A block of mass `M= 4 kg` of height and breath `b` is placed on a rough plank of same mass `M`. A light inextensible string is connected to the upper end of the block and passed through a light smooth pulley as shown in figure. A mass `m =1 kg` is hung to the other end of the string.
a. What should be the minimum value of coefficient of friction between the block and the plank so that, there is no slipping between the block and the wedge?
b. Find the minimum value of `b//h` so that the block does not topple over the plank, friction is absent between the plank and the ground.

Answer

`4//9`

Explanation

Drawing free body diagram of all the three (blocks plank and mass)
a. Since we are taking the limiting case.
Therefore `f=` maximum friction between the block and the plank.
`f=muN=muMg=mu(4)g`
`f=4mug` ............i
Now writing equan of motion for all three
`Mg-Tma(m=1kg)` ........ii
or `g-T=a`
`T-f=Ma(M=4kg)`
or `T-4mug=4a`...........iii
or `f=Ma` or `f=4a` or `4mug=4a`
or `a=mug`
Solving eqn ii, iii and iv for `mu` we get `mu=1/9`
Therefore minnimum value of `mu` should be `1/9` or `muge1/9`
b. At the time or toplling, normal reaction `N` will pass through the front edge as shown below:
So, torque of `T` about `Ple`torque of `Mg` about `P`
or `Thle(Mg)(b/2)` or `Tle((Mg)/2)(b/h)`
From eqn ii, iii and iv we can find that
`T=8/9g`

Therefore `8/9gle((Mg)/2)(b/h)`
or `bhge16/(9M)(M=4kg)`
or `(b/h)ge4/9`
Therefore minimum value of `b/h` so that the block does not topple is `4/9`