What is the maximum coefficient of friction for a composite body on a slope?

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Homework Help Overview

The discussion revolves around determining the maximum coefficient of friction for a composite body on a slope, specifically involving a rod and a pipe. Participants are analyzing the conditions under which the maximum friction occurs and the implications of torque in this scenario.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining torque calculations related to normal and friction forces, questioning the placement of the rod in relation to the pipe's axis, and exploring how these factors influence the coefficient of friction. Some are considering the relationship between the angle of the slope and the coefficient of friction.

Discussion Status

The discussion is active, with participants providing hints and clarifications regarding the setup and calculations. There is acknowledgment of the need to refine the diagram and assumptions to accurately determine the maximum coefficient of friction.

Contextual Notes

Participants are working with specific hints from a textbook and are navigating the complexities of torque and friction in the context of a composite system. There are references to the need for the rod to be positioned correctly for maximum friction, indicating potential constraints in the problem setup.

Rikudo
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Homework Statement
A thin rod of mass m is welded on inner surface of a thin-walled pipe of mass M parallel to the axis of the pipe. The composite body thus formed is placed on a uniform slope as in the figure. Inclination of the slope is adjustable. Find range of coefficient of friction between the pipe and the slope for which the body slides down without rotation.
Relevant Equations
Torque
1657948211713.png


Here is the hint that the book gave me:

"For the maximum value of μ, the rod must be to the extreme right i.e. horizontally rightwards of the axis of the pipe"

I think what it meant is the same as this:
1657948709558.png

Note: in the calculation below, ##r## is the distance from the center of pipe to the CoM of the system and ##r = mR/(m+M)##

So, the torque calculation (with the system's CoM as origin) for maximum μ will be :
##N r = f R##
##Nr=NμR##
##m/(m+M)=μ##

But, the book says that it is ##m/\sqrt{M(M+2m)}##

Where did I do wrong?
 
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I would start with this picture
torque wrt CoM: torque from normal force = Nrsinθ = torque from friction force = μN*L
Find L as an expression of r and θ
1657953100131.png
 
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drmalawi said:
The weight of the rod will also produce a torque
This will be neutralized by torque produced by the pipe's weight. Hence, in the torque equation, we only use friction and normal force.
 
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Rikudo said:
This will be neutralized by torque produced by the pipe's weight. Hence, in the torque equation, we only use friction and normal force.
Yeah I forgot that it should only slip, not rotate at all.
Look at the pic above, try to calculate µ as a function of θ
 
@Rikudo now picture is correct...
 
AH! Now I got it. Thanks!
 
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Rikudo said:
"For the maximum value of μ, the rod must be to the extreme right i.e. horizontally rightwards of the axis of the pipe"

I think what it meant is the same as this:
View attachment 304258
For information, the hint says:
"For the maximum value of μ, the rod must be to the extreme right i.e. horizontally rightwards of the axis of the pipe"​

But in your diagram, the rod (red dot) is not at the extreme right, and therefore not at the same horizontal level (i.e. not at the same height) as the axis of the pipe.
 
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Steve4Physics said:
But in your diagram, the rod (red dot) is not at the extreme right, and therefore not at the same horizontal level (i.e. not at the same height) as the axis of the pipe.
That's OK. One can find ##\mu## at some angle ##\varphi## below the horizontal line and then maximize ##\mu## with respect to ##\varphi##. Just a little bit more work, but that's what I did to explore what ##\mu## looks like as a function of ##\varphi.##
 

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