# Fletcher's trolley apparatus - maximum value of coefficient

1. Jun 17, 2011

### aeromat

1. The problem statement, all variables and given/known data
A Fletcher's trolley apparatus is set up with a block of wood of mass 4.0kg and a suspended mass of 2.0kg.

3. 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:
$$a = \frac{Fg - Ff}{mass1 + mass2}$$

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:

$$Ff = μFn$$
$$Ff = μ(mass1)(g)$$

2. Jun 17, 2011

### Staff: Mentor

Hint: Express the acceleration in terms of μ. (Replace Fg and Ff with what they equal.)

3. Jun 17, 2011

### aeromat

That will give me the following:
$$a = [\frac{(mass2)(g) - μ(mass1)(g)}{(mass1 + mass2)}]$$
But then I have a, which I have nothing to sub in for.

4. Jun 17, 2011

### Staff: Mentor

That's good.
As μ increases, what happens to a?

5. Jun 17, 2011

### aeromat

a increases, but will become negative.. a is proportional to μ ....

6. Jun 18, 2011

### Staff: Mentor

Have another look at the formula. (And if something increases, how can it become negative?)
Not exactly.