Friction On Stacked Block

The only force exerted on the lower block is the friction force from the above block. If the two blocks are to move together, then their accelerations must be equal.

Suppose the frictional force on the lower block is $f$ . The maximum possible value of friction, $f_\text{max}$ between the two blocks is $\mu_s N$ . The normal reaction between the blocks is $N = m_2 g$ , therefore $f_\text{max} = \mu m_2 g$ .

Thus, the maximum possible acceleration of the lower block is $a_\text{max} = \dfrac{f_\text{max}}{m_1} = \dfrac{\mu m_2g}{m_1}$ . This is the limit on the acceleration of the blocks. If both blocks have to move together, then they can't move with acceleration greater than $a_\text{max}$ .

We can show that the maximum force for which the blocks would move together is $F_\text{max} = (m_1 + m_2) a_\text{max}$ :

Suppose the force $F$ were greater than $F_\text{max}$ , then the acceleration of the center of mass of the blocks would be greater than $a_\text{max}$ . The acceleration of the lower block cannot be greater than $a_\text{max}$ , which implies that the acceleration of the upper block is greater than $a_\text{max}$ . We see that both blocks would have different accelerations, which means that both blocks would move separately.

We have $\begin{aligned} F_\text{max} & = (m_1 + m_2) a_\text{max} \\ & = (m_1 + m_2) \frac{\mu m_2g}{m_1} \\ & = (40 + 10) \times 0.6 \times \frac{10}{40} \times 9.8 \\ & = \SI{73.5}{\newton} \end{aligned}$

Firstly it is important to note that the question asks about the blocks moving together. This means that blocks will be at rest with respect to each other and hence there will be no relative motion between the blocks. This means that only static friction will act between the blocks.

Lets take a look at the FBD of the two blocks:

The system of two blocks is moving forward with acceleration a

Along Horizontal, the balance of forces on each mass is given by...

$\begin{aligned} M_2 a &= F - f & (1) \\ M_1 a &= f & (2) \end{aligned}$

Along vertical:

$\begin{aligned} N_1 & =M_2 g & (3) \end{aligned}$

Now substituting value of a from equation (2) in equation (1) , we get $\mathbf{F}$ = $\mathbf{f }$ $\Bigg ($ $\Big($ $M_{1}$ + $M_{2}$ $\Big)$ / $M_{1}$ $\Bigg )$

From above equation we clearly see that F= $F_{max}$ when f= $f_{max}$

$f_{max}$ = $\mu$ $N_{1}$ g , where $\mu$ is coefficient of static friction

Putting

• $\mu$ =0.6
• $M_{2}$ =10
• $M_{1}$ =40

We get $f_{max}$ = 58.8

and $F_{max}$ = $\boxed{73.5}$