91Ó°ÊÓ

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

The net force acting on the block is given by,

$$\begin{gathered} \underset{}{∑}{F}_x=m\stackrel{→}{a} \\ M\stackrel{→}{a}=F_n \end{gathered}$$

From the figure, the frictional force is given by,

$$\begin{gathered} f=Mg \\ {\mathrm{μ}}_s{F}_n=Mg \\ {\mathrm{μ}}_s\left(M\stackrel{→}{a}\right)=Mg \\ \stackrel{→}{a}=\frac{g}{{\mathrm{μ}}_s} \end{gathered}$$

It is known that if the acceleration of the block $\stackrel{→}{a}â‰\yen \frac{g}{{\mathrm{μ}}_s}$, 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 $\stackrel{→}{a}=\frac{g}{{\mathrm{μ}}_s}$, and the will describe that the block was lifted by an unknown upward force, such as a wind force.