# Why doesn't the static frictional force between two blocks, placed on top of one another, cancel out the applied force?

After having attended a specific mechanics course I've struggled with a certain problem, which is related to static friction. First of all, if some force is applied to a block (the source of the applied force is not of importance in this scenario), that's placed on top of another block, it's going to accelerate in case there's no static friction to cancel the applied force. However, as the left picture below suggests, there's a static frictional force between the two blocks, and for simplicity no frictional force between the lower block and the ground. Now the static friction will always be equal in magnitude to the applied force but in the opposite direction, at least to a certain level. As can be seen, $$F_1$$ compensates for $$F_{app}$$, at the same time, Newton's third law states that each applied force has an equal and opposite force which acts on the other object. Hence, the static frictional force $$F_2$$ is applied on the lower block, so I would say that the net force on the above block will become 0N and the lower block will experience a net force of $$F_2$$. As a result the above block will remain stationary and the lower block is going to accelerate, but this final result doesn't seem to make sense to me: the upper block remaining still why the lower block is going to accelerate. I must miss some crucial point about the static frictional force, but what? Thanks. The key mistake you made is in the following statement:

As can be seen, $$F_1$$ compensates for $$F_{app}$$, at the same time, Newton's third law states that each applied force has an equal and opposite force which acts on the other object. Hence, the static frictional force $$F_2$$ is applied on the lower block, so I would say that the net force on the above block will become 0N and the lower block will experience a net force of $$F_2$$

The friction force $$F_1$$ does not cancel out $$F_{applied}$$ for a net force of zero on the upper block. Assuming the maximum possible static friction force is not exceeded, both blocks will stick together and have the same acceleration with respect to the ground.

As my undergraduate mechanics professor used to say over and over again, "Draw a free body diagram!!!". A free body diagram (FBD) should show all the external forces acting on the body.

Fig 1 below is a FBD of the combined top and bottom blocks. It assumes (1) no friction between the lower block Y and the ground and (2) that the maximum possible static friction force between the blocks is not exceeded so that the blocks stick together. Based on the FBD the only net external force acting on the combined blocks is $$F_{applied}$$. Therefore, applying Newton's second the acceleration $$a$$ of the combined blocks with respect to the ground is

$$a=\frac{F_{applied}}{M_{X}+M_{Y}}$$

Fig 2 are FBDs for the individual blocks. For each block the equal and opposite friction force that each exerts upon the other per Newton's third law acts as an external force. Since the blocks move together, each block has the same acceleration $$a$$ as the combined blocks of Fig 1.

From the FBD on the upper block applying Newton's second law we have

$$F_{Xnet}=F_{applied}-f=M_{X}a=\frac{M_{X}}{M_{X}+M_{Y}}F_{applied}$$

Solving the equation for the friction force $$f$$

$$f=\frac{M_Y}{M_{X}+M_Y}F_{applied}$$

From the FBD on the lower block we see that the only external force acting on the block is the friction force of the upper block. Therefore applying Newton's second law to the lower block

$$F_{Ynet}=f=M_{Y}a=\frac{M_{Y}}{M_{X}+M_{Y}}F_{applied}$$

The values of the static friction force is the same, as it should be.

But does that mean my sketched situation is physically impossible? So if 𝐹𝑎𝑝𝑝𝑙𝑖𝑒𝑑=𝑓𝑚𝑎𝑥−𝑠𝑡𝑎𝑡𝑖𝑐 I would say that 𝐹𝑋,𝑛𝑒𝑡=0 and 𝐹𝑌,𝑛𝑒𝑡=𝐹𝑚𝑎𝑥−𝑠𝑡𝑎𝑡𝑖𝑐 because of Newton's 3rd law.

I think I now understand the root cause of your issue. The static friction force is a variable force that matches the applied force up to the maximum static friction force. They do in fact cancel each other but only with respect to the acceleration of the top block relative to the bottom block, but not with respect to the acceleration of the top block relative to the ground.

Up to and until the applied force equals the maximum static friction force when slipping is impending, there is no slippage and the acceleration of the top block is the same as the bottom. But once slipping occurs, the friction force transitions from static to kinetic friction and the upper block moves relative to the lower block. The coefficient of kinetic friction (and thus the kinetic friction force) is generally less than that of the static friction force. At that point the applied force will exceed the kinetic friction force for a net force that causes the upper block to accelerate with respect to the lower block, as well as with respect to the ground.

The acceleration of the lower block with respect to the ground will be dictated by the kinetic friction force, which is generally considered constant regardless of the applied force.

