An elevator cab and its load have a combined mass of 1200 kg. Find the tension in the supporting cable when the cab, originally moving downw

Question

An elevator cab and its load have a combined mass of 1200 kg. Find the tension in the supporting cable when the cab, originally moving downward at 10 m/s, is brought to rest with constant acceleration in a distance of 35 m.

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Delwyn 4 years 2021-08-21T20:44:17+00:00 1 Answers 23 views 0

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  1. Orla
    0
    2021-08-21T20:45:33+00:00
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    Answer:

    10044 N

    Explanation:

    The acceleration of the cab is calculated using the equation of motion:

    v^2 = u^2+2as

    v is the final velocity = 0 m/s in this question, since it is brought to rest

    u is the initial velocity = 10 m/s

    a is the acceleration

    s is the distance = 35 m

    a = \dfrac{v^2-u^2}{2s} = \dfrac{(0 \text{ m/s})^2-(10 \text{ m/s})^2}{2\times (35\text{ m})} = -1.43\text{ m/s}^2

    Since it accelerates downwards, its resultant acceleration is

    a_R = g + a

    g is the acceleration of gravity.

    a_R = (9.8-1.43)\text{ m/s}^2 = 8.37\text{ m/s}^2

    The tension in the cable is

    T = ma_R = (1200\text{ kg})(8.37\text{ m/s}^2) = 10044 \text{ N}

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