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Exercises · 8.9

Q.A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m−210^{8}\ \text{N m}^{-2}, what is the maximum load the cable can support?

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The maximum load is found using stress = force/area. For a steel cable of radius 1.5 cm and maximum allowable stress 108 N/m210^8\ \text{N/m}^2, the maximum load is about 7.07×104 N7.07 \times 10^4\ \text{N}.

The idea here is straightforward: stress is defined as force per unit area. When we say "maximum stress" the cable can handle, we mean the largest force per cross-sectional area it can bear without failing. So the maximum load — the force it can support — is simply that maximum stress multiplied by the cable's cross-sectional area.

The cable is circular in cross-section, so its area is πr2\pi r^2. The radius is given in centimetres, but stress is in SI units (N/m²), so we must convert to metres first. A common slip is to forget this conversion and get an answer off by a factor of 10410^4.

Let's work it through.

  1. Convert radius to metres

    r=1.5 cm=1.5×10−2 mr = 1.5\ \text{cm} = 1.5 \times 10^{-2}\ \text{m}

  2. Find cross-sectional area

    A=πr2=π(1.5×10−2)2A = \pi r^2 = \pi (1.5 \times 10^{-2})^2

    =π×2.25×10−4= \pi \times 2.25 \times 10^{-4}

    =7.0686×10−4 m2= 7.0686 \times 10^{-4}\ \text{m}^2 (approximately)

  3. Apply the stress formula

    Stress σ=FA\sigma = \frac{F}{A}, so F=σ×AF = \sigma \times A

    F=(108)×(7.0686×10−4)F = (10^8) \times (7.0686 \times 10^{-4})

    =7.0686×104 N= 7.0686 \times 10^4\ \text{N} …

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