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From the left end exerts narmz force 01 position when the beam beginswomnan of mass59.5steps onto the lcft end of the beam and begins walkingthe rignt as the figure...

Question

From the left end exerts narmz force 01 position when the beam beginswomnan of mass59.5steps onto the lcft end of the beam and begins walkingthe rignt as the figure below: Ine g032 to iind the woma(a) Sketch free-body diagran_ labeling the gravitational file with maximum size of MB,) Choose File No file chosennorna forces actingthe beam and placing the woman meters the right of the first pivot; which the origin. (SubmiThis ansvr8 Has nol bebn gradud yel;(b) Wherc thc woman when thc normal forccg

from the left end exerts narmz force 01 position when the beam begins womnan of mass 59.5 steps onto the lcft end of the beam and begins walking the rignt as the figure below: Ine g032 to iind the woma (a) Sketch free-body diagran_ labeling the gravitational file with maximum size of MB,) Choose File No file chosen norna forces acting the beam and placing the woman meters the right of the first pivot; which the origin. (Submi This ansvr8 Has nol bebn gradud yel; (b) Wherc thc woman when thc normal forcc greatest? What when ppam abaut tip? (d) Use the force equation af equilibrum find the vaiue Of n3 wnen the beam about Using the result of part (c) and the torque equilibrium equation with torques comdutecd around the second pivot point; find the woman oosirion When the beam aoout (f) Check the answer part (e) bY computing Troes around the first plvot Doint



Answers

A beam resting on two pivots has a length of $L=$ $6.00 \mathrm{~m}$ and mass $M=90.0 \mathrm{~kg}$. The pivot under the left end exerts a normal force $n_{1}$ on the beam, and the second pivot placed a distance $\ell=4.00 \mathrm{~m}$ from the left end exerts a normal force $n_{2} . A$ woman of mass $m=55.0 \mathrm{~kg}$ steps onto the left end of the beam and begins walking to the right as in Figure $\mathrm{P} 8.12 .$ The goal is to find the woman's position when the beam begins to tip. (a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman $x$ meters to the right of the first pivot, which is the origin. (b) Where is the woman when the normal force $n_{1}$ is the greatest? (c) What is $n_{1}$ when the beam is about to tip? (d) Usc the force equation of equilibrium to find the value of $n_{2}$ when the beam is about to tip. (e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. (f) Check the answer to part (e) by computing torques around the first pivot point. Except for possible slight differences due to rounding, is the answer the same?

