Question

A uniform steel bar of cross-sectional area $A$ and length $L$ is suspended so that it hangs vertically. The stress at the middle point of the bar is ( $\rho $ is the density of steel ) ?
(A) $\dfrac{{L\rho g}}{{2A}}$
(B) $\dfrac{{L\rho g}}{2}$
(C) $\dfrac{{LA}}{{\rho g}}$
(D) $L\rho g$

Answer 1

Hint: To solve this type of problem we should know about the basic physical property of material. Be should know about the stress strain and young’s modulus of rigidity.
Stress: the ratio of internal opposing force to the cross section area at which force is exerted. Internal opposing force is generated due to applied force .So, internal opposing force is equal in magnitude to external deforming force but opposite in direction.
 $\sigma = \dfrac{F}{A}$
Strain: Ratio of change in shape to original length
 $\xi = \dfrac{{\Delta L}}{L}$
Formula used:
Mass per unit length $ = \rho A$ .
 $\sigma = \dfrac{F}{A}$

Complete step by step solution:
As given the density of string $ = \rho $ .
Let, $A$ will be a cross-section area.
Mass per unit length $ = \rho A$ .
Mass of string below the middle point $ = \rho \times A \times \dfrac{L}{2}{\sin ^{ - 1}}\theta $ .
So, self-weight the string as force so, $F = \dfrac{{\rho ALg}}{2}$ .
As we know stress, $\sigma = \dfrac{F}{A}$ .
Keeping value in the equation we get,
 \[ \Rightarrow \sigma = \dfrac{{\rho ALg}}{{2A}}\]
 $ \Rightarrow \sigma = \dfrac{{\rho Lg}}{2}$
So, the stress at the middle part will be equal to $\sigma = \dfrac{{\rho Lg}}{2}$ .

Hence, (B) is the correct option.

Additional information:
1. Strain is a dimensionless quantity.
2. Young modulus of rigidity is constant for specific material and it doesn't vary.

Note:
1. If there is no any force is applied and string is of negligible thickness then we don’t take self-weight of string as force . But string has mass and force is given then we will add then take it as combined force.
2. If string is of non-uniform shape then use integration to find mass of string.