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

[haruspex]

I have a general question, if a vertical simple pulley had two blocks both of same mass, would the tension in the string be zero?

If it were, what would be holding the blocks up?

 

[haruspex]

Basically, could I crudely imagine it like this, with the situation drawn upside down (see attachment)? If so, why do we necessarily need the pulley to rotate (as in attachment)? Why couldn't the pulley simply glide forward and have the string roll over its surface without the pulley having to rotate in order to make it roll? Thanks.

If the surface of the pulley is smooth then of course the string could slide over the surface. But with real pulleys that doesn't happen. The friction of pulley on its axle is much less than the friction of string on pulley (usually).

 

[judas_priest]

Exactly what I thought. So would it be mg or 2mg?

 

[dreamLord]

Exactly what I thought. So would it be mg or 2mg?

If it was 2mg, what would happen to the masses individually? If you wrote their force equations, would they be at rest or would they be accelerating?

 

[judas_priest]

T = Mg and T = Mg would be the two equations, but what would the total tension be? On adding the two equation and evaluating, it gives Mg. I don't have an answer with me. Hence clarifying

 

[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.