Coefficient of static friction for tongs.

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Homework Help Overview

The discussion revolves around a problem involving the coefficient of static friction for tongs holding a tube. The original poster expresses confusion regarding the forces and moments acting on the tongs and the tube, particularly in relation to static equilibrium conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze the forces acting on the tongs and the tube, questioning the relationships between these forces and their implications for static equilibrium. Some participants suggest drawing free body diagrams and resolving forces into components to better understand the problem.

Discussion Status

Participants are actively engaging with the problem, offering guidance on how to visualize the forces involved. There is a focus on understanding the geometry of the situation and the relationships between normal and frictional forces. The original poster expresses gratitude for the assistance received, indicating a productive exchange.

Contextual Notes

The original poster indicates uncertainty about calculating the normal force and the coefficient of friction, which are central to solving the problem. There is mention of specific angles and the need to consider equilibrium conditions just prior to slippage.

jonjacson
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Homework Statement



I´ll show the problem with a picture:

2s0ldfp.jpg


Homework Equations



M=rxf

Sum Forces = 0

Sum Moments= 0

The Attempt at a Solution



I recognize that I am completely lost with this problem, so I´ll try to analyze every force acting on every body to show where are my doubts.

When I analyze forces and moments acting at the part T1, I see a red force F2 applied to it, it is vertical, T1 acts on the tube through F3, I think that F3 is much closer to the axis of rotation at point A, so F3 should be bigger than F2, here arises my first question.

The equation of moments for T1 tells us that F3 is bigger than F2, but What happens with the forces?

I assume that there must be an equal and opposite reaction from the tube against T1, but it is difficult to imagine, that force must come from the tong T2 acting through F4, which I assume that is equal to F7, so on the y-axis F3 and F7 should compensate, and on the x-axis the friction should be equal to F7x+F3x. IF that is true, there is a net vertical force acting to T1 which is F2 and that can´t be true in static conditions.

But I don´t know how to calculate the minimum value since Friction=μ N, I don´t know μ but I don´t know N too, I know that F5 and F6 should compensate forces through x-axis but I don´t know how to calculate this.

Any suggestions, comments ... are welcome.
 
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Draw a free body diagram of the tube. The tong surfaces are tangent to the tube. Use geometry to identify the two contact points of the tongs with the tube. These are the only locations where forces are being exerted on the tube. These forces can be resolved into components normal and tangent to the tube.

Chet
 
Well I draw that but since N are orthogonal to the friction force F I cannot understand how to solve the statics equations.

I´m sorry if I am not able to see the solution.
 

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jonjacson said:
Well I draw that but since N are orthogonal to the friction force F I cannot understand how to solve the statics equations.

I´m sorry if I am not able to see the solution.

No problem. I can help you some more. Draw lines from the center of the tube to the tangent points with the tube. What angle do these lines make with a vertical line (y-direction) through the center of the tube? Let the normal component of each contact force of the tongs with the tube be N. When the the tube is just about to slip out of the tongs, the tangential components of the contact forces will be μN. Draw these 4 components of the contact forces on your free body diagram. Note the directions that each of the 4 components is pointing. Resolve each of the 4 components into components in the x- and y-directions. By symmetry, the components in the y-direction should cancel and sum to zero. Sum the components in the x-direction, and set the sum to zero (equilibrium in the x-direction just prior to slippage).

Chet
 
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Chestermiller said:
No problem. I can help you some more. Draw lines from the center of the tube to the tangent points with the tube. What angle do these lines make with a vertical line (y-direction) through the center of the tube? Let the normal component of each contact force of the tongs with the tube be N. When the the tube is just about to slip out of the tongs, the tangential components of the contact forces will be μN. Draw these 4 components of the contact forces on your free body diagram. Note the directions that each of the 4 components is pointing. Resolve each of the 4 components into components in the x- and y-directions. By symmetry, the components in the y-direction should cancel and sum to zero. Sum the components in the x-direction, and set the sum to zero (equilibrium in the x-direction just prior to slippage).

Chet

Thank you very much!

I got the right result, I´ll show the equation:

-N sin(10) +μN cos(10) +μN cos(10) -N sin(10)=0

Finally:

μ= sin(10)/ cos(10) = 0.1763

This problem has helped me a lot to understand how friction forces act, thanks!
 
jonjacson said:
Thank you very much!

I got the right result, I´ll show the equation:

-N sin(10) +μN cos(10) +μN cos(10) -N sin(10)=0

Finally:

μ= sin(10)/ cos(10) = 0.1763

This problem has helped me a lot to understand how friction forces act, thanks!

Wow! You're welcome. It's very gratifying to receive a "thank you" like this.

Chet
 

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