Calculating Masses in a Dynamic Pulley System

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SUMMARY

The discussion revolves around solving the dynamics of a pulley system involving two masses, m1 and m2. The correct equation for m1 is established as m1 = (m2(2g - a))/(4a), contrasting with the incorrect equation m1 = (m2(g - a))/a presented by a participant. Key insights include the necessity of a free body diagram to visualize forces, the massless nature of the pulley, and the clarification that the system involves two pulleys rather than one, which ultimately resolves the confusion regarding the relationship between the accelerations of the masses.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with free body diagrams
  • Basic knowledge of pulley systems and their dynamics
  • Concept of tension in strings and gravitational forces
NEXT STEPS
  • Study the derivation of equations of motion for pulley systems
  • Learn about the effects of multiple pulleys on system dynamics
  • Explore advanced topics in classical mechanics, such as energy conservation in pulley systems
  • Investigate the role of friction in pulley systems and its impact on motion
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Students studying physics, particularly those focusing on mechanics, educators teaching dynamics, and anyone interested in understanding the complexities of pulley systems in motion.

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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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