This post is linked to this. The post has many answers, but none are satisfactory. So I am reposting it, since the original post became dormant. When does the top block start to slide assuming sufficient friction between the two and zero friction between the bottom block and the ground?
Asked
Active
Viewed 502 times
0
-
3Please explain what exactly is unsatisfactory about the answers in the linked post. It is generally not encouraged to repost old questions. Instead, what you could do is edit it or offer a bounty to draw more attention. – jng224 Mar 22 '21 at 18:24
1 Answers
1
Consider 2 blocks, A on top of B, and you apply a force on block B.
The top block will start to slide relative to the bottom block when the force required to accelerate block A is greater than the maximum static friction. I.e. when $$f_{\rm A \ on \ B}>\mu N$$ where $\mu$ is the static friction coefficient between blocks A and B and $N$ is the normal force.
To solve the problem, assume that A does not slide relative to block B (you can therefore impose the condition that $a_A = a_B = a$). Find the acceleration $a$.* You can then find the required force $f_{\rm B \ on \ A}$ to prevent sliding. Compare that force to $\mu N$. When $f_{\rm B \ on \ A}$ is greater than $\mu N$, there will be sliding.
*the acceleration can be calculated as follows: you know that the net force on B is $m_B a = F-f$ where $F$ is the applied force and $f$ is the friction force required for no sliding between A and B. The net force on block A is $m_Aa = f$. Eliminating $f$ we get $m_Ba=F-m_Aa$.
user256872
- 6,601
- 2
- 12
- 33
-
-
-
-
If F-f=ma and Ma=f, then when f=F, the top blockof mass m does not accelerate but the bottom block does. Isn't that counterintuitive ? – Mar 23 '21 at 03:03
-
-
$f$ will change. $f$ is not constant. the value of $f$ will be such that $0\leq f\leq \mu N$ and change such that there is no sliding. the limiting case is when $f$ needs to be larger than $\mu N$ which is when sliding occurs – user256872 Mar 23 '21 at 03:06
-
you can think of $f$ as being (informally) "whatever it needs to be so there is no slipping, unless $f$ needs to be bigger than $\mu N$ in which case there is slipping (because $f$ can't be as big as it needs to be)" – user256872 Mar 23 '21 at 03:08
-
Consider the case when frictional force balances out the applied force. Why does the bottom block move, since the force is being applied block? More specifically what are the forces on the bottom block when force is applied on the top block? – Mar 23 '21 at 07:20
-
the case (w/ no slipping) where the applied force equals the friction force between the 2 blocks would simply not occur, because the friction force changes as a function of the applied force – user256872 Mar 23 '21 at 16:09
-
Suppose I give a force smaller than the max. force than friction will give. Then I slowly increase the force, frictional force increases. The force when the max friction value is achieved is equal to the applied force. – Mar 23 '21 at 16:47
-
-
-
if you start from the premise that $f$ changes such that the blocks do not slip (and is only limited by $\mu N$), the acceleration of the blocks must always be the same. if that's true, there is no way for $f$ to equal $F$ (unless both are zero), since the blocks would have different accelerations which violates our premise. now suppose we get $F$ high enough such that $f$ is maxed out. when we start to decrease $F$, $f$ will also decrease. in this particular case, there is no way (apart from f=F=0) to make f=F and have no slipping. – user256872 Mar 24 '21 at 05:15
-
-
1But if $F-f=ma$ and $f=Ma'$, $a\neq a'$ then can we exert a force $F=f$? And if we can, does the top block remain at rest while the bottom block moves? Didn't see something like that! – Mar 24 '21 at 08:31
-
I mean is it physically possible to exert a force on the top block, but cause the bottom block to move? – Mar 24 '21 at 10:25