Calculating Masses in a Dynamic Pulley System

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Homework Help Overview

The problem involves a dynamic pulley system with two masses, m1 and m2, connected by a thread over a pulley. The original poster seeks to prove a specific relationship between the masses and their acceleration, but encounters difficulties in deriving the correct equation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the masses, including tension and gravitational forces, and question how to set up the equations of motion. There is a focus on the relationship between the accelerations of the two masses and the role of the pulley.

Discussion Status

Participants are exploring various interpretations of the problem, including the implications of the pulley being massless and the motion of m1 on a plane. Some guidance has been offered regarding the setup of equations, but there is no explicit consensus on the correct approach or solution.

Contextual Notes

There is mention of confusion regarding the dynamics of m1 being on a plane and the relationship between the accelerations of the two masses. The original poster also notes a potential error in the problem setup, suggesting that there may be additional pulleys involved.

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Homework Statement



You have a a mass, m1, in a plane. This mass is connected to a pulley by a thread, to a mass m2. Prove that:

\begin{equation}
m_{1} = \dfrac{m_{2}(2g - a)}{4a}
\end{equation}

Homework Equations



0e6a5d2340e77c6e9b97cafbb2f27f0d.png


9e4631226608aba70dca3fc61ca115d1.png


The Attempt at a Solution



I don't know why, but I can get this:

\begin{equation}
m_{1} = \dfrac{m_{2}(g - a)}{a}
\end{equation}

It's a similar equation, but it's not the same. How can I get the right solution?

Thank you!
 
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PhoenixWright said:
I don't know why, but I can get this:

m1=m2(g−a)a​
(2)(2)m1=m2(g−a)a\begin{equation} m_{1} = \dfrac{m_{2}(g - a)}{a} \end{equation}

It's a similar equation, but it's not the same. How can I get the right solution?
how you take a start-give free body diagram of the masses?
 
drvrm said:
how you take a start-give free body diagram of the masses?

m1 is in a plane (it's not inclinated) and m2 is is hanging from a pulley. They are connected by a thread.

In m1 we only consider Tension force (because N = mg), and in m2 we consider Tension force and m_{2}g

Thanks
 
PhoenixWright said:
m1 is in a plane (it's not inclinated) and m2 is is hanging from a pulley. They are connected by a thread.

In m1 we only consider Tension force (because N = mg), and in m2 we consider Tension force and m_{2}g

Thanks
so , how you write the equation of motion? the pulley is massless or not? if m1 is sitting on a plane will there be motion?
 
drvrm said:
so , how you write the equation of motion? the pulley is massless or not? if m1 is sitting on a plane will there be motion?

The pulley is massless. The plane has no motion.

I use this:
\begin{equation}
m_{1}\vec{g} + \vec{N} + \vec{T_{1}} = m_{1}\vec{a_{1}}
\ \\
m_{2}\vec{g} + \vec{T_{2}} = m_{2}\vec{a_{2}}
\end{equation}
Therefore:
\begin{equation}
T_{1} = m_{1}a_{1} \ \\
T_{2} - m_{2}g = m_{2}a_{2}
\ \\
T_{1} = T_{2} (massless)
\ \\
a_{1} = -a_{2}
\end{equation}
Then, I got the equation I put before... But it's not what I must get.
 
PhoenixWright said:
The pulley is massless. The plane has no motion.

i just feel that how the mass m1 can move sitting tight on the plane- if it is raised from the plane by the string then what happens to gravitational pull-sorry i can not visualize it ,pl make the dynamics more clear
 
drvrm said:
i just feel that how the mass m1 can move sitting tight on the plane- if it is raised from the plane by the string then what happens to gravitational pull-sorry i can not visualize it ,pl make the dynamics more clear

This is a pic of the situation:

zxv1g2.jpg
 
PhoenixWright said:
This is a pic of the situation:

in the above case your equations seem to be correct but the accelerations are not related like a(1) = -a(2) as one is horizontal and the other is vertical
 
drvrm said:
in the above case your equations seem to be correct but the accelerations are not related like a(1) = -a(2) as one is horizontal and the other is vertical

Thanks, but, anyway, I can't get the solution if I suppose a1 = a2...
 
  • #10
Problem solved. The exercise was wrong, there were 2 pulleys, not only 1.
 

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