# 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
4 years, 4 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

alt text

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).

- 4 years, 4 months ago

Thanks! That's the actual answer.

- 4 years, 4 months ago

0

- 4 years, 3 months ago

0.866

- 4 years, 4 months ago

Why?

- 4 years, 4 months ago