00:02
So in this question we have been given that a rectangular raft is floating on a lake and the top surface of the raft is a, which is this top surface.
00:15
And we are told that the density of the material of the raft is given to us, row r, 650 kg per meter cube.
00:23
The top area is given.
00:24
The volume of the raft is also given in meter cube.
00:26
And we are also given the density of the water of the lake, which is 1 ,000 kages per meter cube.
00:30
So we are told that initially this raft is floating.
00:34
So in that condition we are first asked to draw the free body diagram of this raft and show the forces on the raft.
00:41
So let's do that first.
00:43
So let's say that this is the raft and let's say this is the water level and which it is.
00:51
So there will be a mass of the raft times gravity weight of the graft acting down on the raft.
01:00
And there will be a point force.
01:03
Acting on top which balances it because of which the body does not sink or does not rise it just floats because these two forces are in balance the point force will be the force which acts on the raft because of the water that is displayed due to archimedes principle so these are the two forces that act on the raft this is your part a of the question then part b asks us to find this point force so point force actually has a formula but we will use that later first part is easy that point force in this case if we just equate these two forces is equal to mass of the raft times gravity and the mass of the raft is nothing but density of the raft times volume of the raft times gravity and we know all these values so let's put them in 650 times 1 .8 meter cube times 9 .81 and this gives us that the point force is 1147 .7 newton so this is the answer for your part b.
02:06
This is the point force.
02:09
Alright so the next part then asks us to find the the length of the side of the raft that is above the water.
02:19
So if you see the original diagram this h is what we are trying to find.
02:23
So let us call this length of the side of the raft as l such that the volume of the raft becomes the area times the length.
02:34
The cross -section area or the face area times the length so that becomes the volume of the raft and h is what we're looking for h which is above the water this part of the length which is about the water so now let's use the archbidispincipal point force formula which says that fb is row of water times g times volume displaced which is the volume of water that is displaced so in this case the volume of water that is displaced or let's just rearrange this formula a little bit the volume of water displaced will be equal to fb by row w by g and this volume of water displaced will be equal to a times l minus h because if you see the volume water displaced is basically this volume down here and if the uh volume of the entire raft is a times l the volume of this portion is so this portion basically becomes equal to l minus h if this is h and that side is l so that's the volume we're talking about so now that we have this we can again rearrange this formula a little bit and from here we get that h is equal to l minus f b by r w g and a and we need to further write this because we don't know l so we can put this in terms of from the equation on top that this becomes equal to vr by a, volume by area of cross section, minus fb by row wga...