physics help A 10.89-km steel cable with a diameter of 3 cm is wrapped around a

Question

physics help

A 10.89-km steel cable with a diameter of 3 cm is wrapped around a huge spool on a sea vessel. The end of the cable is attached to a ton of scientific instruments in a box (mass 1000 kg, displacement volume 0.3 m^3) that will be used to investigate the challenger deep The scientific instruments are slowly lowered into the ocean Some information for the steel cable: density: 8000 kg/m^3: Young's modulus: 100 GPa: breaking strength: 1500 GPa. Assume the densities remain constant, and neglect the compression of the instruments and cable due to the water pressure. However, include the weight of the cable itself in the longitudinal stress, and also consider the volume displaced by the cable (a) Will the instruments reach the ocean floor 10.9 km below the surface will the cable run out first or will the cable break first? (b) If the instruments reach the ocean floor, what length of cable remains on the spool? If the cable runs out first, how far above the ocean floor will the instruments be? If the cable breaks first, how far will the instruments fall to the ocean floor?

Explanation / Answer

given, length l = 10890 m
   diamtere of the wire, d = 3cm = 0.03 m = 0.015 m
   mass of scientific instrument box, M = 1000 kg
   displacement volume, V = 0.3 m^3
   density of steel able, rho' = 8000 kg/m^3
   Youngs modulus, Y = 100 GPa
   breaking strength, G = 1500 GPa
   a) at a depth of 10.9 km
   SO THE ELONGATION OF ANY WIRE UNDER ITS OWN WEIGHT IS GIVEN BY
   dl = rho'*g*l^2/2Y = 8000*9.81*10890^2/2*100*10^9 = 46.5355 m
   so net length of the wire after elongation
   l' = l + dl = 10890 + 46.5355 = 10936.5355

   so the wire can reach the depth of 10.9 km

   so at the top most point
   tension in wire = weight of the wire + weight of the instrument
   T = rho'*(g*l*pi*d^2)/4 + Mg - V*rho*g = [8000*10890*pi*(0.03^2)/4 + 1000 - 0.3*1000]*9.81 = 610981.5071 N
   so stress = T/pi*(d^2/4) = 0.864 Gpa

   hence the wire reaches the bottom of the ocean without breaking

b) length of cable remaining in the spool = l' - 10900 = 10936.5355 - 10900 = 36.5355 m

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