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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. If the three masses are in the ratio of $1:2:3$, what is the ratio of the contact forces of the two heavier blocks, i.e, $F_2:F_3$?

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Answer: $\large\frac{5}{3}$
Lets draw the Free Body Diagrams for the three masses in contact and the resultant forces:
Here, R' is the contact force on m2m_2 and is the reaction force between m1m_1 and m2m_2
Similarly, R is the contact force on m3m_3 and is the reaction force between m2m_2 and m3m_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 $F_3=R= m_3a = F\large\frac{m_3}{m_1+m_2+m_3}$
Therefore we can solve for $F_2=R'= m_2a+m_3a = F\large\frac{m_2+m_3}{m_1+m_2+m_3}$
Given that the masses are in the ratio $1:2:3$, $F_3 = F \large\frac{3m}{m+2m+3m}$$ = \large\frac{3F}{6}$
Given that the masses are in the ratio $1:2:3$, $F_2 = F \large\frac{2m+3m}{m+2m+3m}$$ = \large\frac{5F}{6}$
$\Rightarrow$ Ratio of $F_2:F_3$ is $\large\frac{F_2}{F_3}$$ = \Large\frac{\frac{5F}{6}}{\frac{3F}{6}}$$ = \large\frac{5}{3}$
answered Aug 20, 2014 by balaji.thirumalai
edited Aug 20, 2014 by balaji.thirumalai
 

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