# Newton's laws - Pedagogical machine

1. Aug 13, 2013

### Saitama

1. The problem statement, all variables and given/known data
A "Pedagogical machine" is illustrated in the attachment. All surfaces are frictionless. What force F must be applied to $M_1$ to keep $M_3$ from rising or falling?

Ans. clue. For equal masses F=3Mg

2. Relevant equations

3. The attempt at a solution
I would like to do this problem in both inertial and non-inertial frames but first with the inertial frame.
The net force acting on the system is F. Hence,
$$F=(M_1+M_2+M_3)a \Rightarrow a=\frac{F}{M_1+M_2+M_3}$$
Let a' be the acceleration $M_2$ and $M_3$ with respect to $M_1$.
For $M_1$, a normal force (at the point where it is in contact with $M_3$) and a tension acts on it.
$$F-N-T=M_1a (*)$$
For $M_2$, $T=M_2(a+a')$
For $M_3$, $M_3g-T=M_3a'$ and $N=M_3a$
Substituting T and N in the equation (*)
$$F=M_1a+M_3a+M_2a+M_2a'$$
Solving, $a'=0$. This is obviously wrong.

Any help is appreciated. Thanks!

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2. Aug 13, 2013

### D H

Staff Emeritus
Your problem stems from using $F=(m_1+m_2+m_3)a$. This is only true if a'=0, so of course you derived a'=0 as a consequence.

Suppose the rope is cut. The mass m3 is going to fall to the bottom of the shaft, but while doing so the shaft keeps mass m3 from moving side-to-side with respect to mass m1. Thus from a horizontal perspective, if the rope has been cut then $F=(m_1+m_3)\ddot x_1$, where $\ddot x_1$ is the horizontal component of the center of mass of mass m1. What about mass m2? It's not a part of the system anymore, at least with regard to horizontal motion. The surface is frictionless, and you don't care about the vertical component of the force on mass m1.

Now suppose the rope is intact. What horizontal forces, if any, does this add to the m1+m3 system? Hint: What does the pulley do?

3. Aug 13, 2013

### Tanya Sharma

Why are you solving for a' ? a' is zero if F is such that M3 doesnt rise or fall.

Considering horizontal direction,for M2 →T=M2a

Considering vertical direction,for M3 →T=M3g

a=(M3g)/M2

For equal masses, a=g and F=3Mg

Last edited: Aug 13, 2013
4. Aug 13, 2013

### Saitama

Tension. It would act in the direction to left. Correct?

Also, I would like to derive the acceleration of individual masses, not only for the case when M3 is not allowed to either rise or fall.

5. Aug 13, 2013

### Tanya Sharma

Hi Pranav :)

What is bothering you in this problem ?

Tension will act leftwards on M1 and rightwards on M2.

6. Aug 13, 2013

### Saitama

I don't know but I just can't reach the final answer.
I will write down the equations again.
For $M_1$, $F-N-T=M_1a$.
For $M_2$, $T=M_2(a+a')$
For $M_3$, $N=M_3a$ and $M_3g-T=M_3a'$

Oh well, looks like I get the answer by solving the above equations. I should have been a bit more careful.

Thanks for the help D H and Tanya!

Also, how do I solve this in non-inertial frame?

7. Aug 13, 2013

### voko

First off, what frame do you think is suitable here?

8. Aug 13, 2013

### Tanya Sharma

Apply a pseudo force in horizontal direction(opposite to M1) on M2 and M3 and solve just as you would do normally.

Last edited: Aug 13, 2013
9. Aug 13, 2013

### WannabeNewton

Consider the frame with acceleration $a = \frac{F}{M_1 + M_2 + M_3}$. What is the condition that $M_3$ is neither falling nor rising in this frame? It's just that $M_2$ has to be at rest in this frame. Take it from there; it's thankfully very simple.

10. Aug 13, 2013

### Saitama

That worked. Thanks WBN!