Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg, which in turn is on a horizontal tabletop (Fig. P5.92). There is no friction between block Aand the tabletop, but the coefficient of static friction between blocks Aand Bis 0.750. A light string attached to block Apasses over a frictionless, massless pulley, and block Cis suspended from the other end of the string. What is the largest mass that block Ccan have so that blocks Aand Bstill slide together when the system is released from rest?

The given data can be listed below as:

• The mass of block B is $m_1=5.00\hspace{0.33em}kg$.
• The mass of block A is $m_2=8.00\hspace{0.33em}kg$.
• The static frictional coefficient between blocks A and B is ${\mu }_s=0.750$.

Tension is described as a force that acts on a body by virtue of its elastic nature. Moreover, tension is also considered as a pair of action and reaction forces.

The equation of the frictional force on the block B is expressed as:

$f={\mu }_s{m}_{1}g$

Here, f is the frictional force,${\mu }_s$ is the static frictional coefficient between the block A and B,$m_1$ is the mass of the block B and is the acceleration due to gravity.

Substitute the values in the above equation.

$$\begin{gathered} f=\left(0.750\right)\left(5.00kg\right)\left(9.8m/s^2\right) \\ =\left(3.75kg\right)\left(9.8m/s^2\right) \\ =36.75kg·m/s^2\times \frac{1N}{1kg·m/s^2} \\ =36.75N \end{gathered}$$

The equation of the acceleration of the block B is expressed as:

$a=\frac{f}{m_1}$

Here, a is the acceleration of the block B.

Substitute the values in the above equation.

$$\begin{gathered} a=\frac{36.75N}{5.00kg} \\ =7.35N/kg \\ =7.35N/kg\times \frac{1kg·m/s^2}{1N} \\ =7.35m/s^2 \end{gathered}$$

The equation of the mass of the block C is expressed as:

$$\begin{gathered} Mg-f=Ma+m_{2}a \\ {m}_{2}a+f=M\left(g-a\right) \\ \frac{m_{2}a+f}{g-a}=M \end{gathered}$$

Here, M is the mass of the block C.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{M}=\frac{\left(8.00\mathrm{kg}\right)\left(7.35\mathrm{m}/{\mathrm{s}}^2\right)+36.75\mathrm{N}}{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-\left(7.35\mathrm{m}/{\mathrm{s}}^2\right)} \\ =\frac{\left(58.8\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\right)+36.75\mathrm{N}\times {\displaystyle \frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}}}{\left(2.45\mathrm{m}/{\mathrm{s}}^2\right)} \\ =\frac{\left(95.55\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\right)}{\left(2.45\mathrm{m}/{\mathrm{s}}^2\right)} \\ =39\mathrm{kg} \end{gathered}$$

Thus, the largest mass the block C can have is 39 kg.