Physics - Vectors/Forces

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A 10 kg weight is on an inclined plane 10 ft long and 6 ft high. If there is no friction, what force parallel to the plane will prevent the body from sliding?

  • Physics - Vectors/Forces -

    Given: 10 kg, 10 ft and 6 ft.

    Solution:
    Find the force parallel to the plane

    Sin A = opposite/hypotenuse
    Sin A = 6ft/10ft
    = 0.6 ft.

    Find the A of the sin.

    A = sin^-1(6ft/10ft)
    A = 36.87°

    Find f1 and f2:
    Note: m (mass)
    g (gravity pulls) = 9.81 m/s^2

    F1= sin (teta) x mg
    = sin 36.87° x 10 kg(9.81m/s^2)
    F1 = 58.86 N prevent the body from sliding

    F2 = cos (teta) x mg
    F2 = cos 36.87° x 10 kg(9.81m/s^2)
    F2 = 78.48 N vertical mass

  • Physics - Vectors/Forces -

    (6/10) 10 g = 6 g = 6*9.81 = 58.86 Newtons

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