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

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Homework Help Overview

The problem involves a rod of mass M that is held vertically on the floor by static friction, with the goal of determining the maximum force F that can be applied at its midpoint before the rod slips. The context includes concepts from dynamics and static friction, specifically focusing on the coefficient of static friction, μ_s, and the forces acting on the rod.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need to analyze the forces and torques acting on the rod, with references to free body diagrams (FBD) and the application of Newton's laws. There is exploration of rotational dynamics, including the moment of inertia and angular acceleration, as well as translational motion of the center of mass.

Discussion Status

The discussion is ongoing, with participants providing insights into the relationships between force, torque, and acceleration. Some participants are questioning the validity of their calculations and the implications of their results, particularly regarding the dependence of force on itself and the interpretation of negative values in their equations.

Contextual Notes

There are indications of confusion regarding the setup of the problem, particularly in terms of torque calculations and the application point of the force. Participants are encouraged to revisit their assumptions and calculations to clarify their reasoning.

skateza
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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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