Solved Problem on Dynamics
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A cart moves over a straight and horizontal surface. In the cart there is an inclined plane, it makes an
angle θ with the horizontal plane. On the plane is placed a body, the coefficient of friction
between the body and the plane is μ. Determine the acceleration of the cart, so that the body is
about to rise the plane. Assume g for the acceleration due to gravity.
Problem data:
- Angle of the inclined plane: θ;
- Coefficient of friction between the body and the plan: μ;
- Acceleration due to gravity: g.
We choose a frame of reference with the x-axis parallel to the horizontal plane and in the same
direction as the cart acceleration.
The ground (Earth), assumed without acceleration, is an inertial referential. The cart has acceleration a relative to the ground, non-inertial referential. For the body to remain at rest, on the cart, it must have the same acceleration a of the cart relative to the ground (Figure 1).
The ground (Earth), assumed without acceleration, is an inertial referential. The cart has acceleration a relative to the ground, non-inertial referential. For the body to remain at rest, on the cart, it must have the same acceleration a of the cart relative to the ground (Figure 1).
Solution
Drawing a free-body diagram, we have the forces that act on it (Figure 2).
- \( {\vec F}_{g} \): gravitational force;
- \( \vec{N} \): normal reaction force;
- \( {\vec F}_{f} \): force of friction between the plane and block.
As the body is about to rise, we have the force of friction \( {\vec{F}}_{f} \) between the plane and the body in the downward direction of the plane opposing this movement.
The force of friction \( {\vec{F}}_{f} \) and the normal reaction force \( \vec{N} \) can be decomposed into two components, a parallel component to the x-axis, \( {\vec F}_{fx} \) and \( {\vec{N}}_{x} \) and the other component normal or perpendicular, \( {\vec F}_{fy} \) and \( {\vec N}_{y} \).
The angle of the inclined plane is θ in Figure 3-A, we see that the angle between the x direction and the inclined plane is also θ, are alternate angles.
The normal reaction force is perpendicular to the inclined plane, making an angle of 90°, the angle between the x direction and normal direction is \( \alpha =90°-\theta \) (Figure 3-B).
As the x and y directions are perpendicular to each other, the angle between the normal
force and the y direction is (Figure 3-C)
\[ 90°-\alpha=90°-(90°-\theta )=90°-90°+\theta=\theta \]
The angle between the force of friction and the component of the friction force in the x
direction,
\( {\vec F}_{fx} \),
is θ same angle of the inclined plane, are alternate angles (Figure 4).
We draw the vectors in a coordinate system (Figure 5).
Applying Newton's Second Law
Applying Newton's Second Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec{F}=m\vec{a}} \tag{I}
\end{gather}
\]
In the y direction, there is no movement, gravitational force,
\( {\vec F}_{g} \)
the component of the normal reaction force,
\( {\vec N}_{y} \)
and the component of the friction force in the y direction
\( {\vec F}_{fy} \)
cancel
\[
\begin{gather}
N_{y}=F_{g}+F_{fy} \tag{II}
\end{gather}
\]
the component of the normal force in y direction given by
\[
\begin{gather}
N_{y}=N\cos \theta \tag{III}
\end{gather}
\]
the component of the force of friction in the y direction given by
\[
\begin{gather}
F_{fy}=F_{f}\sin \theta \tag{IV}
\end{gather}
\]
the force of friction is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{f}=\mu N} \tag{V}
\end{gather}
\]
substituting the expression (V) into expression (IV)
\[
\begin{gather}
F_{fy}=\mu N\sin \theta \tag{VI}
\end{gather}
\]
the gravitational force given by is
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{g}=mg} \tag{VII}
\end{gather}
\]
substituting expressions (III), (VI), and (VII) into expression (II)
\[
\begin{gather}
N\cos \theta =mg+\mu N\sin \theta \\[5pt]
N\cos \theta-\mu N\sin \theta =mg \tag{VIII}
\end{gather}
\]
In the x direction, we have the component of the normal force and the component of the friction force,
applying the expression (I)
\[
\begin{gather}
N_{x}+F_{fx}=ma \tag{IX}
\end{gather}
\]
the component of the normal force in the x direction is given by
\[
\begin{gather}
N_{x}=N\sin \theta \tag{X}
\end{gather}
\]
the component of the force of friction in the x direction is given by
\[
\begin{gather}
F_{fx}=F_{f}\cos \theta \tag{XI}
\end{gather}
\]
substituting the expression (V) into expression (XI)
\[
\begin{gather}
F_{fx}=\mu N\cos \theta \tag{XII}
\end{gather}
\]
substituting expressions (X) and (XII) into expression (IX)
\[
\begin{gather}
N\sin \theta +\mu N\cos \theta =ma \tag{XIII}
\end{gather}
\]
Dividing the expression (XIII) by (VII)
\[
\begin{gather}
\frac{N\sin \theta +\mu N\cos \theta }{N\cos \theta-\mu N\sin \theta }=\frac{ma}{mg}\\[5pt]
\frac{\cancel{N}(\sin \theta +\mu \cos \theta )}{\cancel{N}(\cos \theta -\mu\sin \theta )}=\frac{\cancel{m}a}{\cancel{m}g}\\[5pt]
\frac{(\sin \theta +\mu \cos \theta )}{(\cos \theta -\mu \sin \theta )}=\frac{a}{g}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a=g\left(\frac{\sin \theta +\mu \cos \theta }{\cos \theta -\mu\sin \theta }\right)}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .