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 that makes an angle θ with the horizontal, and on the inclined plane is placed a body. Determine the acceleration of the cart for the body to remain at rest on the inclined plane. Neglect the friction between the body and the inclined plane and assume g for the acceleration due to gravity.


Problem data:
  • Angle of the inclined plane:    θ;
  • Acceleration due to gravity:    g.
Problem diagram:

We choose a frame of reference with the x-axis parallel to the horizontal plane and in the same direction as the acceleration of the cart.
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 of the cart relative to the ground (Figure 1).
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 on the block.
Figure 2

The normal reaction force \( \vec{N} \) on the inclined plane can be decomposed into two components, a component in the direction of the x-axis, \( {\vec N}_{x} \), and the other in the direction of the y-axis, \( {\vec N}_{y} \), of the referential frame. The angle of the inclined plane is given as θ, in Figure 3-A, we see that the angle of the x direction and the inclined plane is also θ, are internal angles.

Figure 3

The normal reaction force is perpendicular to the inclined plane, makes an angle of 90°, the angle between the x direction and the normal force is \( \alpha =90°-\theta \) (Figure 3-B). As x and y directions are perpendicular to each other, the angle between the normal force and the y-axis is (Figure 3-C)
\[ 90°-\alpha=90°-(90°-\theta )=90°-90°+\theta=\theta \]
Drawing the forces in a coordinates system (Figure 4), we use the angle θ between the normal force and the y-axis, unlike what is usually done when we use the angle with the x-axis.
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 \( {\vec F}_{g} \), and the component of the normal force in y direction, \( {\vec N}_{y} \), cancel
\[ \begin{gather} N_{y}=F_{g} \tag{II} \end{gather} \]
the component of the normal force in the y direction is given by
\[ \begin{gather} N_{y}=N\cos \theta \tag{III} \end{gather} \]
Figure 4

the gravitational force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{IV} \end{gather} \]
substituting expressions (III) and (IV) into expression (II)
\[ \begin{gather} N\cos \theta =mg \tag{V} \end{gather} \]
In the x direction, we have the component of the normal force in this direction, applying the expression (I)
\[ \begin{gather} N_{x}=ma \tag{VI} \end{gather} \]
the component of the normal force in the x direction is given by
\[ \begin{gather} N_{x}=N\sin \theta \tag{VII} \end{gather} \]
substituting the expression (VII) into expression (VI)
\[ \begin{gather} N\sin \theta =ma \tag{VIII} \end{gather} \]
Dividing the expression (VII) by (V)
\[ \begin{gather} \frac{\cancel{N}\sin \theta }{\cancel{N}\cos \theta }=\frac{\cancel{m}a}{\cancel{m}g} \end{gather} \]
From the Trigonometry    \( \tan \theta =\frac{\sin \theta }{\cos \theta} \)
\[ \begin{gather} \tan \theta =\frac{a}{g} \end{gather} \]
The acceleration will be given by
\[ \begin{gather} \bbox[#FFCCCC,10px] {a=g\;\tan \theta} \end{gather} \]
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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 .