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Suppose you have two blocks, stacked, and the top block, of mass $m_1$, is subject to a force $$\textbf{F} = F_1 \; \textbf{e}_x$$ and there exists friction $u_k$ between the blocks but the lower block is placed upon a frictionless surface. The force on the lower block is given by $$\textbf{F}_f = m_1g\mu_k\; \textbf{e}_x$$

My question is, i'm sure, trivial and one I should know. Why isn't the effective mass of the lower block, $m_2$, the sum the two masses, $m_1$ and $m_2$? That is, why isn't the acceleration of the bottom block given by $$\textbf{a} = \frac{m_1g\mu_k}{m_1+m_2}\; \textbf{e}_x$$ Instead, the actual answer is $$\textbf{a} = \frac{m_1g\mu_k}{m_2}\; \textbf{e}_x$$ which doesn't make intuitive sense to me. This answer implies that the acceleration of the bottom block would be the same even if $F_f$ were not a frictional force but instead applied to the left of the bottom block in the absence of the top block. How can this be?

EDIT

I suppose the title is misleading. In both "answers" $m_1$ is present. My question more has to do with the "effective" mass of the bottom block being $m_2$ or $m_1+m_2$

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    $\begingroup$ What does your $(+\hat{x})$ stand for? Shall this indicate the unit vector in x-direction? $\endgroup$ – Semoi Aug 30 '17 at 19:01
  • $\begingroup$ Yes. My coordinate scheme places +x to the right $\endgroup$ – john morrison Aug 30 '17 at 19:03
  • $\begingroup$ [I've decided that this should be an answer, so I've posted it as such.] Because F=ma for each object individually - where F is the sum of forces ONLY on that object, a is the acceleration ONLY of that object, and m is the mass of that object ALONE. The sum of the masses would come into play with the normal force between the lower block and the table, but there is no friction there so it doesn't affect the motion. $\endgroup$ – Geoffrey Aug 30 '17 at 19:46
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The reason you get the same result as if $F_f$ were simply applied to the bottom block is because $F_f$ is a force applied to the bottom block. Whether it's applied to the top or the side is not relevant.

The mass of the top block will have an effect on the acceleration of the top block which you are assuming is greater than the acceleration of the bottom block. If it weren't and the blocks accelerate at the same rate you wouldn't have kinetic friction anymore and the force on the lower block would be less than $m_1g\mu_k$.

If the magnitude of the force applied on the top block, $m_1$, is less than $$F_{app} = (\frac{m_1}{m_2}+1)F_f$$ the two masses will move together with an acceleration of $$a=\frac{F_{app}}{m_1+m_2}$$ But, for any larger force applied on the top block it just starts to slip across the surface of the bottom block. A larger force will make it slip faster but the force of friction doesn't depend on how fast it's slipping.

Because the bottom block is resting on a frictionless surface increasing the mass of the upper block doesn't increase a restrictive force of friction on the bottom block.

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  • $\begingroup$ My question relates more to why one doesn't divide by the sum of the masses. The reason I bring up the force acting on the side of the bottom block in the absence of the top is because in this case, there is no block to "weigh down" the bottom one. $\endgroup$ – john morrison Aug 30 '17 at 19:20
  • $\begingroup$ I'll edit my answer to expand the explanation a bit... $\endgroup$ – M. Enns Aug 30 '17 at 19:36
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    $\begingroup$ @johnmorrison One doesn't divide by the sum of the masses because you have to ONLY consider the mass of the object ON WHICH the force is being applied. Dividing by the sum of the masses in this case implies that the friction force is acting somehow on a large solid block comprised of the two smaller ones which is incorrect. $\endgroup$ – Ruslan Mushkaev Aug 30 '17 at 20:27
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  • The origin of the force acting on the bottom block doesn't matter. It could be either a spring or what so ever. In your example it is the friction force $F_f = m_1 g \mu_k$.
  • If a force $F$ acts on a body of mass $m_2$, the mass accelerates by $a_2 = F/m_2$.
  • The force is acting to the right, not to the left. Just place two objects on top of each another and do the experiment. In order to have a "frictionless" surface, you could pull on the lower object. Then observe in which direction the upper object is moving. Try different velocities.
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  • $\begingroup$ My question is not related to the origin of the force; see my comment to M. Enns. My question is entirely placed in the denominator of the expression for the acceleration. I understand that F=ma. Given a force, divide by the mass and you obtain the acceleration. Why, however, doesn't the mass of the upper block "add to" the mass of the lower? $\endgroup$ – john morrison Aug 30 '17 at 19:24
  • $\begingroup$ Why should it? As I tried to point out in my answer, you could just flipp the system upside-down: Pull on the lower mass and watch the upper mass. In this situation you probably would not assume $m_1 + m_2$ in the denominator, do you? The top mass in your origin question is only used to generate some friction force. This friction force is then used to accel the bottom mass. $\endgroup$ – Semoi Aug 30 '17 at 19:43
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The equation for motion for the upper block will be given as,

