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 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
System:
Box as point particle.
Interactions: External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).
Model: Point Particle Dynamics.
Approach:
Diagrammatic Representation
We begin with a free body diagram:
Mathematical Representation
With the free body diagram as a guide, we write the equations of Newton's 2nd Law:
\begin
[ \sum F_
= F\cos\theta - F_
= ma_
]
[ \sum F_
= N - F\sin\theta - mg = ma_
] \end
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:
\begin
[ N = F\sin\theta + mg ]\end
Now, we can write the friction force in terms of F and known quantities:
\begin
[ F_
= \mu_
N = \mu_
\left(F\sin\theta + mg\right)]\end
Substituting into the x-component equation yields:
\begin
[ F\cos\theta - \mu_
\left(F\sin\theta + mg\right) = ma_
]\end
which is solved to obtain:
\begin
[ F = \frac{ma_
+\mu_
mg}{\cos\theta - \mu_
\sin\theta} = \mbox
]\end
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
System: Box as point particle.
Interactions: External influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).
Model: Point Particle Dynamics.
Approach:
Diagrammatic Representation
We again begin with a free body diagram:
Mathematical Representation
The diagrammatic representation suggests the form of Newton's 2nd Law:
\begin
[ \sum F_
= F\cos\theta - F_
= ma_
]
[ \sum F_
= N + F\sin\theta - mg = ma_
] \end
Again using the fact that ay is zero if the box is moving along the level floor gives us:
\begin
[ N = mg - F\sin\theta]\end
so
\begin
[ F_
= \mu_
\left(mg - F\sin\theta\right)]\end
which is substituted into the x-component equation and solved to give:
\begin
[ F = \frac{ma_
+\mu_
mg}{\cos\theta + \mu_
\sin\theta} = \mbox
]\end