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Part A
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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 |
Solution
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Box as |
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We begin with a free body diagram:
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With the free body diagram as a guide, we write the equations of Newton's 2nd Law:
Latex |
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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 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/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 and friction). Model: [Point Particle Dynamics]. Approach: We begin with a free body diagram: FBD 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
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use
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the
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fact
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that
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the
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box
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is
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sliding
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over
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level
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ground
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to
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tell
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us
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that
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a
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y =
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0
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(the
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box
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is
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not
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moving
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at
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all
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in
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the
...
y-direction).
...
Thus:
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}\begin{large}\[ N = F\sin\theta + mg \]\end{large}{latex} |
Now,
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we
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can
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write
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the
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friction
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force
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in
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terms
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of
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F
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and
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known
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quantities:
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}\begin{large}\[ F_{f} = \mu_{k}N = \mu_{k}\left(F\sin\theta + mg\right)\]\end{large}{latex} |
Substituting
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into
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the
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x-component
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equation
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yields:
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}\begin{large}\[ F\cos\theta - \mu_{k}\left(F\sin\theta + mg\right) = ma_{x}\]\end{large}{latex} |
which
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is
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solved
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to
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obtain:
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}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta - \mu_{k}\sin\theta} = \mbox{150 N}\]\end{large}{latex} h2. Part B !pushblock2_2.png|width=40%! A person pulls a |
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Part B
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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We again begin with a free body diagram:
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The diagrammatic representation suggests the form of Newton's 2nd Law:
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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/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 and friction). Model: [Point Particle Dynamics]. Approach: We again begin with a free body diagram: FBD Which implies 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
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y is
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zero
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if
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the
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box
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is
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moving
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along
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the
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level
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floor
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gives
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us:
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}\begin{large}\[ N = mg - F\sin\theta\]\end{large}{latex |
so
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} so {latex}\begin{large}\[ F_{f} = \mu_{k}\left(mg - F\sin\theta\right)\]\end{large}{latex} |
which
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is
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substituted
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into
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the
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x-component
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equation
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and
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solved
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to
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give:
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}\begin{large}\[ F = \frac{ma_{x} +\mu_{k}mg}{\cos\theta + \mu_{k}\sin\theta} = \mbox{88 N}\]\end{large}{latex} |
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