00:01
So in this problem, you're asked to design a safety seesaw that's going to permit people of different masses, maybe wildly different masses, to sit on the seesaw and still maintain rotational equilibrium.
00:17
Now, the seesaw is going to consist of uniform board of mass m equals 8 kilograms and a length of 12 meters.
00:25
And we're going to move the pivot point from the center, a distance d.
00:30
And so we can move that.
00:31
Can adjust wherever that's going to be.
00:34
In the design, i'm going to have each person sitting a distance x offset or x off of 29 centimeters from their end of the plank.
00:45
And for the test case, we're going to have one passenger is going to be a child whose mass is 21 kilograms.
00:51
And the other passenger on the seesaw on the other side will be an adult with a mass of ma, which is equal to n times the mass of the child and then in this case it's going to equal five.
01:03
So the adult is going to have a mass of five times the child.
01:08
Now here's a kind of diagram i've made of the seesaw.
01:13
And what we're going to do in order to solve this problem is we're going to identify all the forces that act to rotate this seesaw.
01:23
And then we're going to set the torque on the left hand side equal to the torque on the right hand side in magnitude.
01:29
Now, the forces that are acting on this are the force of the mass on the left -hand side of the seesaw.
01:38
So i've got the mass on the left -hand side of the seesaw, and it's got a gravitational force.
01:47
It's just disappearing.
01:49
And then i've got the mass of the adult on the left -hand side.
01:53
And then on the right -hand side, i've got the mass of the board, the right -hand side of the board over there.
02:01
And then the mass of the child.
02:04
And now i know that if i just do this, just by using the right -hand rule, the torque in general is equal to here is equal to r cross -f.
02:23
So f in this case for each of those gravitational forces.
02:26
So if i take the cross -product of that, that r, which goes out from the pivot point, out to wherever the mass acts, and i take the cross product of that with the downward gravitational force.
02:43
The torque on the left -hand side is going to point out of the page, and the torque on the right -hand side is going to point into the page.
02:52
So one's going to be positive, and one's going to be negative.
02:55
And the whole thing will balance if the magnitude of the torques, two torques are the same.
02:59
So i'm not going to worry right now about the directions because i know what the directions are.
03:05
I don't care about that.
03:06
I'm really only interested in making sure that the torque on the left hand side is equal in magnitude to the torque on the right hand side.
03:16
So the question, main question is what's the distance d of the pivot point from the center in units and meters? and i don't have the actually shown on this diagram, but you can see, it's pretty easy to see, it should be easy to see that the d, is the distance from the center of the board, which is l over 2, minus l1, right? so if i took the center of the board and i subtracted l1, that's the distance d.
03:47
So that's what i'm going to try and find in all of these, in the calculations we're going to do now.
03:53
Now, the thing i want to do then is first of all to identify what's the mass of the board, just the board itself with any people on it, contain on the left -hand side, on the right hand side.
04:06
So the mass on the left hand side is equal to the ratio of how much of the board is on the left hand side.
04:16
And that's l1 over the entire length times the mass of the board.
04:21
And the mass of the right hand side is equal to whatever fraction of that is on the right hand side, which is l2 over big l times the mass of the board.
04:32
M.
04:34
So now i can calculate what the torque is on the left hand side.
04:38
The torque on the left hand side then is equal to the mass of the adult times d1.
04:46
So that's a distance from the pivot point to the location of the adult position times g.
04:57
Now the mass on the left hand side of the board coming from the board itself actually acts right in the center point, in the center point, between the fulcrum and the end of the board.
05:11
So that acts at this point, right between here and here, basically right about in the middle there.
05:18
And the mass on the right hand side acts right in the middle of this point.
05:22
So it actually acts at the distance from the pivot point on the left hand side of l1 over two.
05:30
And on the right hand side, it's equal to l2 over two.
05:34
So if i'm gonna add now in that, contribution from the mass of the board i have the mass on the left hand side times l1 times g over 2 and then the torque on the right hand side is going to equal the mass of the child times d1 or d2 so you need d2 times g plus the mass of the right hand side times l i forgot that yeah times l2 over 2 so times g okay so now and i can put in now these expressions for the mass on the left hand side and the mass on the right hand side and i get that the torque on the left hand side is equal to the mass v adult times d1 times g plus l1 squared times big m g over two big l and the torque on the right hand side is equal to the mass of the child times d2 times g plus l2 squared big m g over two big l okay so that's good now what we're going to do then is going to take these expressions and we're going to put in some some relationships we have here because we all want to we're going to end up solving this whole problem for l1 and once we solve l1 then we can find d just by subtracting it from big l over 2 so i'm going to put in a bunch of relationships here namely that the mass of the adult is equal to n times the mass of the child now we know that n is equal to 5 but i'm not going to put the number in yet because i don't want 2.
07:45
We also know that l2, distance l2 is equal to the length of the whole board minus l1.
07:53
And d1, the distance d1, is equal to l1 minus this offset value x offset, and d2 is equal to l2 minus x offset and if we put in the fact that l2 is equal to l big l minus l1 then d2 is equal to big l minus l1 minus that offset value so now we've got all of those now i'm gonna you have to remember what those equations look like you can copy them down and now we'll go to the next page because what i'm going to do now is i'm going to write in here what the torque on the left and the right hand side are in terms of all of these parameters which are known the only thing we don't know is l1 so here the torque on the left hand side is equal to n times the mass of the child times l1 minus x offset that's d1 essentially plus l1 squared times big m over 2 big l and all of that times g and now the torque on the right hand sign is equal to the mass of the child times big l minus l1 minus the offset x value plus and now i'm going to put in l2 squared which is big l squared plus plus and now i'm going to put in l2 squared which is big l squared plus l squared plus l 1 squared minus 2 l l1 1, 2 l1, big l, little l1, all of that times m, all of that divided by 2l, and then all of that times g...