Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin wedge. One

Question

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin wedge. One edge of the wedge is vertical, and the tip makes an angle of 33. The block that hangs vertically weighs 2.85 kg, and the block on the incline weighs 2.94 kg. If the two blocks do not move, what is magnitude of the force of friction on the second second block

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Minh Khuê 1 year 2021-09-03T02:56:51+00:00 1 Answers 4 views 0

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    2021-09-03T02:58:10+00:00

    Answer:

    13.6 N

    Explanation:

    Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 – 33 = 57.

    Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

    We assume the motion is downwards in the direction of the hanging block, M.

    We now write equations of motion for each block.

    So

    Mg – T = Ma    (1) and T – mgcos57 – F = ma where F is the frictional force on the block on the incline and a is their acceleration.

    Now, since both blocks do not move, a = 0.

    So, Mg – T = M(0) = 0     and T – mgcos57 – F = m(0) = 0

    Mg – T = 0    (3) and T – mgcos57 – F = 0 (4)

    From (3), T = Mg

    Substituting T into (4), we have

    T – mgcos57 – F = 0

    Mg – mgcos57 – F = 0

    So, Mg – mgcos57 = F  

    F = Mg – mgcos57

    F = (M – mcos57)g

    Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

    We find F, thus

    F = (2.94 kg – 2.85 kgcos57)9.8 m/s²

    F = (2.94 kg – 2.85 kg × 0.5446)9.8 m/s²

    F = (2.94 kg – 1.552 kg)9.8 m/s²

    F = (1.388 kg)9.8 m/s²

    F = 13.6024 kgm/s²

    F ≅ 13.6 N

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