enter image description here

Question: What is the acceleration of each body, if they slip along the incline? What is the normal between them? Assume the incline to be frictionless and neglect air resistance.

My teacher solved this by assuming the acceleration of both masses to be same, and so the normal force then came out to be zero. How do we know that this assumption about the accelerations is correct?

My teacher also said that even if we don’t consider both bodies’ acceleration to be the same, Newton’s second law can be applied to each to get equations for each: (the sign convention I’m going to be following is positive in the direction of motion and negative opposite to the direction of motion)

For $m_2$ the acceleration is $a_2$ and the normal force between $m_2$ and $m_1$ be $N$:


For $m_1$ the acceleration is $a_1$:


He then justified the assumption that the acceleration of both bodies is same by saying that if accelerations of the bodies were different then the normal force between them would be zero and by plugging the value of N in the equation for each body, we’ll get $a=g\sin\theta$ for each mass

I get that if $a_2 < a_1$, the bodies would lose contact and no normal would exist between them. But if $a_2 > a_1$, wouldn’t the normal be between them be non-zero? How does my teacher’s justification hold if $m_2$ doesn’t accelerate slower than $m_1$ does?

| cite | improve this question | | | | |
  • $\begingroup$ Try to set up Newton's 2nd law on each of them and you be able to solve for the normal force between them $\endgroup$ – Steeven Mar 21 '19 at 14:39
  • $\begingroup$ Hi, Brenda. We don't solve standard homework-type exercises here, but rather answer conceptual difficulties. Please read the Help section on how to write good questions. $\endgroup$ – Bill N Mar 21 '19 at 14:41
  • $\begingroup$ @Steeven You would have to (correctly) assume that the two blocks accelerate at the same rate as well $\endgroup$ – BioPhysicist Mar 21 '19 at 14:41
  • $\begingroup$ Two objects with different masses fall with the same acceleration. The same is true for a frictionless inclined plane. $\endgroup$ – harshit54 Mar 21 '19 at 15:07
  • $\begingroup$ @Aaron Stevens That’s precisely where I think I’m facing difficulty. How do I know, if I assume their acceleration to be the same, that my assumption is correct? Is there any way to solve this question without having to assume same acceleration for both masses? $\endgroup$ – Brenda Mar 21 '19 at 15:10

The below analysis assumes the blocks are released from rest.

But if $a_2>a_1$, wouldn’t the normal be between them be non-zero?

If this is the case then you have more issues than this. $m_2$ will be traveling at a faster velocity down the incline than $m_1$. In order for this to be true $m_2$ would literally have to be passing through $m_1$ down the incline. We know that the blocks do not do this. Therefore you can throw out the possibility that $a_2>a_1$.

He then justified the assumption that the acceleration of both bodies is same by saying that if accelerations of the bodies were different then the normal force between them would be zero

If they are different, then our only other option, based on the above discussion, is that $a_2<a_1$. But this means that $m_1$ will be traveling faster down the incline than $m_2$. Thus they can't even be in contact anymore, and this must mean $N=0$.

Now it must be pointed out that this does not mean that this is impossible (yet). Right now we have just shown that if $a_2<a_1$ then $N=0$ However let's keep on reasoning. Based on this using the equations you give, you end up with $$m_1a_1=m_1g\sin\theta$$ $$m_2a_2=m_2g\sin\theta$$ i.e. $$a_1=a_2=g\sin\theta$$

Ah so we have arrived at another contradiction! Assuming $a_2<a_1$ leads us to conclude that $a_2$ is not less than $a_1$ Therefore, we cannot say $a_2<a_1$

The only option we have left is that $a_1=a_2$. Therefore the accelerations must be equal. And you can use this to determine what $N$ must be and then what the acceleration is.

| cite | improve this answer | | | | |

In order to determine whether one mass might try to accelerate more than the other, you should determine the individual accelerations. In this situation, you will find that the individual acceleration of each is $g\sin\theta$. So, there's the initial justification that they have equal accelerations.

If you want to assume they have different free accelerations you must come up with a set of forces which makes them have differing accelerations. Differing kinetic frictions would be such a set, but the problem doesn't allow friction. Someone pushing upward on $m_1$ but not directly on $m_2$ would be another. In that situation, $m_2$ would be restricted from passing through $m_1$ so a normal force would then need to exist resulting in the two objects accelerating identically.

From a physics-with-calculus point of view, if the blocks are moving along the plane identically, they will have the same velocity versus time functions as each other. That means that the time derivative (slope of the velocity) of each one is identical to the other. That means the accelerations must be identical.

| cite | improve this answer | | | | |

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