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11 October, 04:54

A 23 kg suitcase is being pulled with a constant speed by a handle that is at an angle of 29° above the horizontal. If the normal force exerted on the suitcase is 160 n, what is the force f applied to the handle?

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Answers (2)
  1. 11 October, 05:07
    0
    329.89N

    Explanation:

    Normal force is a force that is perpendicular to the horizontal surface. The normal force will therefore acts in the positive y direction.

    To get the force F applied at the handle, we will resolve the force to the component of the normal force (y component)

    The component of the force along the positive y axis (Fy) will be + Fsin (theta) where theta is 29°

    Since both the normal force and the resolved force are acting in the same direction, Fy = normal force

    Fsin (theta) = 160N

    Fsin29° = 160N

    F = 160/sin29°

    F = 160/0.485

    F = 329.89N

    Therefore the force F applied to the handle is 329.89N
  2. 11 October, 05:20
    0
    The force F applied to the handle = 330.03 N

    Explanation:

    The force can be resolved in to two, horizontal component and vertical component. If θ is the angle between horizontal and applied force we have

    Horizontal component of force = F cos θ

    Vertical component of force = F sin θ

    In this problem normal force exerted on the suitcase is 160 N, that is vertical component of force = 160 N and angle θ = 29⁰.

    So, F sin 29 = 160

    F = 330.03 N

    The force F applied to the handle = 330.03 N
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