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Pulley acceleration help

  • Thread starter madah12
  • Start date
  • #1
326
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Homework Statement



hi in this picture ( http://img26.imageshack.us/img26/8458/53487724.png [Broken] ) the teacher says that even if m1 is greater than m2 m2 will accelerate down and m1 will accelerate to the right but by newtons third law shouldn't m1 have a leftward force to the rope which should pull m2 up?

Homework Equations



F(m2 to rope ) = -F(rope to m2)
F(m1 to rope ) = -F(rope to m1)

The Attempt at a Solution

 
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Answers and Replies

  • #2
648
2


Newton's 3rd law says that the rope pulls m1 to the right, and m1 pulls the rope to the left (and this in turn pulls m2 in an upwards direction*). There is no force on m1 to the left.
* This upwards force on m2 acts against the weight of m2 downwards.
So as far as m2 is concerned, there are 2 forces, its weight downwards, and the tension in the rope upwards.
As far as m1 is concerned, there is a force from the rope pulling to the right.
 
  • #3
326
1


so why does m2 accelerate downwards? since tension is up and weight is down shouldn't Fnet = 0 ?
 
  • #4
648
2


so why does m2 accelerate downwards? since tension is up and weight is down shouldn't Fnet = 0 ?
It accelerates downwards because
a) its weight mg is greater than T, the tension in the string; and
b) the mass m1 is pulled by a force (T) to the right, and the two masses are connected to the same string.
Think about it; how could m1 accelerate to the right yet m2 not accelerate down?
Because they are connected, both masses must have the same acceleration, a.
The equation for m2 is M2g-T=M2a
The equation for m1 is T=M1a
Given g, m1 and m2 you have 2 unknowns, a and T, and two equations.
Eliminate T and solve for a will give you the common acceleration of both masses.
 

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