The blocks A B have masses of 2 6 kg respectively are kept on a rough horizontal surface of - -0-2- If a horizontal force of 6 N pushes them- Calculate the force that the 2 kg block exerts on the other block-

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The blocks A ...

Question

The blocks A & B have masses of $2$ & $6$ kg respectively are kept on a rough horizontal surface of $μ = 0.2$ . If a horizontal force of $6$ N pushes them. Calculate the force that the $2$ kg block exerts on the other block.

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Solution

Now, the maximum frictional force applied by the surface on the two blocks is as follow:
The maximum frictional force on the block A is,
$F_{f a} = μ (N_{A})$
$F_{f a} = μ (m_{A} \times g)$
$F_{f a} = 0.2 (2 k g \times 9.8 m / s^{2})$
$F_{f a} = 3.92 N$
Here, the maximum frictional force applied by the surface on the block A is $F_{f a}$
The maximum frictional force on the block B is,
$F_{f b} = μ (N_{B})$
$F_{f b} = μ (m_{B} \times g)$
$F_{f b} = 0.2 (6 k g \times 9.8 m / s^{2})$
$F_{f b} = 11.76 N$
Here, the maximum frictional force applied by the surface on the block B is $F_{f b}$
so, now the total maximum frictional force that can be applied by the surface on the two block is $F_{f} = F_{f a} + F_{f b}$
$F_{f} = 3.92 N + 11.76 N$
$F_{f} = 15.68 N$
But, here the external force on the blocks is only $6 N$ hence, the applied force is not enough to move the blocks because, the frictional force will neutralize the force. So, the acceleration of the system is zero.
This is same for the both the figures.
The force applied by the block $2 k g$ on the other block for figure 1 can be found as follows:
since the acceleration of the $6 k g$ block is zero so,
$F_{b} - μ (N_{B}) = 0$
$F_{b} = μ (N_{B})$
here, $F_{b}$ is the force on the block B.

Now again the acceleration of the whole blocks is zero so,
$F - m u (N_{A}) - μ (N_{B}) = 0$
$μ (N_{B}) = F - μ (N_{A})$
$μ (N_{B}) = 6 N - μ (2 k g \times 9.8 m s^{2})$
$μ (N_{B}) = 2.08 N$
The force applied by the block $2 k g$ on the other block for figure 2 can be found as follows:
since the acceleration of the $2 k g$ block is zero so,
$F_{a} - μ (N_{A}) = 0$
$F_{a} = μ (N_{A})$
here, $F_{a}$ is the force on the block A.

Now again the acceleration of the whole blocks is zero so,
$F - μ (N_{A}) - μ (N_{B}) = 0$
$μ (N_{A}) = F - μ (N_{B})$
$μ (N_{A}) = 6 N - μ (6 k g \times 9.8 m s^{2}$
$μ (N_{A}) = 0 N$
so, the force applied by the block of mass $2 k g$ on the other block is
$F_{b} = 0 N$

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The blocks A & B have masses of 2 & 6 kg respectively are kept on a rough horizontal surface of μ=0.2. If a horizontal force of 6 N pushes them. Calculate the force that the 2 kg block exerts on the other block.

The blocks A & B have masses of $2$ & $6$ kg respectively are kept on a rough horizontal surface of $μ = 0.2$ . If a horizontal force of $6$ N pushes them. Calculate the force that the $2$ kg block exerts on the other block.