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{
Wiki Markup
Composition Setup
 Deck of Cards
idpartdeck
 Card
labelPart A
Part A
Image Added
 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
 Toggle Cloak
idsysA
 System: 
 Cloak
idsysA
 Box as .sysA
intA Interactions: intA External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).
 Toggle Cloak
idmodA
 Model: 
 Cloak
idmodA
Point Particle Dynamics.
 Cloak
modA
modA
 Toggle Cloak
idappA
 Approach: 
 Cloak
idappA
 Toggle Cloak
iddiagA
  Diagrammatic Representation 
 Cloak
iddiagA
 We begin with a free body diagram:
Image Added
 Cloak
diagA
diagA
 Toggle Cloak
idmathA
  Mathematical Representation 
 Cloak
idmathA
With the free body diagram as a guide, we write the equations of Newton's 2nd Law:
 Latex
}{composition-setup}

{deck:id=partdeck}
{card:label=Part A}


h3. Part A

!Pushing a Box Some More^pushbox2_1.png|width=400!

{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{excerpt} characterized by a coefficient of kinetic friction of 0.45.  The box accelerates horizontally at a rate of 2.0 m/s{color:black}^2^{color}.  What is the magnitude of _F_?

h4. Solution

{toggle-cloak:id=sysA} *System:*  {cloak:id=sysA} Box as [point particle].{cloak:sysA}

{toggle-cloak:id=intA} *Interactions:* {cloak:id=intA} External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).{cloak}

{toggle-cloak:id=modA} *Model:* {cloak:id=modA}[Point Particle Dynamics].{cloak:modA}

{toggle-cloak:id=appA} *Approach:* 

{cloak:id=appA}

{toggle-cloak:id=diagA} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagA}

 We begin with a free body diagram:

!pushfrictionmore1.png!

{cloak:diagA}

{toggle-cloak:id=mathA} {color:red} *Mathematical Representation* {color}

{cloak:id=mathA}


With the free body diagram as a guide, we write the equations of [Newton's 2nd Law|Newton's Second 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}{latex}


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


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


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}{latex}


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}{latex}


which 
...
 is 
...
 solved 
...
 to 
...
 obtain:


{
 Latex
}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta - \mu_{k}\sin\theta} = \mbox{150 N}\]\end{large}{latex}

{cloak:mathA}
{cloak:appA}
{card:Part A}
{card:label=Part B}

h3. Part B

!pushblock2_2.png|width=400!

A person pulls a box of mass
 Cloak
mathA
mathA
 Cloak
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
 Toggle Cloak
idsysB
 System: 
 Cloak
idsysB
 Box as point particle.
 Cloak
sysB
sysB
 Toggle Cloak
idintB
 Interactions: 
 Cloak
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: 
 Cloak
idmodB
 Point Particle Dynamics.
 Cloak
modB
modB
 Toggle Cloak
idappB
 Approach: 
 Cloak
idappB
 Toggle Cloak
iddiagB
  Diagrammatic Representation 
 Cloak
iddiagB
 We again begin with a free body diagram:
Image Added
 Cloak
diagB
diagB
 Toggle Cloak
idmathB
  Mathematical Representation 
 Cloak
idmathB
The diagrammatic representation suggests the form of Newton's 2nd Law:
 Latex
 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/s{color:black}^2^{color}.  What is the magnitude of _F_?

h4. Solution

{toggle-cloak:id=sysB} *System:*  {cloak:id=sysB} Box as [point particle].{cloak:sysB}

{toggle-cloak:id=intB} *Interactions:* {cloak:id=intB}External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).{cloak:intB}

{toggle-cloak:id=modB} *Model:* {cloak:id=modB} [Point Particle Dynamics].{cloak:modB}

{toggle-cloak:id=appB} *Approach:* 

{cloak:id=appB}

{toggle-cloak:id=diagB} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagB}

 We again begin with a free body diagram:

!pushfrictionmore2.png!

{cloak:diagB}

{toggle-cloak:id=mathB} {color:red} *Mathematical Representation* {color}

{cloak:id=mathB}

The diagrammatic representation suggests the form of [Newton's 2nd Law|Newton's Second 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}{latex}


Again 
...
 using 
...
 the 
...
 fact 
...
 that 
...
a
...
y 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
{latex}

so

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


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


{
 Latex
}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta + \mu_{k}\sin\theta} = \mbox{88 N}\]\end{large}{latex}

{cloak:mathB}
{cloak:appB}
{card:Part B}
{deck:partdeck}

 Cloak
mathB
mathB
 Cloak
appB
appB
 Card
Part B
Part B
 Deck of Cards
partdeck
partdeck