A heavy uniform cable of length 100 m is hanging vertically from a fixed support. The graph of the internal restoring force (tension) F developed in the cable versus the distance x from its lower free end is a straight line passing through the origin (0 m , 0 N ) and reaching a maximum of (100 m , 4000 N ) at the top support. If the cross-sectional area of the cable is 5 10⁻⁴ m ^2 and its Young's modulus is 2 10¹¹ N

Step-by-step solution

The elongation d( l) of a small element of length dx at a distance x from the lower end is given by Hooke's law: d( l) = F(x) dx AY The total elongation of the cable is the integral of this expression over its entire length: l = ₀^ L F(x) dx AY = 1 AY ₀^ L F(x) dx The integral ₀^ L F(x) dx represents the area under the given F-x graph. Since the graph is a straight line from (0, 0) to (100, 4000) , the area is that of a triangle: Area = 1 2 base height = 1 2 100 4000 = 2 10^5 J Now, substitute the values of A and Y