Dynamic problem with pulley

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1. Mar 17, 2016

PhoenixWright

1. The problem statement, all variables and given/known data

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:

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

2. Relevant equations

3. The attempt at a solution

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

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

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

Thank you!

2. Mar 17, 2016

drvrm

how you take a start-give free body diagram of the masses?

3. Mar 17, 2016

PhoenixWright

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

4. Mar 17, 2016

drvrm

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?

5. Mar 17, 2016

PhoenixWright

The pulley is massless. The plane has no motion.

I use this:

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}}

Therefore:

T_{1} = m_{1}a_{1} \ \\
T_{2} - m_{2}g = m_{2}a_{2}
\ \\
T_{1} = T_{2} (massless)
\ \\
a_{1} = -a_{2}

Then, I got the equation I put before... But it's not what I must get.

6. Mar 17, 2016

drvrm

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

7. Mar 17, 2016

PhoenixWright

This is a pic of the situation:

8. Mar 17, 2016

drvrm

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

9. Mar 17, 2016

PhoenixWright

Thanks, but, anyway, I can't get the solution if I suppose a1 = a2...

10. Mar 17, 2016

PhoenixWright

Problem solved. The exercise was wrong, there were 2 pulleys, not only 1.