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Two masses and a pulley

  • Thread starter Remi David
  • Start date
  • #1
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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Answers and Replies

  • #2
jbriggs444
Science Advisor
Homework Helper
2019 Award
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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?
 
  • #3
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
 

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