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Tension in a String

For the syste...

Question

For the system shown in the figure, if $F_{2}$ and $F_{3}$ are forces acting on the masses $m_{2}$ and $m_{3}$ respectively, then the correct relation between the forces is (Assume that the blocks are connected by inextensible strings and the surface is frictionless)

F_{3} = F_{1} + F_{2}

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F_{3} = \frac{m_{1} F_{1}}{m_{1} + m_{2} + m_{3}}

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F_{3} = \frac{m_{2} F_{1}}{m_{1} + m_{2} + m_{3}}

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F_{3} = \frac{m_{3} F_{1}}{m_{1} + m_{2} + m_{3}}

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Solution

The correct option is D $F_{3} = \frac{m_{3} F_{1}}{m_{1} + m_{2} + m_{3}}$
Assuming the connecting strings are inextensible, the whole system will move with common acceleration $a$ . Therefore we can draw FBD of the blocks, treating it as one system.

$F_{1} = (m_{1} + m_{2} + m_{3}) a$
$\Rightarrow a = \frac{F_{1}}{m_{1} + m_{2} + m_{3}}$

Now, draw the FBD of mass $m_{3}$ alone

$F_{3} = m_{3} a - - (1)$

Substituting $a$ in equation (1), we get
$F_{3} = \frac{m_{3} F_{1}}{m_{1} + m_{2} + m_{3}}$

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For the system shown in the figure, if F2 and F3 are forces acting on the masses m2 and m3 respectively, then the correct relation between the forces is (Assume that the blocks are connected by inextensible strings and the surface is frictionless)

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