Question

You have a steel cable 100 m long, that you want to use for a zipline. You set it up in the middle of summer when the temperature is 40◦C. You install it to be completely taut between its endpoints. If the temperature outside can drop to -10◦C during the winter, how much extra force do the joints holding the ends of the zipline have to be able to withstand due to the temperature changes (other than a normal load for the passenger and the force needed to hold up the line itself)? The cable has a roughly circular cross section and has a radius of 1 cm. Steel has a linear thermal expansion coeﬃcient of 12×10−6 and a Young’s modulus of 200×109 Pa.

Answer #1

The expansion in length of cable due to thermal expansion is given by ∆L = a×L×∆T

( a denotes alpha = co-effecient of thermal expansion )

L = cable length

∆T = temperature difference

Therefore (∆L)/(L) = a×∆T = 12×10^{-6}×(40-(-10))

Therefore Strain = 6×10^{-5}

Now we know that Young's modulus Y = stress/strain

Therefore stress = Y × strain =
(200×10^{9})×(6×10^{-5})

Therefore Stress = 1.2×10^{7} N/m^{2}

Cross sectional area of cable
A = π×r² = 3.14×(1×10^{-2})^{2} (r=1cm =
10^{-2}m)

Therefore A=3.14×10^{-4} m²

Stress = Force/(Cross sectional area)

So, Force = Stress × Area =
(1.2×10^{7})×(3.14×10^{-4})

So F = 3769.9 N

This much extra force the joints have to withstand.

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