Solved Problem on Dynamics

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 inclined plane and direction downward.
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 gravitational \( {\vec F}_{g} \) force can be decomposed into two components, one component parallel to the x-axis, \( {\vec F}_{gx} \), and the other component normal or perpendicular, \( {\vec F}_{gy} \). In the triangle in the right in the Figure 3-A we see that the gravitational force \( {\vec F}_{g} \) is perpendicular to the horizontal plane makes a 90° angle, the angle between the inclined plane and the horizontal plane is equal to θ, as the sum of the interior angles of a triangle equals to 180°, the angle α between the gravitational force and the parallel component should be

\[ \alpha +\theta +90°=180°\Rightarrow \alpha=180°-\theta -90°\Rightarrow \alpha=90°-\theta \]

The components of the gravitational force in the x and y directions are perpendicular to each other, in the triangle to the right, we have the angle between the gravitational force \( {\vec F}_{g} \) and the component of gravitational force in the y direction \( {\vec F}_{gy} \) is

\[ 90°-\alpha \Rightarrow 90°-(90°-\theta)\Rightarrow 90°-90°+\theta \Rightarrow \theta \]

Figure 3

The acceleration of the cart can also be decomposed in the x and y directions (Figure 3-B). The angle between the acceleration \( \vec{a} \) and the component of the acceleration in the direction of the inclined plane \( {\vec a}_{x} \) is θ, the same angle of the inclined plane, are alternate angles.

Drawing the forces in a system of coordinates (Figure 4), we can get its components.
The components of the acceleration are given by

\[ \begin{gather} a_{x}=a\cos \theta \tag{I} \end{gather} \]

\[ \begin{gather} a_{y}=a\sin \theta \tag{II} \end{gather} \]

and the gravitational force components are given by

\[ \begin{gather} F_{gx}=F_{g}\sin \theta \tag{III} \end{gather} \]

\[ \begin{gather} F_{gy}=F_{g}\cos \theta \tag{IV} \end{gather} \]

Figure 4

Applying Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{V} \end{gather} \]

In the x direction, we only have the component F_gx of the gravitational force

\[ \begin{gather} F_{gx}=ma_{x} \tag{VI} \end{gather} \]

substituting expressions (I) and (III) into expression (VI)

\[ \begin{gather} F_{g}\sin \theta =ma\cos \theta \tag{VII} \end{gather} \]

the gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{VIII} \end{gather} \]

substituting the expression (VIII) into expression (VII)

\[ \begin{gather} mg\sin \theta =ma\cos \theta \\[5pt] a=\frac{\cancel{m}g\sin \theta}{\cancel{m}\;\cos \theta} \end{gather} \]

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

\[ \begin{gather} \bbox[#FFCCCC,10px] {a=g\tan \theta} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .