# Friction, Dynamics question In the given question $F=2t$ where $t$ is time in seconds for which the force has been applied. Friction exists only between $A$ and $B$ with coefficient of friction $\mu=1/2$. Here we can easily see that the $f_{max}=\mu N$=$10N$ where $f_{max}$ stands for the maximum value of friction, hence therefore the acceleration that can be given to the block B would be $$\frac{f_{max}}{4} =2.5m/s^2,$$ hence both the blocks would move together till $$\frac{F}{m_A + m_B}=2.5$$=$$\frac{2t}{6}=2.5$$ hence the time comes out to be equal to 7.5 seconds and till $t=7.5 sec$ they move together.

But I am myself confused in proving the last line that till $t=7.5 sec$ they move together. For example lets consider time $t=1sec$. At the taken time $F=2N$.

Now when $F=2N$, the a friction of $2N$ would also act in the opposite direction on $A$ as $f_{max}=10N$. From Newton's third law an opposite reaction of frictional force would act on $B$ and $B$ would start moving ahead with an acceleration of $0.5m/s^2$ Hence now the block B has started motion but $A$ according to equations of dynamics is at rest which is actually not possible as till $7.5 sec$ the must move together. Please tell me where I am wrong?

## 2 Answers

Remember that static friction is a constraint force: it enforces the rule "no motion between these surfaces" as long as the force needed to do so does not exceed the maximum.

The force of static friction will have the value needed to prevent relative motion unless that value exceeds the maximum.

So there are two possibilities to what happens here.

1. Friction prevents relative motion and the blocks move as a unit.

2. Friction is insufficient to prevent relative movement and A moves faster than B eventually falling off the edge of the lower block. If there is a different coefficient of kinetic friction we also switch to using it for these cases.

In case (1) the acceleartion at $t = 1 \,\mathrm{s}$ would be $a = (2\,\mathrm{n})/(6\,\mathrm{kg}) = \frac{1}{3}\,\mathrm{m/s^2}$. This will require an actual frictional force of $f = m_A a = (4\,\mathrm{kg})(1/3\,\mathrm{m/s^2}) = 4/3\,\mathrm{N}$ which is lower than the maximum and so is allowed. Case (1) is physically correct at $1\,\mathrm{s}$.

By contrast if we consider the situation at $t = 8 \,\mathrm{s}$ we find $a = 8/3\,\mathrm{m/s^2}$ and $f = 32/3\,\mathrm{N} > 10\,\mathrm{N}$ which violates the upper limit on the frictional force and the problem must be handled as case (2).

• Sorry @dmckee because I post irrevelant comment here, but I wouldn't like to come to h-bar chatroom because of some personal reasons, so I had no ther way. Will you yourself undelete my answer? Or should I remind you? Because you have deleted two answers of mine before this and haven't undeleted them until now! Sorry again. Thank you very much! – lucas Jun 25 '16 at 21:31
• Flag the post when a couple of weeks have passed. Use the custom flag reason and ask to have it undeleted. – dmckee Jun 25 '16 at 21:35
• Thank you. But what about those two answers? I don't know where are they to flag? – lucas Jun 25 '16 at 21:36
• The data explorer is your friend: data.stackexchange.com/physics/query/263141/… Er ... the data explorer should be your friend but that link doesn't seem to be doing the right thing just now. – dmckee Jun 25 '16 at 21:41
• I couldn't do anything with the link recommended by you. Have you seen the answer given to this post? – lucas Jun 30 '16 at 1:32

When $F=2N$, the a friction of $2N$ would also act in the opposite direction on $A$

This is wrong. Friction force isn't equal to force $F$. It is equal to $F_f=m_Ba_B$

Free body diagrams of blocks are as below: When $F=2\;\mathrm N$ then we have:

$$F-F_f=m_Aa\Longrightarrow2-F_f=2a$$ $$F_f=m_Ba\Longrightarrow F_f=4a$$ Hence $$a=\frac 13\;\mathrm{m/s^2}\Longrightarrow F_f=\frac 43\mathrm N$$