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Hey! Can anyone help me out with this one?

Here \(\mu\) is the coefficient of friction. Find the frictional force between the blocks.

Note by Maharnab Mitra
3 years, 3 months ago

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Firstly, there are 2 possibilities for the motion of the 2 blocks :
\(\bullet\) The blocks stick to each other (i.e they do not slide on each other).
\(\bullet\) The blocks slide on each other.

To confirm the possible motion of the blocks, we first assume that the blocks stick to each other. If we encounter a contradiction, we can conclude that the blocks slide on each other.

Using the diagram, let us write down a few known forces on the upper block.

\(\displaystyle N = 2g\cos(30^o) = \sqrt{3}g\)

Therefore,

\(\displaystyle f_{max} = \mu N = \frac{\sqrt{3}}{2} g\)

As assumed, the blocks stay together. So, the net acceleration of the 2-block system is clearly,

\(\displaystyle a_{net} = g\sin(30^o) = \frac{g}{2}\)

Now, note that the acceleration of the individual blocks is the same as the net acceleration (as they are stuck together).

Using Newton's Second Law on the upper block, we get,

\(\displaystyle f + 2g\sin(30^o) = 2\cdot a_{net} = 2\frac{g}{2} = g\)

\(\displaystyle f + g = g\)

\(\displaystyle \Rightarrow \boxed{f=0}\)

Also, note that this doesn't create any contradiction, as the maximum friction is not surpassed by the friction here. In other words,
\(\displaystyle f_{max} > f\) (which must be true always, except that there can also be an equality). Anish Puthuraya · 3 years, 3 months ago

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@Anish Puthuraya Thanks! That's the actual answer. Maharnab Mitra · 3 years, 3 months ago

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0 Shikhar Jaiswal · 3 years, 3 months ago

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0.866 Vaisakh Vrindavan · 3 years, 3 months ago

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@Vaisakh Vrindavan Why? Maharnab Mitra · 3 years, 3 months ago

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