00:01
Let us begin with subpart a.
00:04
So here the first brick is in equilibrium.
00:08
So here brick number one is in equilibrium.
00:14
So the brick length is given as l.
00:18
L.
00:19
L hence the brick will tilt or fall if not at this equilibrium position.
00:28
So here we can write a1 max equals half into l because the center of gravity lies to the right of l by 2.
00:41
So we can write a1 max equals to l by 2.
00:48
This is the answer for subpart a.
00:53
Now moving to subpart b, that is for brick number 2.
01:01
Here also for the same reason that is the brick is in equilibrium hence we can write a 2 max equals half of a 1 max that is a 1 max is calculated as l by 2 before we get l by 4 or a 2 max is equal to l by 4.
01:34
This is the answer for subart b.
01:40
Moving to subart c, that is for brick number 3.
01:46
So here we can apply the equation for center of mass for the third brick, that is, we can write a3 equals m into minus l by 4 plus m into l by 4 plus m into l by 2.
02:08
This divided by the total mass up to third brick, that is 3n...