Question

2500 WATER d 5 m 60°

          2500
WATER
d
5 m
60°
        
2500
WATER
d
5 m
60°

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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The width of the rectangular cover, which makes an angle of 60° with the horizontal, is 2 m perpendicular to the plane of the page, and the 2500 kg mass connected to it passes over a roller and stands as shown in the figure. At what depth d of water can there be a state of equilibrium, as shown in the figure? Show by calculating (p=1000 kg/m³). - Neglect the mass of the lid. The moment of inertia of the lid about the axis passing through the center of mass can be calculated as l-b²l³/12 (b: width (2m), l: height). For any water height value d, the suspended mass does not come into contact with water. 2500 V 5 m d WATER 60
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Transcript

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00:02 On this question, you're giving a symmetrical cylinder, right? so you have a cylinder with water, right? so the situation is you have a cylinder with water.
00:18 So suppose, let me try to draw the cylinder, a cylinder like this, right? you have a cylinder.
00:24 Oh, it's not well -joy.
00:27 So it's going to go down like this.
00:30 It's good.
00:32 And so there's a sininder.
00:33 And in the cylinder, you initially have full water.
00:36 Now, you drill a little bit of hole here at the bottom, and so what is going to, you know, coming out, right? so what is going to come out? come out.
00:48 So the water level is going to drop, right? for example, after some time, the water level will drop to maybe something like here, x, right? so you're asking about the center of mass of the can plus the water, right? so initially, obviously, the center of mass is just at the center, right? it must be somewhere here, just at the center of the cylinder, right? so initially it must be like that.
01:15 But afterwards, when the water drops, and of course, the center of mass is going to drop as well, right? so you can work out an equation that gives you the center of mass.
01:25 Let's suppose the center of mass, i call it cap to x a bar, right? and this, of course, given by the total mass of the cylinder, which is given by m, right? and multiplied by the position of the center of the mass of the cylinder, which of course, i call it for simplicity, let's call it to be, suppose the cylinder has a mass, has a height which i call it, let's call it h, right? and then the, and we call the direction is going up, right? x direction, right? and this is the bottom, and we call it x equals zero, right? so, and the, the, the sense, of the center of the center of the center of the cylinder, of course, is giving is half a hitch, right? and plus the mass of water, which i will call m, which of course is a function of x, right, and times the center of mass of the water cylinder, which of course is x over two, right? as you could see, and divided by the total mass m plus m x, right? that would be what you'll find.
02:33 And then let's substitute expressions, the first one is m plus h over 2, sorry, not plus, is times h over 2, right? and plus mx, what is the next to the density of water, and times the volume, which is given by the base.
02:55 I call it, suppose the base has a volume a, and then it's given by a times the height, which of course is x, right, and times x over two, right? so i'm divided by m plus row, a, x, right? so finally, we get an expression, which is given by m, sorry, capital m, h over 2, and plus row a over 2x squared, and over m plus row a x.
03:25 So this is how the standard of a mass goes with the x, right? and you're actually, you can actually, i think you can actually do a little plot of this function x and the x, right, the central mass.
03:48 So if you look at x equals here, right, if you put x equal zero, of course, they hide somewhere here, right? it will be half a hedge, right? and then it's going down, right? then you will be going down.
04:03 As x going down, it's going to going down, right? and actually, you see that x, x going down faster, right? so finally, when x becomes zero, the point is the interesting thing is that when you, when you decrease x from initially when x is, sorry, initially when x is h, right? in sure, when x is h, you know, that, um, so an initial, we, we, we, we, we x equals initial, the initial value of x, actually, let me try to enjoy it...
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