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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 friction) 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_

Unknown macro: {x}

= F_

Unknown macro: {A}

- N = ma_

]
[ \sum F_

Unknown macro: {y}

= F_

Unknown macro: {f}

- mg = ma_

]\end

Because the box is held at rest, it has no acceleration at all (ax = ay = 0). Setting ax = 0 in the x direction equation gives:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ N = F_

Unknown macro: {A}

= \mbox

Unknown macro: {300 N}

]\end

Part A

A person holds up a 10 kg box by applying a perfectly horizontal force of 300 N to press it against a rough wall. 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:

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.

Unknown macro: {latex}

\begin

Unknown macro: {large}

[\sum F_

Unknown macro: {x}

= F_

Unknown macro: {A}

- N = ma_

]
[ \sum F_

Unknown macro: {y}

= F_

Unknown macro: {f}

- mg = ma_

]\end

Because the box is held at rest, it has no acceleration at all (ax = ay = 0). Setting ax = 0 in the x direction equation gives:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ N = F_

Unknown macro: {A}

= \mbox

Unknown macro: {300 N}

]\end

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