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  • a) A steel wire of mass μ per unit length with a circular cross-section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall.

a) A steel wire of mass μ per unit length with a circular cross-section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg/m3.

b) If the yield strength of steel is 2.5 × 108 N/m2, what is the maximum weight that can be hung at the lower end of the wire?

 

Answers (1)

An element dx is of mass dm taken from a wire of length L from a distance x. mass per unit length here is u.

dm = u.dx, r=0.1 cm = 0.001 m

A = \pi r^{2} = \pi (10)^{-6}

M = 25 kg, L =10m

 

a) The force that acts in the downward direction on dx = wt

T(x) = (x.u) g + Mg

Y = T\frac{\frac{x}{A}}{\frac{dr}{dx}}

dr is the increase in the length of wire

Y = \frac{T(x). dx}{A. dr}

dr = \frac{1}{AY} (xug + Mg) dx

integrating both sides we get,

r (change in length of wire) = 1/AY [ug x2/g + Mgx] L0

extension in wire of length L =

\frac{gL}{2 \pi r^{2}Y}[ul + 2M]

extension=

10\times 10 [0.25 + 2 \times 0.25] / (2\times 3.14 \times 10^{-6} \times 2 \times 10^{11})

extension = 400 \times 10^{-5} m mass m = volume x density =

(A.L) \times P = \pi r^{2} \times 10 \times 7860 = 0.25 kg

b) at x = L, the wire experiences maximum tension

T(x) = ugx + Mg

T(L) = ugL + Mg

T = (m + M) g {as uL = m}

Yield force = Yield strength x A

 = 250 \times 3.14

Max tension = yield force

(m + M) g = 250 \times 3.14

M = \frac{785}{10} - 0.25 = 78.35 Kg

Posted by

infoexpert24

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