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Welcome to new.
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This time we have a system in equilibrium and the system is holding up different types of fish.
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So, you know, the first one is fish a.
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And then the second one is hanging from the second rod.
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This is fish b right here.
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This is the second fish.
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We're going to call it fish b.
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And then there's an extended rod right there.
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We have fish d and e.
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You know, i have to put the i's in the direction, but you don't have to do that.
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It's just optional.
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So this is fish d.
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And then the last one.
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Is fish c so a, b, c, the rods, you know, we can, this distance here will given this distance as being equivalent to 30 centimeters.
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And then this second distance, i should have had this one a little bit to the right.
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So it's proportional.
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This one right here would be proportional.
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So this whole distance is 30 centimeters and then this one is 7 .5 centimeters.
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This one also we need some adjustment so we just make sure that it makes more sense in terms of location of these rods.
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So this one extends all the way up until right there.
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So we can have a proportional variance in the lengths that we're dealing with.
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So this one is given us 15 centimeters.
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And then this one is given us five centimeters.
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Also, this one is five centimeters.
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And that's this distance up until there.
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This distance from d up until that road is 17 .5.
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I need to adjust that, 17 .5 centimeters.
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Okay, that's the information we're given in this problem.
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The other piece of information we're given is that the mass of b is equivalent to 0 .748 kilograms.
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You know, that's the mass of b.
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So the goal of this problem, so this is the information we're given.
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And then we want to find the mass of a, the mass of b were given, so we'll also need to find the mass of c and the mass of d.
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The system is in equilibrium.
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So we're going to use that to our advantage to determine the masses there all.
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The other thing is that the bars are massless, so they don't have any mass.
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I mean, you could say they are weightless, because if you don't have mass, then you also don't have any weights going on.
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So we're going to use torque.
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Remember, talk is defined as the perpendicular distance times the force itself.
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The system is an equilibrium.
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So that means the sum of talks about any points is zero and also the sum of forces is equivalent to zero.
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It's a hanging system with all those units that we're given.
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We can change the dimensions to meters if you want to.
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If you wanted to change it to meters, you'd have to do a unit conversion.
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So that's one meter of a 100 centimeter for example so that would end up being 0 .075 meters but you know we don't have to we don't have to change the we don't have to change the the units because they're going to cancel out anyways the other thing which will strengthen our problem is that we do have a we do have a we're going to label these ones.
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So this one, we're going to label it as i.
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This rod is i.
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So the force that's pulling up there, we're going to call that f sub i.
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And then we're also going to label this one two, roman 1 and roman 2.
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So this force right here will be roman sub -roman 2.
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So sub -roman 1 and sub -roman 2, those are the two forces.
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Also, you know, we have we have another force that we haven't really talked about.
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And that's that's the force at the top.
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Okay, that's the force at the top.
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So this whole thing is hanging.
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This one is hanging right here.
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This one is hanging.
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I mean, if you wanted to, you could call this.
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Is f.
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This one, there's a force there.
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You know, we can call that three, the third one, the last one.
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So this would be f sub three.
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It's a system in equilibrium.
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The fish are hanging from different positions.
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This is a.
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This is b.
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This is c.
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And this is d.
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Those are the positions of the fish.
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The system is in equilibrium.
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So it's not moving and we're supposed to find the three masses so we need to find mc and m d because we're given b that's the information we're given so we'll start by looking at at one we'll start by looking at actually we need to we can rename these subscripts at thing because we're going backwards so yes so these subscripts instead of calling this one two we're going to call this one one yeah that makes sense and then we'll call this one too so this one is two that makes sense so then the the net talk around one is zero and so around 1 is 0 and so in the next page we want to see how we're going to use this so if you think about the way one appears we have d right there and then this is 17 .50 centimeters and then we have f1 holding it i think we have to have a clear diagram this so this is super horizontal so we have to account for that so this is horizontal all the way like that and we're only thinking about this section right here so the initial section so then this one is our one and the force responsible at that point is f1 that that's the force responsible.
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This distance as you can recall is 17 .50 centimeters and this other distance right here is 5 centimeters even though it's not very proportional but that's how it is.
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This is d and this is c so we're going to start with that and we're seeing that the net torque around 1 is 0, roman 1 remember so md md is going counterclockwise, so it's positive talk, and this one is going clockwise, so it's negative talk.
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So mdg times 17 .5 centimeters minus, because this is a negative torque, mcg times 5 .0 centimeters, that's equivalent to zero.
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And so we can solve for mc, the mass of c.
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We can solve for the mass of c.
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So mcg 5 .0 centimeters becomes equivalent to mdg 17 .5 centimeters.
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These two gs cancel out.
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So mc equals to md 17 .5 centimeters.
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All of 5 centimeters that we simplify that so mc becomes equivalent to 3 .5 md 3 .5 md and this is a critical equation that we're going to be using so we need to keep track of that.
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That's the first part of the problem involved the talk.
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In the next page, we're thinking about the sum of forces in the y will be zero.
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And again, we're still at, you know, there's a force here, fi.
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This is d and this is c.
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They're pulling down like that.
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And this one is also pulling down.
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And we're saying that if the net talk along the y is zero the net force not the net torque the net force so fi minus mdg minus mcg equals to zero so down is negative up is positive upward is positive downwards is negative so that's means the force that's holding those two positions it's going to be it's going to be a negative a positive force and the other ones are going to be negative.
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And so, fi becomes equal to mdg plus mcg.
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You can simplify that.
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Fi becomes md plus mc times g times g.
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And recall from the previous page, we already defined mc as 3 .5 md.
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So mc equals to 3 .5 md and that means we can change mc to become 3 .5 md.
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Okay, so just remember that all we're doing is plugging in and this is g.
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So fi becomes 4 .5 md times g because this is one, the coefficient of that is of md.
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The initial md is 1, and then the coefficient of the second md is 3 .5.
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This is the second equation that's going to be helpful in solving the problem.
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And then we'll still go ahead in a new page, but prior to that, we have to move on to the next level.
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This was level 1.
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At level 1, we were able to get force 1.
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Now we are at level two.
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We're looking at the impact of force two.
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So at level two, the diagram changes a little bit.
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You know, at level two, you have, at level two, you have, this force is playing a role, so it's pushing downwards.
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This is f1.
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Remember, you know, we have d and c hanging, from right here.
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So this is d and c.
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So this force right here is f1.
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And then if you go back, you see that the second one is b.
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So this mass right here is b.
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And then of course, f2 is pointing upwards.
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You know, f2 is pointing upwards.
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We can emphasize these two forces and those weights.
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And we can do the same thing on this one.
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These are the weights we're dealing with, d and c, and then they're being supported by fi.
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So coming back to this again, we're seeing that the net talk around the second level is zero...