Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

up vote 1 down vote accepted

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: $$F_f=m_2a=(4\mathrm{kg})\left(\frac{5}{6}\frac{\mathrm{m}}{\mathrm{s^2}}\right)=3.3\mathrm{N}<4\mathrm{N}=\mu m_1 g$$

share|improve this answer
    
Thank you. I understand now. –  xylon97 Jun 18 '13 at 12:56
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.