Question

How to do it?

Answer 1

(A) -- (r)
(B) -- (q)
(C) -- (p)
(D) -- (s)

Explanation:

The explanation below is a bit detailed and lengthy - but most of it can be easily worked out in one's head. Since friction, with its limiting values, is often confusing to students, I have decided to err on the side of verbosity.

The friction between the two blocks will try to make them fall with the same acceleration, if possible. The strategy we shall adopt is to assume that the two blocks fall with the same acceleration, - and check whether it is possible for the friction between the two blocks to supply the force that will be needed for this. If not, we start over.

Since there is no horizontal movement the normal force exerted by the wall on the left block, and the two blocks on each other are each equal to 10 N.

The weights of the two blocks are 10 N each.

Unless the limiting value for the force of this friction were to exceed `20" N"` (not posible in any of the four cases here) , the combined weight of the blocks, they will always move with respect to the wall - so the friction force exerted by the wall is kinetic friction. We neglect the small difference between the coefficients of static and kinetic friction.

Case (A)

Since `mu_1=0` there is no friction between the wall and the left hand block. So, the net vertical force on the two block system is 20 N (the combined weight). Thus, the acceleration of the two block system, if they move together, is `10" m"*"s"^-2`.

For this to be achieved the force of friction needed between the two blocks is zero. Since zero is well withing the value of limiting friction (sic) , this is what will happen!

So, the accelerations are

`color(red)(a_A=a_B=10" m"*"s"^-2)`

Case (B)

The net vertical force on the left block is `10" N"-1" N" = 9" N"`. So, it will accelerate down at `9" ms"^-2`. On the other hand, the only vertical force acting on the right block is its weight, so it will accelerate downwards at `10" ms"^-2`.

`color(red)(a_A=9" m"*"s"^-2, a_B=10" m"*"s"^-2)`

In this case, there being no friction between the blocks, the case was simple - and we had no need to adopt the general strategy.

Case (C)

The force of friction exerted by the wall is

`mu_1N=0.1times 10" N" = 1" N"`

This would mean that the two blocks, if they were to move together, would accelerate down at the rate

`(20" N"-1" N")/(2" Kg") = 9.5" m"*"s"^-2`

For this to happen, the net downward force on the right hand block must be `9.5" N"`, which in turn will be possible if there is a force of friction equal to `0.5" N"` exerted by the left block on the right. This is well within the limiting value of friction between the surfaces, which is `10" N"` in this case. So:

`color(red)(a_A=a_B=9.5" m"*"s"^-2)`

Case (D)

The force of friction exerted by the wall is

`mu_1N=1.0times 10" N" = 10" N"`

This would mean that the two blocks, if they were to move together, would accelerate down at the rate

`(20" N"-10" N")/(2" Kg") = 5" m"*"s"^-2`

For this to happen, the net downward force on the right hand block must be `5" N"`, which in turn will be possible if there is a force of friction equal to `5" N"` exerted by the left block on the right. This is beyond the limiting value of friction between the surfaces, which is `1" N"` in this case.

So, here, the blocks will slip with respect to each other, and so the friction between them will be the limiting value. Since obviously the block on the right will move down faster, the force of friction the left hand block exerts on it will be upwards. This means that the net force on the right hand block is `10" N"-1" N" = 9" N"`, leading to an acceleration of `9" ms"^-2` .

The left hand block will have a net downward force of

`10 " N"-10" N"+1 "N" = 1" N"`

and so its acceleration will be `1 "ms"^-2`. Thus

`color(red)(a_A=1" m"*"s"^-2, a_B=9" m"*"s"^-2)`