How Does Acceleration Relate Between Two Blocks in a Frictionless Pulley System?

[dreamLord]

Yes, T = Mg. If T = 2Mg, then the net force on each mass would not be zero, and you would see both of them moving upwards with an acceleration of g.

 

[judas_priest]

Okay, got it. Perfect. Thanks.

 

[Poetria]

Hi, guys,

Could you tell me if I am on the right track:

in this problem the magnitude of the acceleration of block m1 would be:

g*(m1/m2)?

m1*g-(m2*2*g)/2 [if T=T/2)

 

[haruspex]

Hi, guys,

Could you tell me if I am on the right track:

in this problem the magnitude of the acceleration of block m1 would be:

g*(m1/m2)?

m1*g-(m2*2*g)/2 [if T=T/2)[/QUOTE]
Consider what that would imply if m1<m2.
Instead of guessing, just write out the $\Sigma F = ma$ equation for each mass.

 

[Poetria]

I have already done this but something is messed up.

For the mass 1
m1*a=m1*g-T

For the mass 2
m2*2*a=m2*2g-T/2

I understand that m2 would move to the left even if its mass was bigger than m1. Yes, that's my problem. I don't know how to combine the equations.

 

[haruspex]

m2*2g

Where does that term come from?

 

[Poetria]

I thought that for the mass 2 acceleration =2a.
a - acceleration of the mass 1
Well, it is suspect. I will think of it.

 

[haruspex]

I thought that for the mass 2 acceleration =2a.
a - acceleration of the mass 1
Well, it is suspect. I will think of it.

That's correct. It's the m2*g term that I asked about.

 

[Poetria]

I am somewhat farther but there is a mistake in a physical interpretation. Signs are messed up again.

Here you have my equations:

m(1)*g-T=m(1)*a

T=m(1)*(g-a)

m(1)*g-m(1)*(g-a)=m(1)*a

m(1)*g-m(1)*(g-a) + T/2=m(2)*2*a

m(1)*g-m(1)*(g-a) + (m(1)*(g-a))/2 = m(2)*2*a

2*m(1)*g-2*m(1)*(g-a) + (m(1)*(g-a)) = m(2)*4*a

m(1)*a+m(1)*g=m(2)*4*a

a=(-m(1)*g)/(m(1)-4*m(2))

I tried to take a photo but it wasn't legible. I should learn LaTex

 

[haruspex]

m(1)*g-m(1)*(g-a) + T/2=m(2)*2*a

I don't understand where that comes from.
Let's have an equation that only involves m₂, not m₁.
You don't need LaTeX for this, just use subscripts and superscripts. They're the X₂ and X² symbols above where you type.
If you can't be bothered to use those, use m and M for the two masses.
But whatever, please do not post images of handwritten working!

If you do embark on LaTeX, a useful trick is to right-click on LaTeX someone else has posted and select Show Math As -> Tex. You can then copy the text that appears. Just add a double hash (# # with no space in between) at each end.

 

[Poetria]

Ok, many thanks. I will try to work this out in this way.

I thought m(1)*g-m(1)*(g-a) is a force excerted on the pulley to which m(2) is attached and T/2 - the tension of a string for m(2). But if you don't understand this it is obvously wrong.

I have drawn this: It is strange because apart from minus signs the solution was correct but I don't understand it so it is not useful anyway.