For the system of Fig. 4–32 (Example 4–20), how large a mass would box A have to have to prevent any motion from occurring? Assume${\mu }_s=\mathbf{0}\mathbf{.}\mathbf{30}$.

FIGURE 4-32 Example 4–20.

The static frictional force acts between box A and the table to prevent any movement in the given system.

So, the static coefficient of friction acts between them.

The table applies the normal force on box A. When there is no motion, the weight of box B balances the tension in the cord.

The given data can be listed below as:

• The mass of box B is $m_B=2\mathrm{kg}$.
• The static coefficient of friction is ${\mu }_s=0.30$.
• The acceleration due to gravity is $g=9.81 \mathrm{m}/{\mathrm{s}}^2$.

The free body diagram of box B can be shown as:

Here, $F_T$is the tension in the cord, and g is the acceleration due to gravity.

There is no motion in the system, so the acceleration of the boxes is zero.

At the equilibrium condition, the forces along the vertical direction of box B can be expressed as:

$\begin{array}{c}\sum F_y=0\\ F_T-m_{B}g=0\\ F_T=m_{B}g\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}{F}_T=2\mathrm{kg}\times 9.81 \mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =19.62\mathrm{N}\end{array}$

The free body diagram of box A can be shown as:

Here, $F_N$is the normal reaction force on box A, and $F_{fr}$is the frictional force. The tension is constant along the cord. So, the same tension force acts to the right of block A.

At the equilibrium condition, the forces along the horizontal direction of box A can be expressed as:

$\begin{array}{c}\sum F_x=0\\ F_T-F_{fr}=0\\ F_T=F_{fr}\end{array}$

Substitute the values in the above equation.

$F_{fr}=19.62\mathrm{N}$

The static frictional force can be expressed as:

$F_{fr}={\mu }_s{F}_N$

Substitute the values in the above equation.

$\begin{array}{c}19.62\mathrm{N}=0.30\times F_N\\ F_N=65.4\mathrm{N}\end{array}$

At the equilibrium condition, the forces along the vertical direction of box A can be expressed as:

$\begin{array}{c}\sum F_y=0\\ F_N-m_{A}g=0\\ F_N=m_{A}g\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}65.4\mathrm{N}\left(\frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)=m_A\times 9.81 \mathrm{m}/{\mathrm{s}}^2\\ m_A=6.67\mathrm{kg}\end{array}$

Thus, Box A should have a mass of 6.67 kg to be able to prevent any motion from occurring.