The Young’s modulus of steel is 2.0 x 1011 N/m2 and the tensile strength is 8.0

Question

The Young’s modulus of steel is 2.0 x 1011 N/m2 and the tensile strength is 8.0 x 108 N/m2. The density of steel is 7.9 x 103 kg/m3. (a) What is the maximum length of a steel cable that can hang vertically supported from one end of the cable? Assume the cable is supporting no other weight than its own and that the acceleration of gravity is the same all along the cable. [Note that the answer does NOT depend on the cross sectional area.] (b) What would the total elongation of the steel cable in part (a) be just before it breaks? You may assume for simplicity that the cable is elastic, i.e., it obeys Hooke’s law, until the moment it breaks. [Hint: consider the average weight supported by a section of the cable.]

Explanation / Answer

(a)

mass m = V = AL where A is the area and L the length.
Then 8 x 10^8 N/m² = mg / A = ALg / A = gL = 7900kg/m³ * 9.8m/s² * L
L = 10000 m
after rounding from 10333 m.

(b)

At the top, the elongation per meter is
L = FL / EA = ALg * 1m / EA area A cancels
L = 7900kg/m³ * 10333m * 9.8m/s² * 1m / 2e11N/m² = 0.004 m

At the middle, then, the elongation per meter is 0.002 m
since it is supporting half of the weight.

Then the elongation over the entire length must be
L = 0.002m/m * 10333m = 21 m

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