Question

When a weight W is hung from one end of a wire of length L (other end being fixed) , the length of the wire increases by 1. If the same wire is passed over a pulley and two weights W each are hung at the two ends, what will be the total elongation in the wire?

Answer 1

In case (a), let Y be Young's modulus of the material of the

wire. If a is its area of cross section, then we can write

$Y=\frac{F/a}{l/L}\frac{F/a}{l/L}=\frac{Wl}{al}(\because F=W)$

$Increaseinlengthofwirel=\frac{WL}{aY}$

In case (b), we can treat either segment of length L/2

in Fig. 5] 7(b) as if one of its ends is fixed while the other end is

attached to the load W because the point M remains stationary, i.e.,

fixed all the time due to symmetrical load. Thus, when the wire is

passed over the pulley, let l be the increase in the length of each

segment. Since each segment is of length L/2, we have

$Y=\frac{W\left(L/2\right)}{a{l}^{\prime }}$

Increase in length of wire of one side of pulley

$l^{\prime }=\frac{1}{2}\frac{WL}{aY}=\frac{l}{2}$

Therefore, increase in length of both the segments of the wire

=$l^{\prime }+l^{\prime }=l/2+l/2=l$

So the increase in length remains the same.