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6 June, 16:18

8. A 300g mass hangs at the end of a string. A second-string hang from the bottom of that mass and supports a 900g mass. (a) Find the tension in each string when the masses are accelerating upward at 0.700m/s2. (b) Find the tension in each string when the masses are accelerating downward at 0.700m/s2.

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  1. 6 June, 19:30
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    We are looking for:

    Tension A=?

    Tension B=?

    Then convert grams to kilograms

    1kg = 1000g

    300g = 0.3 kg

    900g = 0.9kg

    Given:

    mA = 0.3kg

    mB = 0.9kg

    how? W=mg

    g = 9.8 m/s^2

    ay = 0.700m/s^2

    WA = 2.94N

    WB = 8.82N

    For A [Tension A-Weight A - Tension B=mAay]

    TA - 2.94 - TB=0.3kg (0.700m/s^2)

    TA - 2.94 - TB = 0.21 N

    TA - TB = 0.21N + 2.94 N

    TA - TB = 3.15 N

    * refer TB from below*

    TA - TB = 3.15N

    TA - 9.45N = 3.15N

    TA = 3.15N + 9.45N

    TA = 12.6N

    For B F = ma

    Summation of forces along y-axis = mBay

    -8.82 N + TB = 0.9 kg (0.700 m/s^2)

    -8.82N + TB = 0.63 N

    TB = 0.63 N + 8.82 N

    TB = 9.45 N
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