that you forgot thet the torque was FL/2 and not FL; and the acceleration was (alpha)L/2 and not (alpha)L. Redo and resubmit, please.
 

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