Question:
A uniform rod of length l, young's modulus Y and cross section A, placed on a smooth horizontal table, is pulled by a force F applied parallel to the rod at one end. The other end of the rod is free. Find the strain energy stored in the rod.
My attempt:
If the table was rough(i.e.,had friction), the static friction force would have adjusted itself to match the external force F and hence two equal and opposite forces would have caused stress to occur in the rod. But here the table is smooth, so I don't know what other force will oppose the external force and how stress will be produced.
I'm not sure of this but if I assume that atomic force of attraction would oppose the external force and the rod is reduced to the center of mass(at a distance of $\frac{l}{2}$), the following strain energy would be stored in the rod:-
$$U=\frac{1}{2Y}(stress)^2A\frac{l}{2}$$
$$U=\frac{1}{2Y}(\frac{F}{A})^2A\frac{l}{2}$$
$$U=\frac{F^2l}{4YA}$$
The answser is given as $U=\frac{F^2l}{6YA}$
Best Answer
There is no other force. The rod is being accelerated. The tension $T=F(x/L)$ in the rod will vary linearly from $F$ at the pulled end where $x=L$ to $0$ at the free end where $x=0$.
This situation is like a train of trucks being pulled by an engine. The forces on each truck are different, although if the mass of each is the same then the net force on each is also the same. However, the elastic energy stored in each truck (or in the coupling springs between trucks) depends on the balanced force at the two ends (ie the tension force in the spring) rather than the net force accelerating the truck/spring.