Fletcher's trolley apparatus - maximum value of coefficient

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  • #1
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Homework Statement


A Fletcher's trolley apparatus is set up with a block of wood of mass 4.0kg and a suspended mass of 2.0kg.

The Attempt at a Solution


a) Calculate the acceleration of the sys. and the tension in the string when the mass is released.

m1
FnetX = [T - Ff = m1(a)]
FnetY = [Fn - Fg = 0]

m2
No FnetX
FnetY = [T - Fg = -m2(a)]

Equation to find the acceleration in this situation:
[tex]a = \frac{Fg - Ff}{mass1 + mass2}[/tex]

The second part of the question is:
b) What is the max. value of the coefficient of friction that will allow the sys. to move?

I am stuck as to how to find this out.
I know:

[tex]Ff = μFn[/tex]
[tex]Ff = μ(mass1)(g)[/tex]
 

Answers and Replies

  • #2
Doc Al
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Hint: Express the acceleration in terms of μ. (Replace Fg and Ff with what they equal.)
 
  • #3
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That will give me the following:
[tex]a = [\frac{(mass2)(g) - μ(mass1)(g)}{(mass1 + mass2)}][/tex]
But then I have a, which I have nothing to sub in for.
 
  • #4
Doc Al
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44,981
1,247
That will give me the following:
[tex]a = [\frac{(mass2)(g) - μ(mass1)(g)}{(mass1 + mass2)}][/tex]
That's good.
But then I have a, which I have nothing to sub in for.
As μ increases, what happens to a?
 
  • #5
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a increases, but will become negative.. a is proportional to μ ....
 
  • #6
Doc Al
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a increases, but will become negative..
Have another look at the formula. (And if something increases, how can it become negative?)
a is proportional to μ ....
Not exactly.
 

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