1.

A steel cable is made up of 120 wires each of 1 mm diameter.The rope's length is 4 m, the distance between the suspension points is 3.8 m (Fig.). A weight of 1 tonne mass is suspended in the middle of the rope. What will be the rope's elongation? What weight will cause the rupture of the rope? The rope is made of soft steel.

Answer»


Solution :The sag in the cable is `h = SQRT((l//2)^2 - (d//2)^3) = 0.625 m`. The force extending the cable is `F = mgl//4h = 1.57 xx 10^(4) N`, the crosssectional area of the cable is `120 pi D^2//4 = 30 pi D^2 = 9.4 xx 10^(-5) m^2`. Hence applying Hooke.s law we find the extension of each section of the cable:
`DELTA l - (FL)/(ES) = (mgl^2)/(4ESh)`
The force capable of rupturing the cable is found from the breaking stress : `F_m = sigma_m S`. The load capable of rupturing the cable is `M = 4 sigma_m Sh//gl`.


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