2 masses and 3 pulleys/accelaration

[menager31]

We must count the accelaration of mass two(M2). Pulleys and ropes are massless. I've made the free body diagram and I've got two equations and three unknowns.

Please help:)

 

[Doc Al]

I assume that you've applied Newton's 2nd law to both masses and that gave you your two equations. But what about the constraint equation that relates the accelerations of M1 and M2? That will be your third equation.

(To get additional help, show exactly what you've done so far.)

 

[menager31]

and two equations to those masses

but how i can write the third?

 

[Doc Al]

That's a good diagram showing the forces. What equations did you get?

 

[menager31]

ok
i understood it with those two tensions
and of course we can write equation for the center( downward) pulley
which is contradicted

 

[Doc Al]

I've made the free body diagram and I've got two equations and three unknowns.

What are your two equations? What are your three unknowns?

The constraint equation that I refer to relates the movement of M1 to the movement of M2.

 

[menager31]

What are your two equations? What are your three unknowns?

The constraint equation that I refer to relates the movement of M1 to the movement of M2.

a1, a2, and tension



but I think that topic is closed(write free body diagram for the upward center pulleys)

 

[Doc Al]

So you've solved the problem?

 

[menager31]

no, sorry for my previous posts, I made a mistake.

This scheme is good but I still don't know how to solve this.

 

[lightgrav]

what does the attachment say?

 

[menager31]

//the name?

ok I wrote Newton's second law for the most left pulley:

0 * a = 2T-T

and for the rest massess if T=0 then accelaration of each mass is g, but they cannot move

is it correct?^^

 

[learningphysics]

//the name?

ok I wrote Newton's second law for the most left pulley:

0 * a = 2T-T

and for the rest massess if T=0 then accelaration of each mass is g, but they cannot move

is it correct?^^

Yes, that's what I get as the answer also. It was a kind of trick question. They can move... there's just no tension in the rope...

 

[lightgrav]

No, the tension is the same all through the single rope ...

If you try to apply Newton's 2nd to a "massless" object,
you run into infinities.

Always apply Newton#2 to NONzero mass! (like m1 or m2 !)

 

[learningphysics]

No, the tension is the same all through the single rope ...

If you try to apply Newton's 2nd to a "massless" object,
you run into infinities.

Always apply Newton#2 to NONzero mass! (like m1 or m2 !)

Suppose we used a tiny nonzero mass for the pulleys... then the result would be that the accelerations of each of the two masses is approximately g... I think these zero mass situations are taken as "limit" behavior of small masses...

 

[menager31]

No, the tension is the same all through the single rope ...

If you try to apply Newton's 2nd to a "massless" object,
you run into infinities.

Always apply Newton#2 to NONzero mass! (like m1 or m2 !)

but have a look at this:
http://books.google.com/books?id=Go...ght+pulley+A"&sig=64eRAvW-1cTzjse6U2iXcTrQJxQ


so what's the solution how to write the third equation?

 

[Doc Al]

Still waiting for you to show your two equations. (For m1 & m2.) Then we can worry about the third.

 

[menager31]

m2a2=m2g-T
m1a1=m1g-2T

 

[Doc Al]

OK. Now you must connect the two accelerations. How are they related? Imagine the system in motion--if m2 drops 1 m, what happens to m1?

Assume that the acceleration of m2 has magnitude "a" and is downward. Figure out the direction and magnitude of the acceleration of m1 in terms of "a".

 

[menager31]

yes I've got problem with this
I also tried to do this using conservation of energy (kinematic and
potential) and then differentiate it and solve equation but I've got problems with what you wrote

 

[Doc Al]

constraint equation

No matter how you try to solve this, you'll still need to understand the constraint equation.

Since m1 and m2 are connected by a string, their accelerations are not independent. If you are having difficulty seeing how they relate, get a piece of string or a strip of paper and figure it out. (I'm serious.)

Here's an example of a constraint equation for the simple case of two masses hanging over a single pulley. If one mass goes up 1 m, then the other must go down 1 m. Thus, if one has acceleration +a, the other has acceleration -a. (Written as an equation, that would be: a1 = -a2.)

 

[menager31]

I've only know what heppens withe the string. In case when m1 goes up, m2 goes down and most left part of the sting gets +x, most right -0,5, that part of the rope which end is on the most left pulley gets +0,5x and the last one gets -x.

 

[Doc Al]

Since both lower pulleys are free to move, and you only get one constraint equation from the inextensible string, there's not enough information to solve this problem uniquely. Too many degrees of freedom. Interesting! (My bad, for suggesting otherwise--I didn't look at the diagram too closely at first.)

 

[menager31]

?
I solved it.

 

[Doc Al]

?
I solved it.

Perhaps I missed something, so please share your solution.

 

[menager31]

Perhaps I missed something, so please share your solution.

h2+2h1=const
so
a2+2a1=0

now we've got three equations and three unknowns(a1,a2,T)

a1=g(2m2-m1)/(4m2+m1)

 

[learningphysics]

h2+2h1=const

how did you get that equation?

 

[menager31]

ok , so d1 is distance from m1 to horizontal plane and d2 from m2 to this plane.

when m2 goes down 1,5x then m1 goes up 0,75x
są now distances from the plane are:
h1=d1+0,75x
h2=d2-1,5x
h2+2h1=d2+2d1=const

 

[learningphysics]

ok , so d1 is distance from m1 to horizontal plane and d2 from m2 to this plane.

when m2 goes down 1,5x then m1 goes up 0,75x

Are you sure? I think m1 and m2 can both go down... the distance between the two lower pulleys will shrink...

 

[menager31]

no, I made an experiment :)

 

[einstein_fan]

And this question is also taken from the Polish Physics Olympiad... If it was a polish forum, menager31 would be banned from it...