Three masses, a string and pulley and a table -- solve for the acceleration

[Rishabh Narula]

Homework Statement

Q) Two particles of masses m and 2m are placed on a smooth horizontal table. A string,
which joins them, hang over the edge supporting a light pulley,
which carries a mass 3m
Two parts of the string on the table are parallel and
perpendicular to the edge of the table. The parts of the string
outside the table are vertical. Show that the acceleration of the
particle of mass 3m is 9g/17.

Relevant Equations

F = ma

Let T be the tension in the string, a be the acceleration of
mass 2m, 2a be the acceleration of mass m
T = (m) (2a) ---eq(1)
The mass 3m will come down with acceleration
a’ = (a+2a)/2 = 3a/2

3mg - 2T = 3m . 3a/2

from equation 1

3mg - 2(2ma) = 3m . 3a/2

thus a = 6/17g
thus acceleration of 3m mass = 3/2 a

A' = (3/2) (6/17g) = 9/17 g

i understand the whole solution except for the part where it says -
"The mass 3m will come down with acceleration
a’ = (a+2a)/2 = 3a/2"

Why does 3m come down with accn. = a + 2a / 2 = 3a/2...WHY IS THIS SO?

 

[haruspex]

Problem Statement: Q) Two particles of masses m and 2m are placed on a smooth horizontal table. A string,
which joins them, hang over the edge supporting a light pulley,
which carries a mass 3m
Two parts of the string on the table are parallel and
perpendicular to the edge of the table. The parts of the string
outside the table are vertical. Show that the acceleration of the
particle of mass 3m is 9g/17.
Relevant Equations: F = ma

Why does 3m come down with accn. = a + 2a / 2 = 3a/2...WHY IS THIS SO?

First, imagine the m accelerates at rate 2a but the 2m is held steady. What would 3m's acceleration be?
Now swap over, holding the m fixed, the 2m accelerating at a. What would 3m's acceleration be now?

 

[Rishabh Narula]

First, imagine the m accelerates at rate 2a but the 2m is held steady. What would 3m's acceleration be?
Now swap over, holding the m fixed, the 2m accelerating at a. What would 3m's acceleration be now?

umm.i know you're trying to get me to think which is a good thing but that's confusing me more...could you please just tell what's your point?

 

[haruspex]

umm.i know you're trying to get me to think which is a good thing but that's confusing me more...could you please just tell what's your point?

I'll make it a bit easier.
Suppose one mass moves distance x towards the edge while the other stays fixed. How far will the 3m descend?

 

[Rishabh Narula]

I'll make it a bit easier.
Suppose one mass moves distance x towards the edge while the other stays fixed. How far will the 3m descend?

man please answer the question straight forward...im already confused about it that's why i put it up here,i.e don't answer a question with more question will ya... :3

 

[haruspex]

man please answer the question straight forward...im already confused about it that's why i put it up here,i.e don't answer a question with more question will ya... :3

I have posed you a very easy question in post #4. If I merely tell you the answer you still won't know why; you have to see it for yourself.
But I can try to walk you through it a little.
(I've edited the next para to make it a bit clearer.)
Suppose the two masses on the table are each distance z from the edge.
The string runs from one, to the edge of the table, then down a distance y, under the pulley from which the mass 3m hangs, back up to the table and along to the other mass.
What is the total length of the string?
Now one mass moves distance x towards the edge. The mass 3m is now distance y' below. What is the total length of the string now?

 

[Rishabh Narula]

okay.for first it should be 2z + y...and second should be (z-x) + z +y'=2z - x + y'.if you mean to say total length doesn't change.the 2z + y = 2z - x + y' implies, y'= y + x. now what next since the two masses have different accelerations...and would the 3m mass again decend y' for the second mass...don't know what to do next..?

 

[haruspex]

for first it should be 2z + y

No. The string goes over the edge of the table, down to a pulley and back up. Any distance from the pulley to the 3m mass is in a different string, so doesn’t count.

 

[Rishabh Narula]

No. The string goes over the edge of the table, down to a pulley and back up. Any distance from the pulley to the 3m mass is in a different string, so doesn’t count.

...please enlighten with the right way if you know it... :3

 

[haruspex]

...please enlighten with the right way if you know it... :3

I've edited post #6 to make it clearer. Try now.

 

[jbriggs444]

Two parts of the string on the table are parallel and
perpendicular to the edge of the table. The parts of the string
outside the table are vertical.

I had quite a bit of difficulty picturing this arrangement. The wording is poor and the scenario is, strictly speaking, impossible. [Some of that assessment may be "sour grapes" on my part]

Apparently the bits of string leading away from the masses are nearly parallel to each other and exactly perpendicular to the edge. Or exactly parallel to each other and nearly perpendicular to the edge.

