# 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?

• I assume \mu should be given? At the moment your problem is underconstrained... Commented Jun 17, 2013 at 16:47
• yes, i'm sorry, I forgot to write that. Its 0.2 Commented Jun 17, 2013 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: $$F-m_2a=m_1a$$ $$F=a(m_1+m_2)$$ $$\frac{F}{m_1+m_2}=a$$ $$\frac{5 \mathrm{N} }{2\mathrm{kg}+4\mathrm{kg}}=\frac{5}{6}\frac{\mathrm{m}}{\mathrm{s^2}}$$
From what we have above, you can calculate the friction force as: $$F_f=m_2a=(4\mathrm{kg})\left(\frac{5}{6}\frac{\mathrm{m}}{\mathrm{s^2}}\right)=3.3\mathrm{N}<4\mathrm{N}=\mu m_1 g$$