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

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SUMMARY

The maximum coefficient of friction (μ) for a composite body on a slope is determined by the position of the rod relative to the axis of the pipe. Specifically, for μ to reach its maximum value, the rod must be positioned horizontally to the extreme right of the pipe's axis. The calculations involve torque equations where the normal force (N) and friction force (f) are balanced, leading to the relationship μ = m/(m + M) and the book's assertion that μ = m/√(M(M + 2m)). The key to solving this problem lies in correctly identifying the torque contributions from both the rod and the pipe's weight.

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  • Knowledge of center of mass (CoM) and its significance in mechanics.
  • Ability to analyze forces acting on a body on an inclined plane.
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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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