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So I've been trying to tackle this for the last few hours, but no dice. (exam practice, by the way)

We got two point masses $m_1$ and $m_2$ lying tangent to each other on top of an inclined plane with angle $\phi$. In this scenario, $m_2$ is above $m_1$. The friction co-efficient between the plane and $m_1$ is $k_1$, while for $m_2$ it is equal to $k_2$, with $k_1 > k_2$.

The question is: find the force $m_2$ exerts on $m_1$ as they both slide down. There is no mention of whether they slide with constant speed, so I assume it's not constant.

I've tried calculating the total force on $m_1$ on the axis of the plane, and after drawing all force vectors I found it equal to $m_2g(sin(\phi) - k_2cos(\phi))$. But the answer key states only that $F=\frac{k_1-k_2}{m_1+m_2}m_1m_2gcos(\phi)$ without telling why is that so and I'm confused.

Any help is welcome.

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  • $\begingroup$ @Aaron Stevens 1) Yes. 2) They slide (my bad). 3) Yes, $m_2$ starts above $m_1$, yet they touch for the whole duration of the descent. 4) The exercise doesn't mention anything about constant speed. $\endgroup$ – lightspot21 Sep 18 '18 at 14:44
  • $\begingroup$ Start with your equation and trace it back to find what went wrong with your units. While $mg$ or $kmg$ are units of force, $m$ and $mk$ are not. $\endgroup$ – npojo Sep 18 '18 at 14:55
  • $\begingroup$ @npojo Whoops, looks like I missed a $g$... Fixed. $\endgroup$ – lightspot21 Sep 18 '18 at 15:15
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    $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – Ben Crowell Sep 18 '18 at 15:16
  • $\begingroup$ @AaronStevens Edited. $\endgroup$ – lightspot21 Sep 18 '18 at 15:21
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$m_2$ is starting above $m_1$ on the incline. So $m_2$ is essentially adding an additional force pushing $m_1$ down. By Newton's third law this also means $m_1$ is pushing up the incline on $m_2$. Furthermore, I will assume kinetic friction here. As always, you should start with free body diagrams: FBD

Exploiting the fact that $f_{k_i}=k_iN_i$, as well knowing we have zero acceleration perpendicular to the plane (i.e. $N_i=m_ig\cos\phi$), we know that $$f_{k_i}=m_ik_ig\cos\phi$$

Therefore, we can write down Newton's second law for forces parallel to the incline, where I will take down the incline to be positive:

$$\sum F_1=F+m_1g\sin\phi-m_1gk_1\cos\phi=m_1a$$ $$\sum F_2=-F+m_2g\sin\phi-m_2k_2g\cos\phi=m_2a$$

Note these equations assume that $a_1=a_2=a$, which just means they slide together in this situation.

So now we have two equations with two unknown values ($F$ and $a$). From here it becomes an algebra problem to solve for $F$ in terms of the given "quantities", which I will leave to you. Based on information provided in the question, it seems like you were looking at the $\sum F_2$ equation and assuming that $a=0$, which is not true since the problem does not say the blocks are moving at a constant speed.

NOTE: This is the standard procedure for any problem like this:

  • Draw the free body diagram for all relevant bodies
  • Define your coordinate system (This could be step 1, as sometimes how you define your coordinate system will help determine which forces you need to break into components.)
  • Write out Newton's Second Law equation for each body
  • Relate the accelerations of each body
  • Do algebra to solve for desired quantity

These steps can be used to solve so many problems introductory physics teachers can throw at you dealing with blocks, planes, pulleys, etc.

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  • $\begingroup$ Finally, I get it now! Many thanks for the clear response! $\endgroup$ – lightspot21 Sep 18 '18 at 15:30

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