Question

?a vertical hanging bar of length l and mass m per unit length carries a load of mass M at the lower end , its upper end is clamped to a rigid support . the tensile stress a distance x from support is

Answer 1

Answer:

The tensile stress at a distance x from the support is

$Stress = \frac{Mg + m(l - x)g}{A}$

Explanation:

As we know that tensile stress is ratio of net tension force and area

So here we know that

$F = Mg + (m)(l - x)g$

here we know that mass per unit length of the wire is given as "m"

so the total length of the wire at distance (l - x) from the bottom of the mass M

So total suspended mass is given as

$m_{net} = M + m(l - x)$

so we will have tensile stress is given as

$Stress = \frac{T}{A}$

$Stress = \frac{Mg + m(l - x)g}{A}$

#Learn

Topic : Tensile stress

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