One end of a uniform wire of length L and of | vedclass

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(D) The wire is uniform,so its weight is distributed uniformly over its length $L$.The weight per unit length of the wire is $w = W/L$.At a height $h

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$\frac{\frac{3}{4}W + W_1}{A}$

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(D) The wire is uniform,so its weight is distributed uniformly over its length $L$.The weight per unit length of the wire is $w = W/L$.At a height $h = 3L/4$ from the lower end,the length of the wire below this point is $3L/4$.The weight of this segment of the wire is $W_{segment} = w 	imes (3L/4) = (W/L) 	imes (3L/4) = \frac{3}{4}W$.The total force acting at this cross-section is the sum of the weight of the segment below it and the suspended weight $W_1$.Total force $F = \frac{3}{4}W + W_1$.Stress is defined as force per unit area,so $	ext{Stress} = \frac{F}{A} = \frac{\frac{3}{4}W + W_1}{A}$.

One end of a uniform wire of length $L$ and of weight $W$ is attached rigidly to a point in the roof and a weight $W_1$ is suspended from its lower end. If $A$ is the area of cross-section of the wire,the stress in the wire at a height $3L/4$ from its lower end is

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