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    Part A

    A person holds a 10 kg box against a smooth (i.e. frictionless) wall (as it slides down) by applying a perfectly horizontal force of 300 N. What is the magnitude of the normal force exerted on the box by the wall?

    Solution

    System:

    Interactions:

    Model:

    Approach:

    Diagrammatic Representation

    We begin with a free body diagram for the box:

    It is important to note that any surface has the potential to exert a normal force and that the normal is always perpendicular to the plane of the surface. If the wall did not exert a normal force, the box would simply pass through it.

    Mathematical Representation

    From the free body diagram, we can write the equations of Newton's 2nd Law.

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    \begin

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    [\sum F_

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    = F_

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    - N = ma_

    ]
    [ \sum F_

    Unknown macro: {y}

    = - mg = ma_

    ]\end

    Because the box is held against the wall, it has no movement (and no acceleration) in the x direction (ax = 0). Setting ax = 0 in the x direction equation gives:

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    \begin

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    [ N = F_

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    = \mbox

    Unknown macro: {300 N}

    ]\end

    Part B

    A person moves a 10 kg box up a smooth wall by applying a force of 300 N. The force is applied at an angle of 60° above the horizontal. What is the magnitude of the normal force exerted on the box by the wall?

    Solution

    System:

    Interactions:

    Model:

    Approach:

    Diagrammatic Representation

    We begin with a free body diagram for the box:

    Mathematical Representation

    From the free body diagram, we can write the equations of Newton's 2nd Law.

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    \begin

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    [\sum F_

    Unknown macro: {x}

    = F_

    Unknown macro: {A}

    \cos\theta - N = ma_

    ]
    [ \sum F_

    Unknown macro: {y}

    = F_

    Unknown macro: {A}

    \sin\theta - mg = ma_

    ]\end

    Because Because the box is held against the wall, it has no movement (and no acceleration) in the x direction (ax = 0). Setting ax = 0 in the x direction equation gives:

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    \begin

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    [ N = F_

    Unknown macro: {A}

    \cos\theta = \mbox

    Unknown macro: {150 N}

    ]\end

    Part C

    A person scrapes a 10 kg box along a low, smooth ceiling by applying a force of 300 N at an angle of 30° above the horizontal. What is the magnitude of the normal force exerted on the box by the ceiling?

    Solution

    System:

    Interactions:

    Model:

    Approach:

    Diagrammatic Representation

    We begin with a free body diagram for the box:

    The ceiling must push down to prevent objects from moving up through it.

    Mathematical Representation

    From the free body diagram, we can write the equations of Newton's 2nd Law.

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    \begin

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    [\sum F_

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    = F_

    Unknown macro: {A}

    \cos\theta = ma_

    ]
    [ \sum F_

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    = F_

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    \sin\theta - mg - N = ma_

    ]\end

    Because Because the box is held against the ceiling, it has no movement (and no acceleration) in the y direction (ay = 0). Setting ay = 0 in the y direction equation gives:

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    \begin

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    [ F_

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    \sin\theta - mg - N = 0 ]\end

    which we solve to find:

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    \begin

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    [ N = F_

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    \sin\theta - mg = \mbox

    Unknown macro: {52 N}

    ]\end

    We can check that the y direction is in balance. We have N (52 N) and mg (98 N) on one side, and FA,y on the other (150 N).

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