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Tension in a string

I have attached a pic with my question. I want to know if T1>T2 or not.

As much as I know, Tension in a string = Earth's Gravitational force. But this question is quite confusing.
 

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T=2m1m2g
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(m1+m2)
 
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t2 would be greater if the mass in t2 is greater than the mass in t1
 

Doc Al

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I have attached a pic with my question. I want to know if T1>T2 or not.
Is the pulley massless and frictionless? That's the key.
As much as I know, Tension in a string = Earth's Gravitational force.
Careful! That's not true. In general you'll have to solve for the tension in the string by applying Newton's 2nd law.
 
T1 has a mass greater than T2.
 

Doc Al

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T1 has a mass greater than T2.
I don't know what you mean. T1 and T2 are tensions. Regardless of the size of the mass on the ends of the string segment, the tension on both sides of a massless and frictionless pulley will be the same.

Since the mass on each side of the pulley is the same (2m on each side, added up), it turns out that in this particular case the tension T1 = T2 = 2mg. (The masses don't accelerate.)
 
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Agree completely with DocAl and will add that the tension in a string is the same everywher.... all the way along it.
There is no resultant force on this system so it is either at rest or moving with constant velocity (speed) in one direction or the other.
The only string with a different tension is between the 2 masses m and m ...the tension in here will be mg
The tensions T1 and T2 will be 2mg
 
I personally think that T2 was meant for the second string connecting the masses on the right side of the pulley. Because usually the pulley doesn't have mass (for beginner class) and the tension is uniform in one continuous string (and this is assumed rather than stated). So... I worked out the problem w these assumptions below. Check it out. So the solution looks like T1>T2.
 

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Agree completely with DocAl and will add that the tension in a string is the same everywher.... all the way along it.
Only if the string is also of negligible mass :biggrin:

And if the pulley has some mass, you can find its moment of inertia and apply the equations of torque:

[itex]T_1 R - T_2 R = I\alpha = \frac{Ia}{R}[/itex]

Here I is moment of inertia of pulley and R is its radius ...

And if pulley has no mass, I=0 ...
 
I apologize for confusing you all. I am sorry. I have cleared my doubts with the question. The question was that there's a pulley and two masses 2m and m are tied to the opposite ends of the pulley as shown in the figure. The tension between 2m and pulley is T1 and between m and pulley is T2. Is the tension T1>T2?

The pulley and string are assumed to be massless and frictionless.
 

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Doc Al

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The tension between 2m and pulley is T1 and between m and pulley is T2.
Are you sure that T2 refers to the tension between m and the pulley and not between m and m?

(Is this your diagram, or the one given with the problem?)
 
Yes, I am sure.
 
I can't figure out where the answer is. Please explain it once again. Thanks
 

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