Two weights are connected by a very light cord that passes over an 80.0 N frictionless pulley of radius 0.300 m. The pulley is a solid unifo

Question

Two weights are connected by a very light cord that passes over an 80.0 N frictionless pulley of radius 0.300 m. The pulley is a solid uniform disk and is supported by a hook connected to the ceiling. What force does the ceiling exert on the hook

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Thành Công 6 months 2021-07-21T02:06:50+00:00 1 Answers 2 views 0

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    2021-07-21T02:08:31+00:00

    Answer:

    The force the ceiling  exert on the hook is 321.67 N

    Explanation:

    Here we have the weights being 75 N and 125 N

    The weight of the pulley = 80 N

    Total weight hung from the ceiling = 75 + 125 + 80 = 280 N

    Therefore since the forces acting on the ceiling tends to act downwards as they are the forces due to weights, we have

    The force on the

    T₁ – 125 = m₂a =

    m₂ = 125/9.8 = 12.76 kg

    Similarly T₂ – 75 = m₁a, m₁ = 7.65 kg

    acceleration is given by

    a_y =\frac{W_1 -W_2}{m_1 +m_2 +0.5M} =\frac{125 -75}{12.76 +7.65 +4.077} = 2.042 m/s²

    Therefore

    T₁ – 125 = m₂a = 12.76 × 2.042 = 26.04 N and

    T₁ = 125 + 26.04 = 151.04 N

    Similarly T₂ – 75 = m₁a =  7.65 × 2.042  = 15.63 N

    T₂ = 15.63  + 75 = 90.63 N

    Force on hook = Force exerted by ceiling on hook = T₁ + T₂ + Weight of pulley = 151.04  + 90.63 + 80  = 321.67 N

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