In any case, the places where the two segments hang over the edge are close to one another, separated by the width of the pulley below.

 

[haruspex]

I had quite a bit of difficulty picturing this arrangement. The wording is poor and the scenario is, strictly speaking, impossible. [Some of that assessment may be "sour grapes" on my part]

Apparently the bits of string leading away from the masses are nearly parallel to each other and exactly perpendicular to the edge. Or exactly parallel to each other and nearly perpendicular to the edge.

In any case, the places where the two segments hang over the edge are close to one another, separated by the width of the pulley below.
View attachment 245630

I assumed the table to be rectangular. Doesn't that make it more reasonable?

 

[Rishabh Narula]

I have posed you a very easy question in post #4. If I merely tell you the answer you still won't know why; you have to see it for yourself.
But I can try to walk you through it a little.
(I've edited the next para to make it a bit clearer.)
Suppose the two masses on the table are each distance z from the edge.
The string runs from one, to the edge of the table, then down a distance y, under the pulley from which the mass 3m hangs, back up to the table and along to the other mass.
What is the total length of the string?
Now one mass moves distance x towards the edge. The mass 3m is now distance y' below. What is the total length of the string now?

is it 2z +2y for first one?and 2z -x + 2y' for second?could you just tell the right way and explain it please...this is taking so long... :3 :3

 

[haruspex]

is it 2z +2y for first one?and 2z -x + 2y' for second?could you just tell the right way and explain it please...this is taking so long... :3 :3

Right, but of course the string does not change length, so what equation do you get?

 

[Rishabh Narula]

I have posed you a very easy question in post #4. If I merely tell you the answer you still won't know why; you have to see it for yourself.
But I can try to walk you through it a little.
(I've edited the next para to make it a bit clearer.)
Suppose the two masses on the table are each distance z from the edge.
The string runs from one, to the edge of the table, then down a distance y, under the pulley from which the mass 3m hangs, back up to the table and along to the other mass.
What is the total length of the string?
Now one mass moves distance x towards the edge. The mass 3m is now distance y' below. What is the total length of the string now?

2z + 2y = 2z - x + 2y'...or 2y'= 2y + x...what next?

 

[haruspex]

2z + 2y = 2z - x + 2y'...or 2y'= 2y + x...what next?

Right, but put it in the form of how the change in the mass on the table's position (x) relates to the change in height of the hanging mass.
Next can you figure out what that means about how their accelerations are related?
Final step is to incorporate the movement of the other tabletop mass.

 

[Rishabh Narula]

Right, but put it in the form of how the change in the mass on the table's position (x) relates to the change in height of the hanging mass.
Next can you figure out what that means about how their accelerations are related?
Final step is to incorporate the movement of the other tabletop mass.

change in height = y' - y = x/2...if we take double deravitive (accn.) on both sides here...we'll get accn. of 3m mass = 1/2 the accn.of the mass you chose to move x dis.confused with what to do next...do we take a different change in height if the other mass is made to move?and since the other mass's accn. is twice, wouldn't it move 2x towards the edge?...this is tough to figure. :3

 

[ehild]

change in height = y' - y = x/2...if we take double deravitive (accn.) on both sides here...we'll get accn. of 3m mass = 1/2 the accn.of the mass you chose to move x dis.confused with what to do next...do we take a different change in height if the other mass is made to move?and since the other mass's accn. is twice, wouldn't it move 2x towards the edge?...this is tough to figure. :3

See the picture, it makes you imagine the situation easier. Assume mass a displaces to the right by x, mass b displaces to to the right by y. As the length of the string is constant, the vertical path becomes longer by x+y. As the parts of the string between the edge of the table and the edge of the pulley are equal, the pulley (and mass c) moves down by z=(x+y)/2. The accelerations are double derivative of the displacements...

 

[Rishabh Narula]

See the picture, it makes you imagine the situation easier. Assume mass a displaces to the right by x, mass b displaces to to the right by y. As the length of the string is constant, the vertical path becomes longer by x+y. As the parts of the string between the edge of the table and the edge of the pulley are equal, the pulley (and mass c) moves down by z=(x+y)/2. The accelerations are double derivative of the displacements...
View attachment 245659

nice i get it now.the vertical path gets increased to x+y but it happens through two equal string parts.so each string part goes down by x+y/2...i.e so do pulley and mass 3m.thanks.

 

[ehild]

nice i get it now.the vertical path gets increased to x+y but it happens through two equal string parts.so each string part goes down by x+y/2...i.e so do pulley and mass 3m.thanks.

It is (x+y)/2. Do not forget the parentheses.

 

[Rishabh Narula]

It is (x+y)/2. Do not forget the parentheses.

yeah i meant (x+y)/2.