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    2 boxes A ( 110kg) and B (190kg) sit on the floor of a lift ( 1700kg). Box A sits on Box B. The lift is supported by a light inextensible cable and is descending at a constant acceleration 1.8m/s^2
    The tension at the cable ?
    Force exerted by box B on box A ?
    I got the tension = to 1600N but the MS of the textbook says 3600 ! And i think im lost for the second question. Can someone help?
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    (Original post by Physics Enemy)
    m = 2000 kg, accelerating downward reduces tension by 1.8m. So T = mg - 1.8m = 8m = 16,000 N

    For the last bit, force by B on A equals force by A on B i.e) A's effective weight, so 110 x 8 = 880 N
    U mean 110x 1.8? *
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    (Original post by Tirawi19)
    ...
    (Original post by Physics Enemy)
    m = 2000 kg, accelerating downward reduces tension by 1.8m. So T = mg - 1.8m = 8m = 16,000 N

    For the last bit, force by B on A equals force by A on B i.e) A's effective weight, so 110 x 8 = 880 N
    I re-did the Q by drawing a diagram and doing it properly, to see if I made a mistake earlier.

    Forces on the lift: W - T = ma
    => 2000g - T = 2000 x 1.8
    => T = 2000 x 8
    => T = 16,000 N

    Forces on box a: W_a - R_a = m_a x 1.8
    => 110g - 110 x 1.8 = R_a
    => R_a = 110 x 8
    => R_a = 880 N

    So I stand by my answers from before
 
 
 
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