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Part A

A person holds a 10 kg box against a smooth 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_

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= - 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_

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= F_

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\cos\theta - N = ma_

]
[ \sum F_

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= F_

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\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_

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\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_

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\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

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]\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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[Examples from Dynamics]

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The root page Examples from Dynamics could not be found in space Modeling Applied to Problem Solving.
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