# Why same force acts on two different strings?

Let us take two different strings and arrange them vertically. The first string is attached to the ceiling and connected in series with the second string and a mass $$m$$ is connected at the bottom of the second string. Why does the same force $$mg$$ act in both strings?If we consider the free body diagram on the mass,since there are two ropes,so there will be two different tensions $$T_1, T_2$$. So shouldn't $$T_1+T_2=mg?$$

• The pictures show the force the string exerts on your hand, and not the force that your hand exerts on the string. Jan 7 at 14:49
• Thanks,what about my first question? Jan 7 at 14:54
• I don't understand your first question. If the strings are tied together, one after the other, the tension must be same in both surely? I must be misunderstand the setup/ Jan 7 at 15:01
• I have added picture for clarity,i don't understand why both $T_1$ and $T_2$ will be $mg$. Jan 7 at 15:07
• If $T_1\ne T_2$ there will be a net force of $T_1-T_2$ on the knot joining the strings. Is the knot accelerating? Jan 7 at 15:15

A force (actually a pair of forces) is a measure of interaction between two objects. There is no interaction between the top rope and the block so there are no forces between these two. Of course there is the gravitational interaction between them but here you don't consider this, just contact forces, right? At least not for ropes. Free body diagrams are useful to better understand the situations like this. If you take the bottom part of the rope, it interacts with two other objects, the top rope and the block. So there will be two forces acting on it: one due to ineraction with the top rope (let's call it $$T_{up}$$) acting upwards and one due to the interaction with the block ($$T_{down}$$), downwards. None of these is a weight (not due to gravity). They are both contact forces. There is also the weight of the rope but here you neglect it. The difference $$T_{up}-T_{down}$$ should be equal to the weight of the rope but if you neglect this one then the two have the same magnitude and is what you call $$T_2$$ For the block you have the interaction with the bottom rope and the gravitational interaction with the earth, which is measured by its weight. So you have two forces, the weight (mg acting downwards) and the $$T_{down}$$ acting upwards (the action-reaction pair of $$T_{down}$$. For equilibrium we need the two forces to balance so you have for magnitudes $$T_{up} = T_2 = mg$$. For completeness, for the top part of the rope you have two forces again. One is due to the interaction with the bottom rope, $$-T_{up}$$, which is the action-reaction pair of $$T_{up}$$ and some upward force due to interaction with the ceiling. Again, if we neglect the weight of the rope, the two forces should have same magnitude and this will be what you labelled $$T_1$$. But we also have $$T_2=T_{up}$$ so we find that $$T_1=T_2$$ which shows that you have the same tension all along the ropes. This is true only because you neglect the weight of the ropes.