# Elasticity of a body on which variable force is applied

Let us consider a rod of some cross sectional area $A$ and length $L$. At one end a longitudinal foce of 5x acts on the rod. At a distance of $\frac{L}{2}$ from that point longitudinal force of 5x acts on the rod in the opposite direction. My doubts are:

Since the body is in equilibrium: the tension center of rod and the end where force is applied must be 5x? At the other end of the rod will any tension be present? (The end where no force is applied) does the entire rod elongate or only a part? If its a part, which part? Please provide some intuition.

Finally let us consider the same rod but this time- At one end 5x newtons act. At center of the rod 4x newtons act in the opposite direction. And at the other end x newtons act in the opposite direction. Thus the rod is in equilibrium:

Is the elongation in the rod uniform?

How is tension distributed along the rod? Will it exist such that every point on the rod is in equilibrium? Is it only the tension that will cause elongation? What will be or how should we find tension at a distance of $\frac{2L}{3}$ from the end where 5x newtons is applied?

Rod is of uniform density and Young's modulus $Y$ It follows hookes

• Are you saying that the force depends on the distance x from the end of the rod? In other words, F = F(x) ? – John M Nov 29 '14 at 5:40
• no, there are 2 different forces at 2 different distances – Sashurocks Nov 29 '14 at 6:28
• Right, I'm just confused as what you mean by "5x" – John M Nov 29 '14 at 19:19
• Um sorry, please just consider it as 5 newtons if it helps. – Sashurocks Nov 30 '14 at 9:31 