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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Three blocks of masses $m_1, m_2$ and $m_3$, with $m_1 \lt m_2 \lt m_3$, are placed in contact with each other on a frictionless, horizontal surface, as shown. A constant horizontal force $F$ is applied to $m_1$ as shown. What is the contact force on $m_2$?

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Answer: $F\large\frac{m_2+m_3}{m_1+m_2+m_3}$
Lets draw the Free Body Diagrams for the three masses in contact and the resultant forces:
Here, R' is the contact force on $m_2$ and is the reaction force between $m_1$ and $m_2$
Similarly, R is the contact force on $m_3$ and is the reaction force between $m_2$ and $m_3$
For block-1, $F - R' = m_1a$ and for block-2, $R'-R = m_2a$ and for block-3, $R = m_3a$
Adding the three equations, $F - R' + R' - R + R = F = (m_1+ m_2+m_3)a$
$\Rightarrow a = \large\frac{F}{m_1+m_2+m_2}$
Therefore we can solve for $R' = m_2 a+R = (m_2+m_3)a=F\large\frac{m_2+m_3}{m_1+m_2+m_3}$
answered Aug 20, 2014 by balaji.thirumalai
edited Aug 20, 2014 by balaji.thirumalai
 

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