# Mechanics question (a block on top of a block) [closed]

This question is very similar to this one here.

A block of mass $m_1$ is placed on another block of mass $m_2$ lying on a smooth horizontal surface.
The coefficient of friction (static and kinetic) between $m_1$ and $m_2$ is $\mu$.

Find the acceleration of the blocks if the force applied to $m_1$ is $5N$, given that $m_1 = 2kg$ and $m_2 = 4kg$ and $\mu=0.2$.

I can prove the result in the link, and obtain the critical force as $6N$, so for $5N$, they will have same acceleration, which will be equal to $\frac{5}{6}$.

But when I take the the general case, draw free body diagrams, I get these equation -

$$m_1 a_1 = F - \mu m_1 g \\ m_2 a_2 = \mu m_1 g$$

This gives the answer as $a_1 = \frac{1}{2}$ and $a_2 = 1$

The only place where I think I could have made a mistake would be in case of force of friction. I am taking the maximum value of friction ($4N$), but since the force applied is greater, I presume this can happen.

Can anyone tell me where I go wrong?

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## closed as too localized by Manishearth♦Jun 22 '13 at 15:47

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I assume \mu should be given? At the moment your problem is underconstrained... –  Kyle Jun 17 '13 at 16:47
yes, i'm sorry, I forgot to write that. Its 0.2 –  xylon97 Jun 17 '13 at 16:53

Because you are not pulling with the critical force, $6N$, then static friction $F_f<\mu m_1g$, where $\mu=0.2$.

The equations you get from Newton's second law are: $$F-F_f=m_1a$$ $$F_f=m_2a$$

Substitute $m_2a$ into $F_f$ in the first equation: $$F-m_2a=m_1a$$ $$F=a(m_1+m_2)$$ $$\frac{F}{m_1+m_2}=a$$ $$\frac{5 \mathrm{N} }{2\mathrm{kg}+4\mathrm{kg}}=\frac{5}{6}\frac{\mathrm{m}}{\mathrm{s^2}}$$

The applied force is allowed to be greater than the friction force between the blocks because they are not moving relative to each other. Notice that no matter what the force you apply on the top block, both blocks will have non-zero acceleration, because the horizontal surface is frictionless.

From what we have above, you can calculate the friction force as: $$F_f=m_2a=(4\mathrm{kg})\left(\frac{5}{6}\frac{\mathrm{m}}{\mathrm{s^2}}\right)=3.3\mathrm{N}<4\mathrm{N}=\mu m_1 g$$

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Thank you. I understand now. –  xylon97 Jun 18 '13 at 12:56