A vertical hanging bar of length l and mass m per unit length carries a load of mass M at lower end, its upper end is clamped to a rigid support. The tensile stress a distance x from support is ( A arrow area of cross-section of bar)

Question

A vertical hanging bar of length l and mass m per unit length carries a load of mass M at lower end, its upper end is clamped to a rigid support. The tensile stress a distance x from support is ( A arrow area of cross-section of bar) (A) (M g+m g(l-x)/A) (B) (M g/A) (C) (M g+m g l/A) (D) ((M+m) g x/A l)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/64ac71166fdd6d8d696aa095a27c7dcf-.png" /> Tensile stress =( text Tension at point / text Area ) Tension at distance x from top would be the amount of force acting due to all the weight below it = Mass per unit length of rod × length of rod +M g =m ×(l-x) g+M g So Tensile stress =(m(l-x) g+M g/A)

Q. A vertical hanging bar of length $l$ and mass $m$ per unit length carries a load of mass $M$ at lower end, its upper end is clamped to a rigid support. The tensile stress a distance $x$ from support is ( $A \to$ area of cross-section of bar)

$\frac{M g + m g ( l - x )}{A}$

$\frac{M g}{A}$

$\frac{M g + m g l}{A}$

$\frac{( M + m ) gx}{A l}$

Tensile stress $= \frac{Tension at point}{Area}$
Tension at distance $x$ from top would be the amount of force acting due to all the weight below it
$=$ Mass per unit length of rod $\times$ length of rod $+ M g$
$= m \times (l - x) g + M g$
So Tensile stress $= \frac{m ( l - x ) g + M g}{A}$

Q. A vertical hanging bar of length l and mass m per unit length carries a load of mass M at lower end, its upper end is clamped to a rigid support. The tensile stress a distance x from support is ( A→ area of cross-section of bar)

AMg+mg(l−x)​

AMg​

AMg+mgl​

Al(M+m)gx​