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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, 8 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).

- 3 years, 8 months ago

- 3 years, 8 months ago

0

- 3 years, 7 months ago

0.866

- 3 years, 8 months ago

Why?

- 3 years, 8 months ago