What Initial Velocity Ensures a Glass Slides Off a Table with Coulomb Friction?

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SUMMARY

The discussion focuses on determining the initial velocity required for a glass to slide off a table under the influence of Coulomb friction. The primary equation used is F(friction) = -c(dx/dt), where c is the friction coefficient. Participants clarify that while the problem states only friction acts on the glass, an initial push is necessary to set it in motion. The conversation also addresses the distinction between Coulomb friction and linear damping forces, emphasizing the need for a clear understanding of these concepts in solving the problem.

PREREQUISITES
  • Coulomb friction principles
  • Basic differential equations
  • Understanding of linear damping forces
  • Newton's laws of motion
NEXT STEPS
  • Study the mathematical solutions for problems involving Coulomb friction
  • Learn about linear damping forces and their applications
  • Explore the derivation of equations of motion under friction
  • Investigate the differences between Coulomb friction and viscous drag
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Students in physics or engineering, particularly those studying dynamics and friction, as well as educators looking to clarify concepts related to motion and forces.

kristiakemi
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Homework Statement


Consider a glass of mass m starting at x=0 and sliding on a table of LENGTH H. For what initial velocites V(naught) will the glass fall off the table if the only force is Coulomb friction?


Homework Equations


Fcoulomb = -c(dx/dt) where c >0 and is defined as the friction coefficient.


The Attempt at a Solution



I assumed that we need to find the min initial velocity such that the glass just barely falls off. Thus the initital velocitIES will be all those values greater than V(naught). Sum of the Forces in the x direction are : coulomb friction + F(the initial push). Can I solve this system by saying m*d^2x/dt^2) -c(dx/dt) = 0 with differential equations?

Thanks!
 
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Yes, one of the solutions is c_{1}, and the other is c_{2} e^{\frac{c}{m}t}

Why do you have F(initial push)? even thought you didn't use it, the problem states only one force acting on the glass.
 
Last edited:
I think this is where I'm confused: the problem says the only force acting on it is from friction but there must be another force acting on it in the x direction in order for it to begin moving, right? Since the friction is working against the motion, I know that the F(friction)<Fpush...if there is a push?
 
You don't care about what gave the glass the magnitude of v(0), what you care is to which value of the initial speed, the glass will fall off the table.

Now set the ODE for the v case, too.
 
By the way, i dosed for a moment, but cv is not Coulomb's friction.
 
its not? Thats how its defined in my book...it says exactly:

F(friction) = -c(dx/dt) where c is a positive constant (c>0) referred to as the friction coefficient. This force-velocity relationship is called a linear damping force. We can assume that the frictional force is linearly dependent on the velocity. In later sections, exercises will discuss the mathematical solution of problems involving this type of friction, called coulomb friction.

is there another equation that i should know?
 
kristiakemi said:
its not? Thats how its defined in my book...it says exactly:

F(friction) = -c(dx/dt) where c is a positive constant (c>0) referred to as the friction coefficient. This force-velocity relationship is called a linear damping force. We can assume that the frictional force is linearly dependent on the velocity. In later sections, exercises will discuss the mathematical solution of problems involving this type of friction, called coulomb friction.

is there another equation that i should know?

As far as i know, that's the viscous force (drag for laminar flows). Anyway, since I'm only familiar Coulomb friction (dependent on forces rather than velocity) in the "dry sense" i am not going to argue the book, but it looks odd to me. I'll let this argument rest for other members.
 

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