Question

A rod of length L, mass M and cross section area A is lying on the smooth horizontal surface is as shown in figure. Young's modulus is Y. Force F is applied on one end of the rod. Find stress of point P and elongation in rod.

Answer 1

Stress = $\frac{F}{A}$, Elongation = $\frac{FL}{AY}$

Answer 2

The problem asks for the stress at point P and the elongation of the rod. Assuming the rod is treated as massless or the question is simplified to consider the stress at the point of force application and the elongation of a massless rod:

Stress: The stress ($\sigma$) in a material is defined as the force applied per unit cross-sectional area. When a force F is applied to one end of the rod, the stress at any point within the rod (assuming uniformity or considering the point of application) is given by: $\sigma = \frac{F}{A}$ where F is the applied force and A is the cross-sectional area of the rod.

Elongation: The elongation ($\Delta L$) of a rod under tensile stress is given by the formula derived from Hooke's Law: $\Delta L = \frac{FL}{AY}$ where F is the applied force, L is the original length of the rod, A is the cross-sectional area, and Y is the Young's modulus of the material.

Therefore, the stress at point P (assuming it refers to the stress due to the applied force F) is $\frac{F}{A}$, and the elongation in the rod is $\frac{FL}{AY}$.