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Part A

A person holds a 10 kg box against a 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?

System: Box as point particle subject to external influences from the earth (gravity), the wall (normal force) and the person (applied force).

Model: Point Particle Dynamics.

Approach: 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.

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 A

A person moves a 10 kg box up a 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?

System: Box as point particle subject to external influences from the earth (gravity), the wall (normal force and friction) and the person (applied force).

Model: Point Particle Dynamics.

Approach: We begin with a free body diagram for the box:

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

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