A block is placed against the vertical front of a cart (Fig. P5.95). What acceleration must the cart have so that block A does not fall? The coefficient of static friction between the block and the cart is ${\mathit{\mu }}_{\mathbf{s}}$. How would an observer on the cart describe the behavior of the block?

According to Newton’s second law, the linear force is given by,

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here, is linear force, is the mass of an object, and is the acceleration of the object.

Let the mass of the block be M .

The free-body diagram is as follows.

Let $\overrightarrow{a}$ be the acceleration that the cart has so that block A does not fall.

The net force acting on the block is given by,

$$\begin{gathered} \sum _{}{F}_x=m\overrightarrow{a} \\ M\overrightarrow{a}=F_n \end{gathered}$$

From the figure, the frictional force is given by,

$$\begin{gathered} f=Mg \\ {\mu }_s{F}_n=Mg \\ {\mu }_s\left(M\overrightarrow{a}\right)=Mg \\ \overrightarrow{a}=\frac{g}{{\mu }_s} \end{gathered}$$

It is known that if the acceleration of the block $\overrightarrow{a}\ge \frac{g}{{\mu }_s}$, then the block is prevented from falling, which means the block was lifted by an unknown upward force, such as a wind force.

Therefore, the acceleration of the block is $\overrightarrow{a}=\frac{g}{{\mu }_s}$, and the will describe that the block was lifted by an unknown upward force, such as a wind force.