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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestalt text

LINK:hereFirstly, there are

2possibilities 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

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

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

individualblocks is the same as the net acceleration (as they are stuck together).Using Newton's Second Law on the

upperblock, 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 frictionis 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 · 2 years, 11 months ago

Log in to reply

– Maharnab Mitra · 2 years, 11 months ago

Thanks! That's the actual answer.Log in to reply

0 – Shikhar Jaiswal · 2 years, 11 months ago

Log in to reply

0.866 – Vaisakh Vrindavan · 2 years, 11 months ago

Log in to reply

– Maharnab Mitra · 2 years, 11 months ago

Why?Log in to reply