Hope this helps. • Thanks Bob for this clear and comprehensive elaboration. Now what I seem to understand from your explanation is that you first have to assume that $M_Y$ is going to get the same acceleration as the $M_X$, which is due to $F_{applied}$. When the static friction, needed to accelerate $M_Y$, does not exceed the maximum static friction then $M_X$ and $M_Y$ will have the same acceleration. But does that mean my sketched situation is physically impossible? So if $F_{applied}= f_{max-static}$ I would say that $F_X, net = 0$ and $F_Y, net = F_{max-static}$ because of Newton's 3rd law. Dec 13, 2020 at 21:02
• @Jelle3.0 Everything up to "But.." is correct as long as the maximum static friction force is not exceeded. Please clarify what you mean by $f_{max-static}$. Is $max-static$ a subscript of $f$? Dec 13, 2020 at 21:57
• Exactly, I mean by $F_{max-static}$ the maximum static friction possible between two surfaces. It just seems a little odd that the friction doesn't behave the way as expected. Is there perhaps a reason for why the situation I described earlier doesn't arise? Thanks in advance. Dec 13, 2020 at 22:43
• @Jelle3.0 Got it. See update to my answer. Dec 13, 2020 at 22:56
• I read your updated answer and I think I finally understand how to go about such a problem! Though, as you mention at the start "The friction force $F_1$ does not cancel out $F_{applied}$ for a net force of zero on the upper block. Assuming the maximum possible static friction force is not exceeded, both blocks will stick together and have the same acceleration with respect to the ground." why do we assume this? Is that just how static friction works, from empirical data we know that the 2 objects will always stick together unless the maximum static friction is exceeded? Thanks for the effort. Dec 14, 2020 at 8:13

In this case the static friction is not equal to the force F. You can conclude this only when the body has zero acceleration. As you have no friction at the bottom, both blocks will start to move with some acceleration. You just have to draw free body diagram for each body and apply Newton's second law for each body to find the static friction and the common acceleration. You can also find at what value of F the two bodies start to move relative to each other and the friction becomes kinetic.

• Why doesn't the static friction equals the applied force? I mean it is similar to a block that's on the ground, as long as your applied force is lower than the maximum static force, the block will always remain at rest. Now in this situation it seems to me you can just keep the top block still while the lower block starts accelerating and slippage probably occurs. This doesn't seem intuitive, instead the both blocks (assuming both have the same mass) will accelerate, could this be due to the lower kinetic friction that arises when the lower block start slipping? Dec 12, 2020 at 9:57
• It's not similar in the fact that in the first case the block is at rest in respect to the ground and in the second it is accelerating in respect to the ground. If the two forces were equal, what accelerates the block on top?
– nasu
Dec 12, 2020 at 16:48
• That's exactly the point I was confused about, the top block stays stationary because the static friction is equals to the applied force. But because Newton's 3rd law applies the static friction also applies a force to the lower block, and therefore the lower block starts moving relative to the ground and the top block. Of course, we are assuming the friction between the ground and the lower block is zero. So what will actually happen? Dec 13, 2020 at 17:16
• Both blocks will move with the same acceleration. The applied force is not equal to the static friction in this case.
– nasu
Dec 13, 2020 at 19:41

Now the static friction will always be equal in magnitude to the applied force but in the opposite direction, at least to a certain level.

This is where you're running into trouble. There's no particular requirement that the static frictional force be any particular value, so long as it doesn't exceed $$\mu_s$$ times the normal force. Your assumption is true if the object is at rest; but in this case, the upper block is accelerating, and so the net force on it will not be zero.

A better way to think about static friction problems is to assume that the static friction is less than or equal to its critical value, and then see what constraints this imposes on the situation. If the constraints are violated, then we know that there will be slippage.

In this case, we have from the two free-body diagrams $$F_\text{app} - F_\text{fr} = m_1 a \qquad \text{ and } \qquad F_\text{fr} = m_2 a,$$ where $$m_1$$ and $$m_2$$ are the masses of the top & bottom blocks, respectively. Implicitly, we have assumed that the accelerations of the two blocks are equal. We also know from looking at the vertical forces that the normal force exerted between the two blocks is $$m_1 g$$, so $$F_\text{fr} \leq \mu_s m_1 g$$. This inequality lets us (in principle) place bounds on the value of $$a$$ and/or $$F_\text{app}$$ if we want the blocks to move without slipping; intuitively, it should make some sense that such bounds exist. But more importantly for your question, you can see that $$F_\text{fr} \neq F_\text{app}$$ unless $$m_1 a = 0$$.

• Indeed, I luckily comprehend what you said, regarding the assumption. Although at last you stated that $F_{fr}≠F_{app}$ unless $m_1 a =0$, now why can't that be true? Is this simply how the static friction works, like you can't have the maximum static friction cancel out the applied force and cause the lower block to accelerate due to Newton's 3rd law, you understand what I mean? Dec 14, 2020 at 8:06
• @Jelle3.0: I actually don't understand what you mean. Can you rephrase your question? Dec 14, 2020 at 12:56
• thanks, but after some thinking I've resolved the problem and finally understand it. Dec 14, 2020 at 22:18

You are correct. If your applied force is equal to maximum value of static friction, then lower block will accelerate while the upper block will remain stationary.Thus slipping will occur.

The reason you might feel strange is because in real life the friction between lower block and ground prevent this from happening.

• I suppose you assume that the kinetic friction is the same as the maximum value of the static friction. Usually this not the case.
– nasu
Dec 11, 2020 at 22:11
• @nasu Yes. I was going to mention that but then decided not too... Dec 11, 2020 at 22:12
• The general case would be that the two bodies move with different accelerations. Why mention a very special case and ignore the general one?
– nasu
Dec 11, 2020 at 22:15
• I fail to see how applying a force to the upper block could cause the lower block to accelerate while the upper block remains stationary. Dec 11, 2020 at 22:17
• @nasu I will edit my answer. Generally unless given in the highschool problem coef of static friction is assumed to be equal. But yes you are theoretically correct . The lower block will have some acceleration as $\mu_s>\mu_k$. Dec 11, 2020 at 22:19