Solved Problem on Dynamics

Determine the acceleration that the cart, shown in the figure, must have so that the block does not fall. Assume g for the acceleration due to gravity and μ for the coefficient of friction between the block and cart.

Problem data:

Acceleration due to gravity: g;
Coefficient of friction between the block and cart: μ.

Problem diagram:

If the cart is moving with constant speed (Figure 1-A) only force acting on the system is the gravitational force, the cart does not exert force on the block, and it will fall. To exist force, there must be acceleration (Figure 1-B), but if this force is not enough to balance the forces that act on the block, this will also fall.

Figure 1

So there is a minimum acceleration so that the force of friction is balanced with the gravitational force that acts in the vertical direction on the block to keep it stuck in the cart (Figure 1-C)

Solution

Drawing a free-body diagram, we have the forces that act on the block.

\( {\vec F}_{g} \): gravitational force on the block;
\( {\vec F}_{C} \): force that the accelerated cart exerts on the block;
\( {\vec F}_{f} \): force of friction between block and cart.

Figure 2

Applying Newton's Second Law

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

In the horizontal direction, the only force that acts on the block is the normal reaction force \( \vec{N} \) exerted by the cart, applying the expression (I)

\[ \begin{gather} F_{C}=N=ma \tag{II} \end{gather} \]

In the vertical direction, we have the gravitational force \( {\vec F}_{g} \) and the force of friction \( {\vec{F}}_{f} \), as there is no motion in this direction. the forces cancel

\[ \begin{gather} F_{f}=F_{g} \tag{III} \end{gather} \]

the force of friction is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{at}=\mu N} \tag{IV} \end{gather} \]

the gravitational force is given by

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

substituting expressions (IV) and (V) into expression (III)

\[ \begin{gather} \mu N=mg \tag{VI} \end{gather} \]

substituting the expression (II) into expression (VI)

\[ \begin{gather} \mu \cancel{m}a=\cancel{m}g\\[5pt] \mu a=g\\[5pt] a=\frac{g}{\mu } \end{gather} \]

this is a minimum value of the acceleration so that the block does not fall, for any value of the acceleration greater than this, the block does not fall

\[ \begin{gather} \bbox[#FFCCCC,10px] {a \ge \frac{g}{\mu }} \end{gather} \]

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