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Q 1. Two bodies, $m_1$ and $m_2$, in contact with each other, are under constant acceleration $a$ in the same direction. Find the force exerted by $m_1$ on $m_2$.

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A similar question with my answer

Two bodies, $m_1$ and $m_2$, are in contact with each other. A force $F$ is exerted on $m_1$. What will the force exerted by $m_1$ on $m_2$

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Answer

$a_1=\frac{F}{m_1}$; $a_{net}=\frac{F}{m_{1}+m_{2}}$; $\therefore F_{12}=\frac{Fm_2}{m_{1}+m_{2}}$

($a_1$ = acceleration of $m_1$ when $m_2$ was not present.)

Here there is a contact force because their acceleration is caused by an external force

Back to the original question

In my opinion, the contact force, or force exerted by $m_1$ on $m_2$ must be zero, as they are inherently accelerating and the force is not acting on a particular object as in case 2. Is this reasoning right? What will be the contact force in Question 1?

EDIT

@garyp and @Vladimir : I forgot to mention, $a_1$ in the second question is acceleration of $m_1$ when $m_2$ is absent. When $m_2$ is present, $a_{1}=a_{net}$ as you both said.

@christian and @garyp : In question 1 the external force is not on a particular body as in question 2. Consider two bodies in question 1 to be in free-fall. Force on $m_2$ is $m_{2}a$ and force on $m_1$ is $m_{1}a$. But the force exerted on $m_2$ by $m_1$ must be zero, as the force $m_{1}a$ is not contributing to the acceleration of $m_2$, nor is $m_2$ causing any deceleration in $m_1$

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  • $\begingroup$ Downvoters - Please tell the reason for down voting so that I can modify the question. Give reason $\endgroup$ – user49111 Nov 2 '14 at 10:29
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    $\begingroup$ Thanks for the edit. With those changes you can ignore all of my comments. $\endgroup$ – garyp Nov 2 '14 at 14:09
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Q1 does not allow for a conclusive answer. Without knowing whether the force pushes $m_1$ or pulls $m_2$, no statement regarding the contact force between the two masses can be made. The bodies may even be in free fall, then there would be no contact force.

Assuming that the force acts on $m_1$, the contact force between $m_1$ and $m_2$ is compressive. You have to make a sketch, employing the principle of sections. The overall acceleration is $a=\frac{F}{m_1+m_2}$, the contact force is $m_{2}a$.

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  • $\begingroup$ Yes. There might be an unstated assumption that there is no other force on $m_2$. Some might say that it's a reasonable assumption to make in the context of the problem, but it would be better to say it explicitly. Now I have to make an assumption that the force you mention at the start of the second paragraph is some unstated external force, and that it is the only external force in the situation. $\endgroup$ – garyp Nov 1 '14 at 15:50
  • $\begingroup$ Your conclusion $F_12$ is correct then. Since there is only one acceleration, you should not have $a$ indexed. $\endgroup$ – Rainer Glüge Nov 1 '14 at 16:00
  • $\begingroup$ The force is acting separately on both bodies. It is different from question 2, where force on $m_1$ gets transferred to $m_2$ $\endgroup$ – user49111 Nov 2 '14 at 10:00
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The original question asks about the force on $m_2$. The system is explicitly defined to be object 2. Thus, the contact force is not an internal force.

You are given the acceleration, so the force of $m_1$ on $m_2$ is $m_2a$.

By the way, in your answer you say $a_1=F/m_1$. That is not correct as it does not take into account the force that 1 applies on 2. In fact, since the objects are in contact, $a_1 = a_\mathrm{net}$

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I got the answer. There are two separate cases.

CASE 1

If $m_1$ is exerting a force on $m_2$, which is in turn causing constant acceleration in both the bodies, then $$a_{net}=\frac{F}{m_{1}+m_{2}} ;~~~~~~\therefore F_{12}=\frac{Fm_2}{m_{1}+m_{2}}$$

CASE 2

If the acceleration on both the bodies is caused by an external force separately, as in the case of gravity, (gravitational force on freely falling bodies), then the force exerted on $m_2$ by $m_1$ must be zero, as the force $m_1a$ is not contributing to the acceleration of $m_2$, nor is $m_2$ causing any deceleration in $m_1$

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@imakesmalltalk: You are right, except for $a_1$. It is equal to $a_{\rm{net}}$.

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  • $\begingroup$ No, if there were no force on $m_2$, then $m_2$ would not accelerate. $\endgroup$ – garyp Nov 1 '14 at 15:23
  • $\begingroup$ @garyp - There is no force mentioned in the question. but they both are accelerating constantly. Therefore there is a force acting on the two bodies, but I think, separately, unlike in the case of question 2 $\endgroup$ – user49111 Nov 2 '14 at 2:27

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