One end of a wire of length L and weight W is | vedclass

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Question

(C) The total force acting at a cross-section at a distance $x$ from the lower end is the sum of the weight $W_1$ and the weight of the wire segment b

Vedclass · Accepted Answer

$\frac{W_1 + (3W/4)}{S}$

Vedclass · Answer

(C) The total force acting at a cross-section at a distance $x$ from the lower end is the sum of the weight $W_1$ and the weight of the wire segment below that point.Since the wire has a uniform weight $W$ and length $L$,the weight per unit length is $W/L$.The weight of the wire segment of length $3L/4$ is $(W/L) 	imes (3L/4) = 3W/4$.Therefore,the total force $F$ at this cross-section is $F = W_1 + 3W/4$.Stress is defined as force per unit area,so $	ext{Stress} = F/S = \frac{W_1 + 3W/4}{S}$.

One end of a wire of length $L$ and weight $W$ is attached to a ceiling,and the other end is attached to a weight $W_1$. If the cross-sectional area of the wire is $S$,what is the stress at a distance $3L/4$ from the lower end of the wire?

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