In Fig. 6-34, blocks A and B have weights of $\mathbf{44}\mathbf{}\mathbf{N}$and $\mathbf{22}\mathbf{}\mathbf{N}$, respectively. (a) Determine the minimum weight of block C to keep A from sliding if ${\mathbf{\mu }}_{\mathbf{k}}$between A and the table is $\mathbf{0}\mathbf{.}\mathbf{20}$. (b) Block C suddenly is lifted off A. What is the acceleration of block A if ${\mathbf{\mu }}_{\mathbf{k}}$between A and the table is $\mathbf{0}\mathbf{.}\mathbf{15}$?

Weight of Block A, ${\mathrm{F}}_{\mathrm{gA}}=44\mathrm{N}$

Weight of Block B, ${\mathrm{F}}_{gB}=22\mathrm{N}$

Coefficient of static friction between the block and table, ${\mathrm{\mu }}_a=0.20$

Coefficient of kinetic friction between the block and table, ${\mathrm{\mu }}_{\mathrm{k}}=0.15$

To find the mass of the block of C and the acceleration of block A, draw the free body diagram for each block and then apply Newton's 2nd law of equation. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

${\mathrm{F}}_{\mathrm{net}}=\sum _{}\mathrm{ma}$

where, F is the force, m is mass and a is an acceleration.

Free Body Diagram of the body A and body B:

(Block A + Block C) (Block B)

(Block A + Block C) does not slide. So, its acceleration is 0.By using Newton’s 2nd law of motion along the vertical direction (along y) to the system, (positive x axis along the incline and the positive y along the vertical direction)

$$\begin{gathered} {\mathrm{F}}_{\mathrm{N}}={\mathrm{F}}_{\mathrm{gac}}-0 \\ {\mathrm{F}}_{\mathrm{N}}={\mathrm{F}}_{\mathrm{gac}} \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} \mathrm{f}={\mathrm{\mu }}_{\mathrm{g}}{\mathrm{F}}_{\mathrm{N}} \\ ={\mathrm{\mu }}_{\mathrm{g}}{\mathrm{F}}_{\mathrm{gAC}} \end{gathered}$$ (i)

Now, applying Newton’s 2nd along horizontal direction to the Block(A+C) having massm,

$$\begin{gathered} \mathrm{T}-\mathrm{f}=0 \\ \mathrm{T}=\mathrm{f} \end{gathered}$$ (ii)

Similarly, apply to the Block B having massm’,

$$\begin{gathered} \mathrm{T}-{\mathrm{F}}_{\mathrm{gB}}=0 \\ \mathrm{T}={\mathrm{F}}_{\mathrm{gB}} \end{gathered}$$ (iii)

From equation (i), (ii), and (iii),

${\mathrm{F}}_{\mathrm{gB}}={\mathrm{\mu }}_{\mathrm{k}}{\mathrm{F}}_{\mathrm{gAc}}$

$$\begin{gathered} {\mathrm{F}}_{\mathrm{gAc}}=\frac{{\mathrm{F}}_{\mathrm{gB}}}{{\mathrm{\mu }}_{\mathrm{s}}} \\ =\frac{22}{0.20} \\ =110\mathrm{N} \end{gathered}$$

$$\begin{gathered} {\mathrm{F}}_{\mathrm{gAc}}={\mathrm{F}}_{{\mathrm{g}}^4}+{\mathrm{F}}_{\mathrm{gc}} \\ {\mathrm{F}}_{\mathrm{gc}}={\mathrm{F}}_{\mathrm{gAc}}-{\mathrm{F}}_{\mathrm{gA}} \\ =110-44 \\ =66\mathrm{N} \end{gathered}$$

Thus, the weight of block C is 66 N.

If Block C suddenly lifted off A, then, by using the Newton’s 2nd law of motion along vertical direction on block A only,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{N}}-{\mathrm{F}}_{\mathrm{gA}}=0 \\ {\mathrm{F}}_{\mathrm{N}}={\mathrm{F}}_{\mathrm{gA}} \end{gathered}$$

Similarly, along horizontal direction,

$\mathrm{T}-\mathrm{f}=\left(\frac{{\mathrm{F}}_{\mathrm{gA}}}{\mathrm{g}}\right)\mathrm{a}$ (iv)

Newton’s 2nd law for block B,

${\mathrm{F}}_{\mathrm{gB}}-\mathrm{T}=\left(\frac{{\mathrm{F}}_{\mathrm{gB}}}{\mathrm{g}}\right)\mathrm{a}$ (v)

Adding equation (iv) and (v)

$$\begin{gathered} {\mathrm{F}}_{\mathrm{gB}}-\mathrm{T}=\left(\frac{{\mathrm{F}}_{\mathrm{gB}}}{\mathrm{g}}+\frac{{\mathrm{F}}_{\mathrm{gA}}}{\mathrm{g}}\right)\mathrm{a} \\ \mathrm{a}=\frac{{\mathrm{F}}_{\mathrm{gB}}-\mathrm{f}}{\left(\frac{{\mathrm{F}}_{\mathrm{gB}}}{\mathrm{g}}+\frac{{\mathrm{F}}_{\mathrm{gA}}}{\mathrm{g}}\right)}=\frac{22-\left(0.15\right)\left(44\right)}{\left({\displaystyle \frac{44}{9.81}}+{\displaystyle \frac{22}{9.81}}\right)} \\ =2.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the acceleration of the block A when block is removed is $2.3\mathrm{m}/\mathrm{s}$