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A statics problem with a mass on the end of a string

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I have this problem.
I know G, alpha and mu. I need to know theta for equilibrium.
My problem is that in place of the thread I need to have a tension and when i write the equations of forces on all 3 axes and then I apply the conditions that Ff<=mu*N (Ff-friction force, N-normal on plane, mu-friction coefficient) I end up with the tension and that does not appear in the answer.
I don't know the physics terminologies that good in English.

[Moderator's note: Moved from a technical forum and thus no template.]
 
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berkeman

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Welcome to the PF. :smile:

It's a bit difficult for me to interpret your diagram. I'm guessing that μ is a mass on the end of a string stretched to "O"? What are the other lines and angles meant to represent? And how can that system be in equilibrium if gravity is acting "downward" on the mass μ and the string is pulling up and to the right?
 

μ is not the mass, is the friction coefficient.

G is the gravity force, T is the friction force(with Tx and Ty its components), N is the normal force(on the inclined plane) and S is the tension force on the string.

I wrote the force equations on all 3 axes along with the equilibrium condition T≤μN.

And I end up at cos(θ) inequality instead of sin(θ) as in the given answer: sin(θ)≤μ/tan(α).

I think it has something to do with the tension force but I can't figure it out.
 

TSny

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Please type your equations, if you can. It makes it easier to read and easier for helpers to quote individual equations. If I'm reading your equations correctly, they are

##G \sin \alpha - T_x -S \cos \theta = 0##
##S \sin \theta - T_y = 0##
##N - G \cos \alpha = 0##

where, as you said, ##T## is the force of friction, ##S## is the tension in the string, and ##G## is the weight of the particle.

I believe these equations are correct. But you will need something else. If ##\theta## is at the maximum value such that the particle is just about to slip, what is the direction of the friction force at that position?
 
The direction of the force of friction would be given by the vector summation of its components along x and y axis, using parallelogram law. I draw it on my sketch in top left corner.
 

TSny

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The direction of the force of friction would be given by the vector summation of its components along x and y axis, using parallelogram law. I draw it on my sketch in top left corner.
But you should be able to be more specific. What is the angle between the friction force and the y-axis when the particle is on the verge of slipping?
 

Red is the angle with x-axis, green is with y-axis and blue with z-axis.
 

jbriggs444

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Welcome to the PF. :smile:

It's a bit difficult for me to interpret your diagram. I'm guessing that μ is a mass on the end of a string stretched to "O"? What are the other lines and angles meant to represent? And how can that system be in equilibrium if gravity is acting "downward" on the mass μ and the string is pulling up and to the right?
As I interpret the diagram, we have an inclined plane at angle ##\alpha## above the horizontal.
We have a string nailed to the plane.
A puck on the surface of the plane is prevented from sliding by tension in the string.
The puck is moved to one side at an angle ##\theta## away from directly down-slope.
We are given the coefficient of friction ##\mu##
We want to know the maximum that ##\theta## can be without the puck slipping.
 
As I interpret the diagram, we have an inclined plane at angle ##\alpha## above the horizontal.
We have a string nailed to the plane.
A puck on the surface of the plane is prevented from sliding by tension in the string.
The puck is moved to one side at an angle ##\theta## away from directly down-slope.
We are given the coefficient of friction ##\mu##
We want to know the maximum that ##\theta## can be without the puck slipping.
It's given also the weight G and angle α
 

jbriggs444

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It's given also the weight G and angle α
Yes, we want the solution to be expressed as a function of the givens, ##\alpha## and ##\mu##.

By inspection, the mass, the weight and the acceleration of gravity are irrelevant. They should cancel out of any solution.
 

TSny

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Red is the angle with x-axis, green is with y-axis and blue with z-axis.
The friction force will act parallel to the sloping surface. When the string is at its maximum angle ##\theta## before slipping occurs, what is the specific direction of the friction force? It might be easiest to state the answer to this question by specifying the angle between the friction force and the string.

Hint: Suppose ##\theta## is at its maximum value so the particle is ready to start sliding. If you were to slightly disturb the system by tapping lightly on the sloping surface, the particle will slip on the surface. In what direction would the particle start to slide? Does this help in seeing the direction of the friction force just before slipping?
 

berkeman

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A puck on the surface of the plane is prevented from sliding by tension in the string.
Oh! It's a puck and not a ball. I couldn't figure out why a ball mass would be stable at an angle, thanks. :smile:
 

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