Question

A cable of length 2.5m and diameter 2 mm is supporting a weight of 100kg calculate internal restoring force and stress developed internally.

Answer 1

$Given:$

$Young modulus Y for the wire = 12.5 \times {10}^{11} dyne {cm}^{ - 2}$

$12.5 \times {10}^{10} {Nm}^{ - 2}$

$Diameter = D = 2.5mm = 2.5 \times {10}^{ - 3}$

$Force = F = 100kg f$

$= 100 \times 9.8N$

= 980N

Used the formula:Y=FL/AAL

Percentage increase in length= AL/L*100

$(F/AY) \times 100$

$= (F/\pi {r}^{2}Y) \times 100$

$[980/3.142 \times (1.25 \times {10}^{3} {)}^{2} \times 12.5 \times {10}^{10}] \times 100 = 15.96 \times {10}^{ - 2} = 0.16 percent$

Therefore, the percentage increase in the length of a wire is 0.16%

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