A block of mass m_t = 4.0kgis put on top of a block of mass m_b = 5.0kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12Nmust be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table (Fig. 6-47). Find the magnitudes of

(a) the maximum horizontal force f that can be applied to the lower block so that the blocks will move together and

(b) the resulting acceleration of the blocks.

First, calculate the coefficient of static friction for the surface between the two blocks. When the force applied is at a maximum, the frictional force between the two blocks must also be a maximum. Using Newton's second law, we can solve the given problem.

The system consists of two blocks, one on top of the other. If we pull the bottom block too hard, the top block will slip on the bottom one. We're interested in the maximum force that can be applied such that the two will move together. The free-body diagrams for the two blocks are shown below.

Since F_t = 12N of force has to be applied to the top block for slipping to take place, using F_t = f_s,max

$\begin{array}{rcl}{F}_t& =& {\mu }_s{F}_{N,t}\\ F_t& =& {\mu }_s{m}_{t}g\\ {\mu }_s& =& \frac{F_t}{m_{t}g}\\ & =& \frac{12N}{\left(4.0kg\right)\left(9.8m/s^2\right)}\\ & =& 0.31\end{array}$

Using the same reasoning, for the two masses, the force would be,

$F={\mu }_s\left(m_t+m_b\right)g$

Substituting the value of${\mu }_s$, the maximum horizontal force has a magnitude that can be calculated as:

$F={\mu }_s\left(m_t+m_b\right)g$

Substitute the values, and we get,

$$\begin{gathered} F=0.31\left(4.0kg+5.0kg\right)\left(9.8m/s^2\right) \\ F=27N \end{gathered}$$

Thus, the maximum horizontal force is 27N.

The maximum acceleration is,

$\begin{array}{rcl}{a}_{max}& =& \frac{F}{m_t+m_b}\\ a_{max}& =& {\mu }_{s}g\\ a_{max}& =& 0.31\times 9.8m/s^2\\ & =& 3.0m/s^2\end{array}$

Thus, the resulting acceleration of blocks is 3.0m/s².