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.

  • $\begingroup$ @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. $\endgroup$ Sep 18, 2018 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, 2018 at 14:55
  • $\begingroup$ @npojo Whoops, looks like I missed a $g$... Fixed. $\endgroup$ Sep 18, 2018 at 15:15
  • 1
    $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$
    – user4552
    Sep 18, 2018 at 15:16
  • $\begingroup$ @AaronStevens Edited. $\endgroup$ Sep 18, 2018 at 15:21

1 Answer 1


$m_2$ 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.

  • $\begingroup$ Finally, I get it now! Many thanks for the clear response! $\endgroup$ Sep 18, 2018 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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