Two masses and a pulley

Hello !

Homework Statement

Two blocks M1 and M2 are connected by a string of negligible mass. If the system is released from rest, find how far block M1 slide in time $t$. Neglect friction.

Diagram:

See Attached Image

Clue given in the manual:

If M1 = M2, then solution is $x(t)= \frac{gt^2}{4}$

Homework Equations

[/B]
Newton's Laws:

$\sum_{}^{} \vec{F} = m\vec{a}$
$\vec{F}_{m1/m2} = -\vec{F}_{m2/m1}$

Relation between $x(t)$, $\vec{v}(t)$, $\vec{a}(t)$:

$\vec{v}(t) = \frac{dx}{dt}$
$\vec{a}(t) = \frac{dv}{dt}$

The Attempt at a Solution

So I started the exercise by separating the two masses M1 and M2, and I did a diagram with all the forces exerting on each masses.

Then I used my two newton's laws, to find out the relation between:

1) The tension generated on each masses (Newton third law)
2) Acceleration of M2 based on the tension generated by M2

Here is my approach:

I wrote down:

$\sum_{}^{} \vec{F}_{M2} = m_{M2}\vec{a}_{M2}$
$\sum_{}^{} \vec{F}_{M1} = m_{M1}\vec{a}_{M1}$

So, with $\vec{T}$ being the tension applied to the string, and $\vec{R}$ the resistance from the table:

$\vec{T}_{M2} - \vec{W}_{M2} = m_{M2}\vec{a}_{M2}$
$\vec{R}_{M1} - \vec{W}_{M1} + \vec{T}_{M1} = m_{M1}\vec{a}_{M1}$

I simplified the equation for M1, by suppressing $\vec{R}_{M1}$ and $\vec{W}_{M1}$ as $\vec{R}_{M1} - \vec{W}_{M1} = 0$
I concluded that tension generated by M2 was the tension applied on M1

So,
$\vec{T}_{M2} = - \vec{T}_{M1}$
$\vec{T}_{M1} = - m_{M2}\vec{g}$

using that relation, we can write:

$\vec{a}_{M1} = \frac{m_{M2}}{m{_M1}}g$

Differentiating , I found that:

$\vec{v}(t) = v_{0} + \frac{m_{M2}}{m{_M1}}gt$ with $v_{0} = 0$
$\vec{x}(t) = x_{0} + \frac{m_{M2}}{2m{_M1}}gt^2$ with $x_{0} = 0$

As per the clue given in the manual, I should have a "4" instead of a "2" on the last equation... Looks like my mistake is coming for my differentiation, but I have no idea why... The "4" will come when differentiating a $x^3$.

Don't give me the answer, I just want to have clues as I'm looking to improve (I'm not involved in any kind of scholarship at this time, and I'm doing that for my personal pleasure :D).

Sorry if my English is not the best, English is not my first language (but I'm trying to improve as well)

Thanks
Have a good day
Rémi

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jbriggs444
Homework Helper
2019 Award
It appears that you have assumed that the force $\vec{T_{M2}}$ acting on $M_2$ from tension in the string is given by $g\ m_{M_2}$.

How did you arrive at that?

Hello,

You are right ! I just assumed that the magnitude of $\vec{T_{M2}}$ and $\vec{W_{M2}}$ were equal.
But, now I realize that it's a nonsense as we have an accelerated movement towards the ground.

I will start the exercise again, and see how I can figure out $\vec{T_{M2}}$.
I'll will come back later with the solution

Thanks
Rémi