Q95P - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Describe the Newton’s second law

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.

Step 2: Find the acceleration, and describe the behavior of the block

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.