What is the Tension in Block m2's String in a Pulley System?

[tzonehunter]

Homework Statement

The diagram shows a system of two blocks suspended by ideal strings and pulleys.

If the tension of the string attached to block m₁ is T, what is the tension of the string attached to block m₂?

Homework Equations

Due to conservation of string, a₁/2 = a₂
Force balance on mass 1:
∑F₁ = m₁a₁ = m₁g - T

Force balance on mass 2:
∑F₂ = m₂a₂ = m₂g - T₂

The Attempt at a Solution

Substituting a₁/2 = a₂ into the ∑F₂ expression, I get

a₁ = 2g - 2T₂/m₂

Substituting the above into the ∑F₁ expression, I get

T₂ = m₂g/2 + m₂T/(2m₁)

The correct answer is given as 2T. I'm struggling with this, because I don't see how the pulley could accelerate upwards if this is true, as all forces are balanced.

 

[Vibhor]

The tension in the string is same since the string is assumed massless .It doesn't depend on the acceleration of the blocks .Since two strings are pulling block 2 upwards the upwards force on block 2 is 2T.

 

[STEMucator]

It may also be useful to note the tension across a frictionless pulley is uniform. So if you knew the tension on the left was $T$, the tension in the two other strings is also $T$. Since there is two of them, it is $2T$.

 

[ehild]

Homework Statement

T

Homework Equations

Due to conservation of string, a₁/2 = a₂

That is true for the magnitude of the accelerations. But they are of opposite directions. If m1 moves downward, m2 moves upward.

Force balance on mass 1:
∑F₁ = m₁a₁ = m₁g - T

So you chose m1 moving downward.

Force balance on mass 2:
∑F₂ = m₂a₂ = m₂g - T₂

You have to change the sign of the right-hand side as m2 accelerates upward: m₂a₂ = T₂-m₂g

The tensions also act on the moving pulley, 2T upward and T2 downward. But the pulley is massless, so ma =0. The net force on the pulley must be zero.

ehild

 

[tzonehunter]

Thank you to those that replied. I thought about this over night... I think I'm getting caught up in the idea that the pulley is a real object, but in the problem we're asked to assume an "ideal pulley".

How does this explanation sound:
The moveable pulley (subscript P) must have the same acceleration as m₂. The acceleration of the moveable pulley is:
a_P = F_Net,P / m_P = a₂
F_Net,P = 2T - T₂

As long as m₁ and m₂ are constant, then a₂ is constant and a_P is constant.

An ideal pulley is massless, so as m_P --> 0 , F_Net,P --> 0. The lighter the pulley, the smaller the difference between 2T and T₂. Less net force is required to produce the same acceleration of a less massive object.

This leaves us with

T₂ --> 2T for the idealized case where m_P --> 0.

 

[haruspex]

Thank you to those that replied. I thought about this over night... I think I'm getting caught up in the idea that the pulley is a real object, but in the problem we're asked to assume an "ideal pulley".

How does this explanation sound:
The moveable pulley (subscript P) must have the same acceleration as m₂. The acceleration of the moveable pulley is:
a_P = F_Net,P / m_P = a₂
F_Net,P = 2T - T₂

As long as m₁ and m₂ are constant, then a₂ is constant and a_P is constant. Indeed, if the pulley had non-negligible mass then you could not take the two upward tension forces as equal.

An ideal pulley is massless, so as m_P --> 0 , F_Net,P --> 0. The lighter the pulley, the smaller the difference between 2T and T₂. Less net force is required to produce the same acceleration of a less massive object.

This leaves us with

T₂ --> 2T for the idealized case where m_P --> 0.

Yes. More generally, if an object is taken as massless then you can assume there's no net force on it (or its acceleration would be infinite). Similarly torques.