In the shown figure. If $F = 20 N, m_{1} = m_{2} = 3 k g$ and the acceleration is $0.5 m / s^{2}$ . What will be the tension in the connecting cord if the friction forces on the two blocks are equal? How large is the frictionless force on either block.

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Solution

Let us first consider the 2 masses as a single object. Total mass = $3 + 3 = 6 k g$
Resultant force, $F = m a$
$F \to 6 \times 0.5 = 3 N$
Since an external force of $20 N$ is being applied, the frictional force must be $20 - 3 = 17 N$
( $8.5 N$ on each mass).
Now consider $m_{1}$ by itself. The resultant force on it $F = m a = 3 \times 0.5 = 1.5 N$
Let $T = t e n s i o n i n r o p e j o i n i n g m_{1} a n d m_{2}$
The only 2 horizontal forces on $m_{1}$ are $T$ (to right) and $8.5 N$ (friction to the left)
Resultant horizontal force on $m_{1} = T_{1} - 8.5$
$T - 8.5 = 1.5$
$T = 10 N$

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