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

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The discussion focuses on determining the maximum force F that can be applied to a vertically resting rod before it slips, considering static friction. Participants emphasize the importance of analyzing the torque and resulting acceleration of the rod, as well as applying Newton's second law for translational motion. The correct torque should be calculated using the force applied at the midpoint of the rod, which affects the tangential acceleration of the center of mass. Adjustments to the initial calculations are necessary, particularly regarding the torque and acceleration formulas. The final consensus is that the approach needs to be refined to accurately reflect these factors for a correct solution.
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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 \mu_{s}, 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:

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

\alpha = LF/I = LF/(1/3)ML^{2}
than
a= \alpha L
than
F = m[ \alpha L + \mu_{s}g]
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

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

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:

F = (-2 \mu_{s}Mg)

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

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

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:

F = (-2 \mu_{s}Mg)

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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