How to I find what needs to be the mass of an object in a friction problem?

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The discussion revolves around solving a physics problem involving static and kinetic friction to determine the mass required to prevent motion and to maintain constant speed. The coefficients of friction are provided, and the user is unsure how to proceed without knowing the normal force (N). They have attempted to set up free body diagrams (FBDs) and derived equations for both static and kinetic scenarios but are confused about the role of tension in the system. Clarifications are sought regarding the equations and the relationship between tension, mass, and acceleration. Understanding the tension's role when the system is at rest or moving at constant speed is crucial for solving the problem effectively.
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Homework Statement



In the figure the coefficient of static friction between mass mA and the table is 0.50, whereas the coefficient of kinetic friction is 0.42.
(a) What minimum value of mA will keep the system from starting to move?
(b) What value of mA will keep the system moving at constant speed?

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Homework Equations


So I know that

Fs <= uN
Fk = uN

But I don't know what to do when I lack the Newtons.

Can anyone give me a hint on where to start?


The Attempt at a Solution

 
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If the system isn't moving, what must be the tension in the rope?
 
Would it be 0 because it'd be at equilibrium when not moving?
 
If someone asked you to support that hanging mass mB by holding the upper end of the rope, what force would your arm have to exert? It would not be zero.
 
A force greater or equal to that of the force that is pulling down mb?
 
A force of how many Newtons?
 
Can I multiply 2 * 9.8(m/s^2) to get the mkg/s^2 = N unit?
 
Mustachio said:
Can I multiply 2 * 9.8(m/s^2) to get the mkg/s^2 = N unit?
Yes.
 
So I made 2 FBDs.

...... ^ N
......|
......|
mSn...mKn...|.... T
<---<--------|---------->
......|
......|
......| Ma(g)
...... vT
|
|
o
|
|
2g

And I derived 2 sets of equations.For the first FBD I got
X: T - mkN - msN = (Ma)a
Y: N - (Ma)g = 0

Second one I got

Y: T - 2g = 2a.

Do I have the problem set it up right?
 
  • #10
Mustachio said:
For the first FBD I got
X: T - mkN - msN = (Ma)a
What do mkN and msN stand for? If those are supposed to be the friction forces, realize that only one acts at any given time; it's either static or kinetic friction.

Y: N - (Ma)g = 0
OK.

Second one I got

Y: T - 2g = 2a.
OK. (What's a?)
Do I have the problem set it up right?
To solve the problem, answer the question I asked before:
If the system isn't moving--or is moving at constant speed--what must be the tension in the rope?
 
  • #11
I don't understand the question. :\
 
  • #12
Mustachio said:
I don't understand the question. :\
What does your equation for MB tell you? (What's the acceleration?)
 
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