匿名
匿名 發問於 科學及數學物理學 · 5 年前

Physics Question?

An engineer designs a lift that takes passengers up a slope along a straight line. The floor of the lift is kept horizontal.

Suppose a man of mass 65 kg takes the lift up the slope. Find the normal force N and friction f (magnitude and direction )acting on the man by the floor of the lift when the lift accelerates at 0.5 m/s^(2)

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2 個解答

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  • 天同
    Lv 7
    5 年前
    最愛解答

    Resolve the acceleration into vertical and horizontal directions.

    Vertical acceleration = 0.5.sin(20) m/s^2 = 0.171 m/s^2

    Horizontal acceleration = 0.5.cos(20) m/s^2 = 0.4698 m/s^2

    Consider vertical motion, use: net-force = mass x acceleration

    N - 65g = 65 x 0.171 where N is the normal reaction on the man, and g is the acceleration due to gravity (= 9.81 m/s^2)

    i.e. N = (65 x 0.171 + 65g) N = 653 N

    Consider the horizontal motion, use again: net-force = mass x acceleration

    f = 65 x 0.4698 where f is the frictional force acting on the man

    i.e. f = 30.5 N

    The frictional force is acting horizontally in the forward direction, i.e. towards the right.

  • 匿名
    5 年前

    Take upwards and the direction to the right as positive

    consider the vertical direction.

    apply F=ma

    N-65 x 9.81 = 65 x 0.5 sin20

    N=649N

    The normal reaction is 649N upwards.

    Consider the horizontal direction.

    f=65 x 0.5 cos20

    f=30.5 N

    The friction is 30.5 N towards the right.

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