Is Tension2 always smaller than Tension1 in a tension and pulley system?

[Deadawake]

Homework Statement

M = M
There is another mass "m" attached to mass "M" with a rope
There is an acceleration (clockwise)
Does Tension₂ smaller,bigger or equal to Tension₁?

Homework Equations

F = ma

The Attempt at a Solution

So this is what I understand just by the logic:
In any case - T₁ > T₂ ,

I tried to put it into equations:
T₂ = mg - ma
T₁ = T₂ +Mg - Ma

→→ T₁ = mg - ma + Mg - Ma = m⋅(g-a) + M⋅(g-a)

"(g-a)" is constant value. Hence no matter what is the mass correlation between M & m , T₂ will always be smaller than T₁

Am I right?

Thanks a lot.

 

[TSny]

I tried to put it into equations:
T₂ = mg - ma
T₁ = T₂ +Mg - Ma

→→ T₁ = mg - ma + Mg - Ma = m⋅(g-a) + M⋅(g-a)

"(g-a)" is constant value. Hence no matter what is the mass correlation between M & m , T₂ will always be smaller than T₁

Am I right?

I think you essentially have it, except for maybe one loose end. As you said, (g-a) is a constant. What must be true about this constant in order for your argument to hold?

 

[Deadawake]

mm..
If g=a we get 0 . which means there is no tension in the ropes?

 

[TSny]

Yes, but could a = g?

Maybe I'm being too picky. But is seems to me that your argument requires that (g-a) be a positive constant.
Your equation T₁ = T₂ + Mg - Ma can be written as

T₁ = T₂ +M(g-a).

So, T₁ is greater than T₂ as long as (g-a) is positive. It's not hard to find a reason why (g-a) is positive.

 

[Deadawake]

Yes, but could a = g?

Maybe I'm being too picky. But is seems to me that your argument requires that (g-a) be a positive constant.
Your equation T₁ = T₂ + Mg - Ma can be written as

T₁ = T₂ +M(g-a).

So, T₁ is greater than T₂ as long as (g-a) is positive. It's not hard to find a reason why (g-a) is positive.

If "m" is big enough to create 9.8 m/s² acceleration. so a=g . Logically,I think, it could be the case.
Actually I can't find a reason why "(g-a)" must be positive when "a" is bigger than g... maybe that's the direction of the accelaration? if it minus this term will be always positive. but we already defined the direction of the acceleration within the equation.
so I don't know

 

[TSny]

Tension in a string cannot be negative. So, your equation T₂ = mg - ma implies g - a cannot be negative. So, you just need to rule out a = g which would imply T₂ = 0.

If T₂ = 0, then the free body diagrams for the two blocks of mass M would be identical. Yet you know that one block must accelerate upward while the other accelerates downward.

[EDIT: Another approach is to treat the system as a whole. Taking clockwise motion around the pulley as positive, the net external force acting on the system is F_net = mg + Mg - Mg = mg. Thus the acceleration is F_net divided by the total mass of the system. You can easily check that this is less than g.]