A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The book has a mass

Question

A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The book has a mass of m = 1.7 kg and the coefficient of static friction between her hand and the book is μs = 0.52.What is the minimum force she must apply with each of her hands Fmin in Newtons, to keep the book from falling?

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Farah 4 years 2021-08-15T23:07:40+00:00 2 Answers 1063 views 0

Answers ( )

  1. lethu
    -3
    2021-08-15T23:08:54+00:00
    Reply

    Answer:

    8.66 N

    Explanation:

    Parameters given:

    Mass, m = 1.7 kg

    Coefficient of static friction, μs = 0.52

    The minimum force required to keep the book from falling is given as:

    Fmin = μs * N

    Where N = normal force

    N = m * g

    => Fmin = μs * m * g

    Fmin = 0.52 * 1.7 * 9.8

    Fmin = 8.66 N

  2. kimchi2
    2
    2021-08-15T23:09:05+00:00
    Reply

    Answer:

    4.3316 N

    Explanation:

    Using,

    F’ = μsR………………. Equation 1

    Where F’ = Friction force, μs = Coefficient of static friction, R = normal reaction.

    But,

    R = mg……………….. Equation 2

    Where m = mass of the book,  g = acceleration due to gravity.

    substitute equation 2 into equation 1

    F’ = μsmg……………… Equation 3

    Given: m = 1.7 kg, μs = 0.52, g = 9.8 m/s²

    Substitute into equation 3

    F’ = 1.7(9.8)(0.52)

    F’ = 8.6632 N

    Since she is using both hands, we assume that the force from each hands are equal,

    The force applied on each hand = 8.6632/2 = 4.3316 N.

    Hence the minimum force applied on each hand to keep the book from falling = 4.3316 N

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