Many masses on pulleys

1. Sep 13, 2016

Karol

1. The problem statement, all variables and given/known data

All N+2 masses are m. what is the acceleration of the 2 masses at the ends.

2. Relevant equations
The rope on a weightless pulley has constant tension.

3. The attempt at a solution
The situation is the same as:
$$\left\{\begin{array}{l} Nmg-2T=Nma \\ T-mg=ma \end{array}\right.$$
$$\rightarrow~a=\frac{N-2}{N+2}g$$
$$N=1\rightarrow~a=-\frac{1}{3}g$$
And it's wrong

2. Sep 13, 2016

Staff: Mentor

I wonder about your assumed equivalent system. Try the following thought experiment with the original system: Replace the two end masses with fixed supports so that the rope is fixed at those points and the system will be static. What would be the tension in the rope? Is it Nmg? Nmg/2? Something else?

Last edited: Sep 13, 2016
3. Sep 13, 2016

Staff: Mentor

If the accelerations of the end masses is a, kinematicly, what is the acceleration of each of the N middle masses?

4. Sep 14, 2016

Karol

I will denote as x and a the vertical displacement and acceleration of the 2 masses at the ends, and with y the displacement of each of the N middle masses:
$$2x=2Ny~~\rightarrow~~y=\frac{x}{N}~~\rightarrow~~\ddot y=\frac{a}{N}$$
$$\left\{\begin{array}{l} Nmg-2T=Nm\frac{a}{N} \\ T-mg=ma \end{array}\right.~~\rightarrow~~a=\frac{N-2}{3}g$$

5. Sep 14, 2016

Staff: Mentor

That's not what I get. $$mg-2T=m\frac{a}{N}$$

6. Sep 14, 2016

Staff: Mentor

Hi Karol. Instinctively it is tempting to think that the N masses must all contribute to the net force that the end masses experience. But if you look closely you'll see that by symmetry most of the N masses are balancing each other's contributions over the pulleys between them, and these upper pulleys are supported by the "ceiling". So in effect the "ceiling" is doing most of the "lifting".

I suggest that you focus on one general representative of the N masses to write your equation. I concur with @Chestermiller 's analysis.

7. Sep 14, 2016

Karol

The positive direction is upwards, a, the edge's acceleration is positive and ay, the acceleration of the other masses is negative:
$$\left\{\begin{array}{l} 2T-mg=-ma_y=-m\frac{a}{N} \\ T-mg=ma \end{array}\right.~~\rightarrow~~a=-\frac{N}{2N+1}g$$
Not good, a<0.

8. Sep 14, 2016

Staff: Mentor

Surprisingly, this is actually correct. The two end masses move down, and all the other masses move up. This is a "lever" effect.

9. Sep 15, 2016

Karol

Thank you Haruspex and gneill. so only the 2 masses adjacent to the 2 outer masses contribute. the outer masses must hold half of the adjacent masses, $\frac{1}{2}mg$, plus the acceleration of all the inner N mases

10. Sep 15, 2016

Staff: Mentor

You're welcome. But I think you meant to thank Chestermiller rather than Haruspex. Although Haruspex is often very helpful, too!

11. Sep 15, 2016

Karol

Sorry, true... but thank you all....