what will be tension on string pulled by equal but opposite forces???
Newton's thirld law. If you pull a string fixed at the other end it will have the same tension as the if the string is pulled by equal but opposite forces.
but in both cases there will be two but opposite forces on string so net force will be zero on string why the tension is not zero ???
When you draw a free body diagram be careful not draw both forces in a action-reaction pair. In a action-reaction pair the forces dont act on the same body. If you and I both pull the string with equal but opposite forces, I am the body which you apply force to and you are the body I apply force to. The sketch I made wasn't a FBD as it contained both forces.
The two forces acting at the two ends of the string are not an action-reaction pair. They correspond to two different interactions.
At one end (let's say where you pull by hand) there is the interaction hand-string. You pull with F1, the string pulls with -F1.
At the other end you have interaction between the string and wall and again you have F2 on the string and -F2 on the wall.
If the center of mass of the string does not accelerate, you will have F1=F2 (in magnitude) and the net force is zero. This does not prevent tensions developing in the string.
The fact that the net force is zero just tell you that the acceleration is zero. But they don't have to be equal.
Now, usually the string is assumed massless in elementary problems so even when we have acceleration the two forces are taken as equal. And the tension along the string is assumed to be constant and equal to either one of these forces.
So, to go back to the point, you can and many times do, create tension in a body by applying equal forces at different points of the body.
Do you really expect an elastic band not to stretch if you pull it from both ends? Its centre of mass won't move because there is no nett force, but it's certainly going to stretch.
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