$m_1 a_1=F-F_f$

And that for the lower block will be given as,

$m_2 a_2 =F_f$.

$a_2=\frac{\mu_{k}m_1g}{m_2}$

However this is the case when there is slipping between the two blocks and $F_f$ is at its maxima.

But, when the block move together i.e, $a_1 = a_2 =a$, you will have totally different equations of motion.

$m_1 a=F-F_f$

$m_2 a =F_f$

$a=\frac{F}{m_1+m_2}$ $\tag 1$

Lets take a look at the extreme case when there is no sliding between the two blocks.

Before I start forming the equations, it should be understood that the maximum value of frictional force, in the given condition, is $\mu m_1g$.

Depending on the value of $F$ : the boxes could either stick together and hence, move with same acceleration or slip over each other and have different values of acceleration.

In the extreme case, when the blocks just have the tendency to slip, the frictional force will be at its maxima. Friction force is self adjusting and it tries to be as higher (in magnitude) as it can be, depending on the values of F.

In the extreme case, acceleration of the blocks is the same and frictional force is at its maxima, and $F_{max}$ is that maximum value of $F$ for which no sliding occurs.

$m_1 a = F_{max}-F_f$ $\tag2$

$m_2 a=F_f$ $\tag 3$

$a=\frac{F_{max}}{m_1 + m_2}$ $\tag4$

From the equations $(3)$ and $(4)$ ,

$F_f=\frac{m_2.F_{max}}{m_1 + m_2}$,

$F_{max}=\mu \frac{m_1}{m_2} (m_1+m_2)g $ $\tag 5$

If the value of $ F > F_{max}$, the blocks will slip and have different values of acceleration. For any value of $ F (> F_{max})$ frictional force will be maximum i.e, $\mu m_1g$.

However if $ F < F_{max}$ the blocks will always move together no matter the value of $F$. The value of $F_f$ will shift accordingly in such a way that both blocks move with same acceleration. That's the beauty of frictional force !

Conclusion -

The term for acceleration depends on the value of $F$, the mass of the upper block and the value of $\mu$.

Since you have used $\mu_{k}$ in your equation, the blocks should be sliding over each other. Therefore, they cannot have same acceleration. That's also the reason why you won't get the value of acceleration given by equation $(1)$.

Please note -

For making the equations of motion I have considered constant value of $\mu$. To be specific when there is no tendency between the block to slip the nature of frictional force will be static, otherwise kinetic.

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  • $\begingroup$ @johnmorrison , I was late to answer your question but this answer may clear up some things. $\endgroup$ – Mitchell Aug 30 '17 at 20:11
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You are only dividing by the mass of the lower block because $F=ma$ for each object individually - where $F$ is the sum of forces ONLY on that object, $a$ is the acceleration ONLY of that object, and $m$ is the mass of that object ALONE. The sum of the masses would come into play with the normal force between the lower block and the table, but there is no friction there so it doesn't affect the motion in this problem.

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enter image description here

The picture has a force diagram for one of your masses $m_2$ The magnitude of acceleration in the $+x$ direction is indicated by $a$.

Forces on $m_2$ in $+x$ direction,

$m_2a =N\mu$

Since, the system is at equilibrium along the vertical,

$N= m_1g$

Now, the mass of the top block does have an effect on the net acceleration. The normal force exerted by the top block on the bottom block affects the force of friction in between the surfaces. The frictional force is why the top block would try to "drag" the bottom block to the right.

The top block does have an effect on the bottom block's acceleration. If it wasn't for the friction the top block wouldn't drag the bottom block in he positive $x$ direction.

The acceleration of the bottom block in the $+x$ direction will be the sum of all the forces in that direction times the reciprocal of the mass of the block. $a=\frac{m_1g\mu}{m_2}$.

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protected by Qmechanic Aug 30 '17 at 19:58

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