00:01
So in this problem, we want to find some expressions for describing this situation.
00:05
And we have a block supported by some tension.
00:10
And it is partially submerged with a volume submerged of v sub, a total mass of the volume, a mass of m volume of v.
00:20
We have this water or any liquid that is being subversion, which density row.
00:27
So firstly, we need to draw the free body diagram.
00:32
So represent the object as a point.
00:36
Firstly, we have its weight, which we know to be m times g.
00:43
We have the tension that is applied upwards.
00:47
And we also have the buoyant force, which we'll call b, applied upwards.
00:55
So that is the first part, and we'll use this to continue developing the problem.
01:03
So in part two, we want to find an equation.
01:09
We want to find an equation for the net forces.
01:13
This body is at equilibrium.
01:17
As it's saying, it's moving.
01:20
We'll say that it's equilibrium because it's stationary at this current position, although it's moved in between trials.
01:26
So because it's at equilibrium, the sum of forces is equal to zero.
01:32
We can pick that, for example, this is the y direction, positive and so it becomes the sum of forces in the y is equal to zero so we have buoyancy and tension in the positive direction and the weight m g in the negative direction is equal to zero the buoyancy by archimedes principle is equal to the density of the fluid the volume displaced the sub and the gravity xcelerous unit to gravity g and so if we put that in, we'll v sub g plus t minus m g is equal to zero.
02:19
And so that's the force at any point on this block.
02:25
In part three, we want to find the graph of t versus v sub.
02:34
Where we want to find the equation of it.
02:37
So this is given as y versus x.
02:41
And so you would see the t and v.
02:45
Sub...