Hey! Can anyone help me out with this one?

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

#Physics #Mechanics

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Comments

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LINK: here

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 - 7 years, 4 months ago

Thanks! That's the actual answer.

Maharnab Mitra - 7 years, 4 months ago

0.866

Vaisakh Vrindavan - 7 years, 4 months ago

Why?

Maharnab Mitra - 7 years, 4 months ago

Shikhar Jaiswal - 7 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$