This would be our free body diagram or the sketch of the system essentially yet for party and additionally, we have MG. This would be equaling 90.0 kilograms multiplied by 9.80 meters per second squared and this is giving us 882 Newton's. We then have lower case M G. The smaller mass of 55 kilograms multiplied by 9.80 meters per second squared and this is giving us 539 Newton's Now, at this point, we can say for part B. We know that, um when woman is at X equals zero and someone is greatest. We know that women woman is exerting Max counterclockwise torque again at X equals zero. So essentially and someone will be exerting its maximum clockwise to work at that about the center in order to hold the beam in rotational equilibrium or staticky kilograms. So get this would be given that the torque is equaling zero, so four parts see, then we know that as the woman walks to the right along the beam, she'll eventually reach a point where the beam will actually start to rotate clockwise, so her counter clockwise rotation. Ah, diminish her counterclockwise. Torque diminishes as she as she walks right along the beam. So here, uh, once bebe begins to rotate, uh, we can say the beam will start two. Lift. Uh, it's left most we could say a tipping point. And essentially, what that means is that the normal force will will eventually decrease to zero. So answered one well equals their own new tenants. Four part D. We know that when the beam is right about the tip, the son of the forces in the UAE direction is equaling zero, and the normal force of one is equaling zero. So this is essentially giving us that ends up to minus mg minus lower case M G's equaling zero therefore, and Sub two is equaling MG plus mg. And this is giving us 882 Newton's plus 539 Newton's. This is giving us 1420 Newton's approximately. We have to round 23 significant figures because we can't be more accurate than three significant figures. Four Part e men. We know that at the right most pivot, we could say that some of the two works at the right, most of it will equal zero. And so at this point we know that Ah again and someone is equaling zero and we can say that then. Negative 4.0 meters brother minus x multiplied by M G minus 4.0 meters minus 3.0 meters multiplied by capital M G is equaling zero the larger mess. And so we could then say that M g X is equaling 1.0 meters multiplied by G. And then this would be plus 4.0 meters multiplied by the smaller mass times acceleration due to gravity. We can then say that axe is gonna be equaling 1.0 meters multiplied by AM over Emma plus 4.0 meters. And so this is gonna be equaling 1.0 meters multiplied by 90.0 kilograms, divided by 55.0 kilograms plus 4.0 meters. And we'll put it here. X is gonna equal 5.64 meters again. We have to round 23 significant figures, So that would be your final answer for Part E and finally, for part F. We know that when an equals went and someone equal zero and Step two is equaling, we could say 1.42 times 10 to the third Newton's or again 1420 Nunes either one. And we know that. Then the sum of the torque at the left end is equaling zero and this will give that zero minus 539. Newton's well supplied by X minus 882 Newton's multiplied by three meters plus and that here is 1.42 times 10 to the third. New tenants multiplied by 4.0 meters. This all is equally zero and so here X is equaling negative 3.3 times 10 to the third Newton meters. And this would be divided by, of course, 500 rather negative. 539 Newtons. This is equaling 5.62 meters. This would be our final answer. Four part F. And this 5.62 meters, as you can see, is very close to 5.64 meters. So with rounding errors, it's pretty much the same answer the answer to party is going to be the same answer. To part that is the end of the solution. Thank you for watching.

So we're told that a woman is walking along this beam and it has supported at one end and then in a little in a distance from the other. We're given the distance between these two supports is six meters. The total length of the beam here is outside. The distance between the sports is four meters total length of the being to six meters, The beam weighs 90 kilograms and a woman has a massive 90 kilograms. And the woman has a massive five kilograms. And what we're asked to do is to find well, first of all, what is the maximum force at this? At what displacement with the force be a maximum at this point? Well, that would be a maximum. When the woman is standing at that point, you can see that. So if we draw are forced by a diagram, we have the weight of the beam, the weight of the woman and the reaction forces from the supports. The support reaction forces have to equal the total weight that's acting on the beam. And then if we take moments about this point, you have the moment from the the weight of the beam, the moment from the way of the woman and then the moment from the reaction for us that this that this support and that needs to be zero. What? We can see that if we saw for F one, we're going to see that it's just a decreasing function of X. So when When the woman is at this point when a woman is at X equals zero, um, then the force here will be a maximum, and it turns out that that is 3020 news. Now they ask for when how far the woman can walk before this beam starts to tip. Well, at the point where the beam starts to tip, this load is going to go to zero. So as this woman walks a lot here, the load is gonna change. And eventually it will go to zero as she's somewhere out here, and she then this would then start to rock up. And so we can. If we set F one equals zero, we can then solve for X, and we find that that is 4.61 meters so she would be out. So this is about five years about about here she gets about here. The beam is gonna no longer be an equilibrium because assuming that there's nothing holding to be down onto the support that is just resting on their and the force, we can then figure out the force here. And that is 1420 Newtons. And they asked us to some moments about this point to see if we get the same answer. And what we can do is we can just some of the moment and then plug in all of our values that we got from before X and two and we'll see that we get zero. So we as we sure so that the moment balance is also valid for taking if you take moments about this point.

