# Why is the tension on mass m1 not 2T here? I have named the strings as S1,S2 and S3 on diagram. T is the tension in the string.

If I draw the Free body diagram for m1, I think it should be 2T force on it but on internet , it is just T. So , why do I think it is 2T?

Just imagine that when m2 is pulled , the pulley connecting mass m1 also gets pulled with 2T tension. Then , We have a fixed wall. So , the fixed wall can’t and would never move. So , it applies a T force on the string which travel to mass m1 and pulls it towards m2 direction. Now , till here the force is applied by string S1.

Then , when the pulley gets pulled. The string S2 also applied another T on the mass m1 which pulls it too. So , in conclusion . Total tension on the mass m1 is 2T.

So , where is it that I am going wrong ?

EDIT: Example of double tension or 2T . So ,here. There is a 2T force applied on 6kg mass.

• What's $T$? The only force I see is $F$.
– Wasabi
Apr 28, 2021 at 17:00
• @Wasabi T is the tension in the string Apr 28, 2021 at 17:01
• Which string? If you think it should be the same on all strings, then what do you mean by $2T$?
– Wasabi
Apr 28, 2021 at 17:02
• @Wasabi Ohk. So , yes tension is same in all strings. Apr 28, 2021 at 17:03
• 2T means double tension. I will given an example. Apr 28, 2021 at 17:03 I just drew the forces in the direction that they will allow the rope/wire to be taut.

In your original sketch the directions of the wire forces from the wall and m1 and the wall were the opposite way. The problem there is that if the forces are drawn in your original way then the rope would collapse. Wires can transfer tension forces but not compressive. So ropes can only transfer tensile forces. In the above example the tension force that you have is at every point of the rope exactly the same. Unless you provide an axial force (so unless another person steps in there is no reason for the force to increase. And you will notice (in the image below) that the tension forces drop at the right hand side (the rough sketch underneath is the magnitude of the axial/tensile force) To come back to your original example, I've highlighted the rope from $$m_1$$ with a green thick line. You see that it (ultimately) attaches to the wall at the other end. On that continuous piece of rope the force should be always equal. Now the reason that the force on the right is 2T, is because the pulley is in equilibrium, so the algebraic sum of the forces should be zero. Since, you know the force on the left hand of the image, then you can easily derive that the force to $$m_2$$ is $$2T$$

you are right if you draw the FBD of the pulley, we have 2 ropes carrying the same T to the left and the other rope connecting to mass m2 with a tension that must be equal and opposite to the sum of tension pointing left which is 2T pointing right.

# Edit

After OP's comment about 2T on mass m1.

The tension imparted on the pulley is 2T. Then this tension is divided by two shared between the two ropes on the left equally.

Therefore the tension on m1 is just T. • I meant that on mass m1. There is a 2T Apr 28, 2021 at 18:53
• You can check the discussion chat of mine and also the Q description for what I meant to say. Apr 28, 2021 at 18:54