Two blocks A and B of mass 10kg and 20 kg respectively, are arranged as shown in Fig. In the figure given a constant force $F_{0} = 120 N$ acts on block A and a force F applied horizontally on block B is gradually increased from zero. Discuss the direction and nature of friction force and the accelerations of the block for different values of F.

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Solution

In the above situation, we see that the maximum possible value of friction between the blocks is
$f_{m a x} = μ_{s} m_{A} g = 0.3 \times 10 \times = 30 N$ .

Case I: When F=0.
Considering that there is no slipping between the blocks the acceleration of system will be
$a = \frac{120}{20 + 10} = 4 m s^{- 2}$
From FBD of $B, f = m_{B} a = 20 \times 4 = 80 N$
But $f_{m a x} = 30 N$
We can conclude that the blocks do not move together.
Now drawing the FBD of each block, for finding out individual accelerations.
Acceleration of A,
$a_{A} = \frac{120 - 30}{10} = 9 m s^{- 2}$ towards right
Acceleration of B,
$a_{B} = \frac{30}{20} = 1.5 m s^{- 2}$ towards right

Case II: F is increased from zero till the two blocks just start moving together.
As the two blocks move together, the friction is static in nature and its value is limiting. FBD in this case will be
$a_{A} = \frac{120 - 30}{10} = 9 m s^{- 2}$
$\Rightarrow a_{B} = \frac{F + 30}{20} = a_{A}$
$\frac{F + 30}{20} = 9$ 0r F=150N
Hence, when 0< F < 150N, the blocks do not move together and friction is kinetic. As F increases, acceleration of block B increases from $1.5 m s^{- 2}$ to $9 m s^{- 2}$
As F = 150N, limiting static friction starts acting and the two blocks start moving together.

Case III: When F is increased above 150N. In this scenario, the static friction adjusts itself so as to keep the blocks moving together. The value of static friction starts reducing but the direction still remains same. This happens continuously till the value of friction becomes zero. In this case, the FBD is as follows:
$a_{A} = a_{B} = \frac{120 - f}{10} = \frac{F + f}{20}$
Therefore, when the friction force f gets reduced to zero, the above accelerations become
$a_{A} = a_{B} = \frac{120 - f}{10} = \frac{F + f}{20}$
Therefore, when the friction force f gets reduced to zero, the above accelerations become
$a_{A} = \frac{120}{10} = 12 m s^{- 2} \Rightarrow a_{B} = \frac{F}{20} = a_{A} = 12 m s^{- 2}$
$∴ F = 240 N$
Hence, when $150 N \leq F \leq F \leq 240 N$ , the static friction force continuously decreases from maximum to zero at F=240N. The accelerations of the blocks increase from $9 m s^{- 2}$ to $12 m s - 2$ during the change of force F.

Case IV: When F is increased again from 240N, the direction of friction force on the block reverses but it is still static. F can be increased till this static friction reaches its limiting value. FBD at this junction will be:
The blocks move together, therefore,
$a_{A} = \frac{120 + 30}{10} = 15 m s^{- 2}$
$\Rightarrow a_{B} = \frac{F - 30}{20} = a_{A} = 15 m s^{- 2}$
$∴ \frac{F - 30}{20} = 15 m s^{- 2}$
Hence, F = 330N.

Case V: When F is increased beyond 330N. In this case, the limiting friction is achieved and slipping takes place.

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