00:01
In this question, we have this block that sits on the top of identical block of mass m.
00:06
And the lower block is sit on the rough table.
00:11
And they have given that the friction between these two blocks is muon and the friction between this lower block and the surface is mu two.
00:20
So we have to calculate the minimum force to be applied on the top block to produce motion in the system of the block.
00:30
We are applying this force f on the top of the block.
00:35
Now we have to find the minimum force, right? so we can say that in this direction, due to this coefficient of friction, there will be a frictional force that i am representing with fs, right? and there will be a frictional force in lower block also that i am representing with fl.
00:55
So if i find the fs, that is the frictional force in upper block, that will be equal.
01:00
To mu 1 multiply by n that is m g now they have given that the mass is equal to 5 kilogram and the value of mu 1 is equal to 0 .6 now if i put this values in this equation so i will get 0 .6 multiply by mass that is 5 multiply by g that we have to take 10.
01:28
So the fs will be equal to 30 newton.
01:35
This is not the maximum or i can say the minimum force.
01:39
We have to find another force that is the frictional force in lower block...