Here in this solution In destruction, we have part of rigid object. Did O J instead IQ include medium instructing it really medium in Barbie, we have the distance. Total distance is here we have this distance is three meters and here empty is in dollar or direction and don't isn't upward. And this total distance is from here to head It's four meter Listen, anyone in a poor direction And this is X. This is a MD now I m g music was true 90 90 Getty mask multiply with G Figured it. 82 northern Nugent and m G. You think well with the five que no ground g which is it was 5 39 I knew it'd No, in part C in party Know that about the Wright powered? Only an exert applause Quite pork. All other forces exert counterclockwise talks except for and to reject the deal. Oh, torque the woman is it the woman is add X is equal to zero when anyone is greatest. With this location off women, the counterclockwise talk about the sender off the beam is a maximum. Those and must be exerting is its maximum clockwise dork about the center toe. Word of being in rotational new video in part B. We have, and one is It was true, Tito. As the woman walks through the right along the beam, she will eventually reach a point where the being will start to rotate clockwise about the right, most by word. On this highest, the beam is starting toe uplift. Uplift off the most left, most powered and normal force exerted by that by word will have diminished to to the now in part E when the team is about to and when is it was 20 And we know that permission off FBI is with Geo gives zero place and do, and to minus empty minus MTV. Today it was 20 or we can find, and to Israel was toe mg. Less anti is it was toe each 82 Newton blessed 5 39 Newton. We get one point for two into 10 days to poverty. Newton in part F, requiring the Internet. Torque is zero about the Wright powered when the beam is about to tip and one is it was 20 gives permission off. How is it? Was toe end to multiply zero plus four meter minus X. I'm into MT plus four meters minus three reader into capital. Energy is it was through the roof. Does we? After putting the values we can say this X is equals to one meter multiply 90 kilogram divided by 55 kilograms. Smart plus four meter We did I want to export meat it now in part G. We have when and one is a postal dedo and one is zero and enter the close to 1.41 to 10 days to poverty. Nugent for the work of all the left people, people say Submission Tower The post video minus 5 39 Newton Multiply X minus 8 82 Newton Multiply three meter less plus one point one point for two Multiply tenders to power three Newton multiply with four Mutual. We get submission. Tao is it was 20 now or we can say X is It was two minus 3.3 mortar pretenders to power three Newton meters divided by minus 539 Beautiful Between the bulls, 25962 meter which wouldn't limit off. Rounding added is the same altar in the part off

There's no drawing for # 14. So I guess I'm going to draw myself. There's a beam That rests on two pivots. One of them is at the left end. Got it. The other one is a distance l from the left end. Seems just like the problem we saw before. Okay so we're really just repeating number 12 but symbolically so um I'm gonna draw the woman here because it's seriously not gonna tip until she's like over here. So M. G. Is here MG is here and one is here And two is here. The distance to the woman is X. And then the total distance is L. All right So some of the forces in the Y direction is zero gives us. Um Well first of all we know that when it's about the tip and one is zero. So and two minus M. G minus M. G equals zero. So N two is going to be M plus M. G. Okay so that's the answer to a B. Her position. Okay so her position um I think it was actually easier to do the net torque around one. So that is and two times its distance which was L minus capital MG at el over to minus MG at X. And that is zero. But N. is I can substitute for N. M. Plus M. G. L minus MG. Hell over two minus MG. X equals zero. Now I notice that G cancels out. And so what do I want here X. I need to figure out X. So X is going to equal M plus M L minus M. L. Over two over lower case app. Um I could probably simplify that a little bit more but I'm going to let it go like that so now she wants to reach the end of the being and we need the minimum value of L. So X would have to equal L. In order to just not tip. So X would have to equal capital L. So find the minimum L. Lower case L. All right so M plus M L minus Ml over to has to equal lower case M. L. And so um M plus M L. Would have to equal okay Mm over two plus M. L. So lower case L. Is going to be Mm over two plus M over M plus M L. No I could simplify that a little bit more but it doesn't say we have to simplify. So um because I could just multiply by two. The numerator and the denominator make it a little simpler. Mm Were nevertheless that's good. Thank you for watching


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