An elevator cab that weighs 27.8 kN moves upward. What is the tension in the cable if the cab’s speed is (a) increasing at a rate of 2 1.22

Question

An elevator cab that weighs 27.8 kN moves upward. What is the tension in the cable if the cab’s speed is (a) increasing at a rate of 2 1.22 / msand (b) decreasing at a rate of 2 1.22 / ms?

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Đan Thu 1 year 2021-09-02T14:28:51+00:00 1 Answers 36 views 0

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  1. nguyetanh
    0
    2021-09-02T14:30:50+00:00
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    Answer:

    a)T = 8.63  × 10 ⁴ N, b)T = -3.239 × 10 ⁴ N

    Explanation:

    Given:

    W = 27.8 KN = 27.8 × 10 ³ N,

    For upward motion: Fnet is upward, Tension T is upward and weight W is downward so

    a) a=21.22 m/s² ( not written clearly the unit. if it is acceleration?) then

    Fnet = T – W

    ⇒ T = F + W = ma + W

    T = (W/g)a + W                           (W=mg ⇒m=W/g)

    T = (27.8 × 10 ³ N / 9.8 ) 21.22 m/s² + 27.8 × 10 ³ N

    T = 86,263.265 N

    T = 8.63  × 10 ⁴ N

    b) For Declaration in upward direction a = -21.22 m/s²

    Fnet = T – W

    ⇒ T = F + W = ma + W

    T = (W/g)a + W                        

    T = (27.8 × 10 ³ N / 9.8 ) (-21.22 m/s²) + 27.8 × 10 ³ N

    T = -32395.5 N

    T = -3.239 × 10 ⁴ N

    as Tension can not be negative I hope the value of acceleration and deceleration is correct.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )