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Composition Setup
Deck of Cards
idpartdeck

Card
labelPart A

Part A

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Excerpt

A person pushes a box of mass 15 kg along a floor by applying a force F at an angle of 30° below the horizontal. There is friction between the box and the floor

characterized by a coefficient of kinetic friction of 0.45. The box accelerates horizontally at a rate of 2.0 m/s2. What is the magnitude of F?

Solution

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idsysA
System:
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idsysA

Box as .sysA

intA Interactions: intA External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).

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idmodA
Model:
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idmodA

Point Particle Dynamics.
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modA
modA

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idappA
Approach:

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idappA

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iddiagA
Diagrammatic Representation

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iddiagA

We begin with a free body diagram:

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diagA
diagA

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idmathA
Mathematical Representation

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idmathA

With the free body diagram as a guide, we write the equations of Newton's 2nd Law:

Latex
\begin{large}\[ \sum F_{x} = F\cos\theta - F_{f} = ma_{x}\] 
\[ \sum F_{y} = N - F\sin\theta - mg = ma_{y}\] \end{large}

We can now use the fact that the box is sliding over level ground to tell us that ay = 0 (the box is not moving at all in the y-direction). Thus:

Latex
\begin{large}\[ N = F\sin\theta + mg \]\end{large}

Now, we can write the friction force in terms of F and known quantities:

Latex
\begin{large}\[ F_{f} = \mu_{k}N = \mu_{k}\left(F\sin\theta + mg\right)\]\end{large}

Substituting into the x-component equation yields:

Latex
\begin{large}\[ F\cos\theta - \mu_{k}\left(F\sin\theta + mg\right) = ma_{x}\]\end{large}

which is solved to obtain:

Latex
Wiki Markup
h2. Part A

!Pushing a Box Some More^pushbox2_1.png|width=40%!

A person pushes a box of mass 15 kg along a smooth floor by applying a force _F_ at an angle of 30° below the horizontal..  The box accelerates horizontally at a rate of 2.0 m/s{color:black}^2^{color}.  What is the magnitude of _F_?

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

Model: [Point Particle Dynamics].

Approach:  Before writing [Newton's 2nd Law|Newton's Second Law] for the _x_ direction, we break the applied force _F_ into x- and y-components:

This implies:

{latex}\begin{large}\[ \sum F_{x} = F\cos\theta = ma_{x}\] \end{large}{latex}

Solving for _F_:

{latex}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta - \mu_{k}\sin\theta} = \mbox{34.6150 N}\]\end{large}{latex}

h2. Part B

!pushblock2_2.png|width=40%!

A person pulls a box of mass 15 kg along a smooth floor by applying a force _F_ at an angle of 30° above the horizontal..  The box accelerates horizontally at a rate of 2.0 m/s{color:black}^2^{color}.  What is the magnitude of _F_?

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

Model: [Point Particle Dynamics].

Approach:  Before writing [Newton's 2nd Law|Newton's Second Law] for the _x_ direction, we break the applied force _F_ into x- and y-components:

This implies:

{latex}\begin{large}
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mathA
mathA

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appA
appA

Card
Part A
Part A

Card
labelPart B

Part B

Image Added

A person pulls a box of mass 15 kg along a floor by applying a force F at an angle of 30° above the horizontal. There is friction between the box and the floor characterized by a coefficient of kinetic friction of 0.45. The box accelerates horizontally at a rate of 2.0 m/s2. What is the magnitude of F?

Solution

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idsysB
System:
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idsysB

Box as point particle.
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sysB
sysB

Toggle Cloak
idintB
Interactions:
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idintB

External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).
Cloak
intB
intB

Toggle Cloak
idmodB
Model:
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idmodB

Point Particle Dynamics.
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modB
modB

Toggle Cloak
idappB
Approach:

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idappB

Toggle Cloak
iddiagB
Diagrammatic Representation

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iddiagB

We again begin with a free body diagram:

Image Added

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diagB
diagB

Toggle Cloak
idmathB
Mathematical Representation

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idmathB

The diagrammatic representation suggests the form of Newton's 2nd Law:

Latex
\begin{large}\[ \sum F_{x} = F\cos\theta - F_{f} = ma_{x}\] 
\[ \sum F_{xy} = N + F\cossin\theta - mg = ma_{xy}\] \end{large}

Again using the fact that ay is zero if the box is moving along the level floor gives us:

Latex
\begin{large}\[ N = mg - F\sin\theta\]\end{large}

so

Latex
\begin{large}\[ F_{f} = \mu_{k}\left(mg - F\sin\theta\right)\]\end{large}

which is substituted into the x-component equation and solved to give:

Latex
}

Solving for _F_:

{latex}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta + \mu_{k}\sin\theta} = \mbox{34.688 N}\]\end{large}{latex}

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Part B

Deck of Cards
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partdeck