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1/7 points Previous Answers SerPSET9 12.P.038_ My Notes Ask Your Teacher uniform beam resting on two pivots has length 6.00 mand mass M The pivot under the left end...

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1/7 points Previous Answers SerPSET9 12.P.038_ My Notes Ask Your Teacher uniform beam resting on two pivots has length 6.00 mand mass M The pivot under the left end exerts normal force n1 the beam, and the second pivot located distance 00 m from the left end exerts normal force n2' woman of mass m 65.9 kg steps anto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman position when the beam begins to tip_What is the appropriate analy

1/7 points Previous Answers SerPSET9 12.P.038_ My Notes Ask Your Teacher uniform beam resting on two pivots has length 6.00 mand mass M The pivot under the left end exerts normal force n1 the beam, and the second pivot located distance 00 m from the left end exerts normal force n2' woman of mass m 65.9 kg steps anto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman position when the beam begins to tip_ What is the appropriate analysis model for the beam before it begins to tip? rigid object under net force particle under net force particle in equilibrium rigid object in static equilibrium Sketch force diagram for the beam, labeling the gravitational and normab forces acting on the beam and placing the woman distance the right of the first pivot, which is the origin. Choose File No file chosen This answer has not been graded yet: Where is the woman when the normal force n1 is the greatest? What is n1 when the beam is about to tip? Usc Fext to find the value of n2 when the beam is about to tip_ Using the result of part (d) and about to tip_ ext 0, with torques computed around the second pivot, find the woman'$ position when the beam is (g) Check the answer to part (e) by computing torques around the first pivot point: Need Help? Fuheti



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A uniform beam resting on two pivots has a length $L=$ $6.00 \mathrm{m}$ and mass $M=90.0$ kg. The pivot under the left end exerts a normal force $n_{1}$ on the beam, and the second pivot located a distance $\ell=4.00 \mathrm{m}$ from the left end exerts a normal force $n_{2} .$ A woman of mass $m=55.0$ kg steps onto the left end of the beam and begins walking to the right as in Figure P10.28. The goal is to find the woman's position when the beam begins to tip. (a) What is the appropriate analysis model for the beam before it begins to tip? (b) Sketch a force diagram for the beam, labeling the gravitational and normal forces acting on the beam and placing the woman a distance $x$ to the right of the first pivot, which is the origin. (c) Where is the woman when the normal force $n_{1}$ is the greatest? (d) What is $n_{1}$ when the beam is about to tip? (e) Use Equation 10.27 to find the value of $n_{2}$ when the beam is about to tip. (f) Using the result of part (d) and Equation 10.28 , with torques computed around the second pivot, find the woman's position $x$ when the beam is about to tip. (g) Check the answer to part (e) by computing torques around the first pivot point.

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.

Well there's a lot of words for this one but the drawing isn't all that much. Looks like there's a beam And it's resting on two points with a distance L. Between them looks like there's a girl walking on it with a mess em and the distance to her is axe and then capital L. Is the full length of the whole being. So that was lower case L. Here. Alright, simple enough diagram. Let's see why the words are so long. So it's telling us that L. is six and M massive. The beam is 90. Well that's going to have a force at el over to of M times G. Okay. The left one As a force and one on the B. And the normal force here is going to be end too. Whoops. Lower case. N. Strange thing to use. Okay. Mhm. Okay. I've already done a which is drawing a free body diagram except I should show lower case M. G. Here and I think that's it. That's all there is to be Where is the woman when N. Is the greatest. It would seem intuitively obvious that she would be at the beginning. All the way to the left. And the reason is because um the two downward force is that the one acts in the center of the beams that doesn't move at all. So the other one we want to put at and one okay. See what is N. One when the beam is about to tip, it's about to tip. Well then and one would be zero because it's about the tip. There'd be no force there. Um D use the force equation. All right, So since N one is 0 then we've got and two minus M G minus MG equals zero. Oh, okay. And I need to figure out the value of end to sow and two is just going to be m plus M. G. So putting that in a calculator and I'm going to declare everything that I know. Um hmm Mm. Little M is 55. I didn't write that down yet. Mhm Capital L. Lower case L. Is four. Mhm. Which I didn't write down yet either. Yeah. 123 looks like they give us four numbers. Okay, let's go ahead and do this. G equals 9.81 1st 9.80 1st of all. Uh Capital L. Is six. Capital M is 90 cm. It's 55. L. Is four. Okey dokey moving down here and two is M plus capital M G. So that would be one .4 to kill. Oh, Newton's I got 1420. Okay. E no, we need to use a torque equation, torques computer around the second pivot point. Uh huh. What's the position? Okay. Mhm. The net torque around the second point is going to be and no, no, no, it will be M G. And the distance is going to be L minus X. Plus capital M G. And the distance is going to be L Minour Capital L over two minus and one L. Mhm But wait a minute the beam won't tip until X is way over here. So this should give me a negative value for X. And by the way and one is zero. Mhm. So this should be fine but I need to understand that X is going to be I guess not a negative value but it's going to be greater than uh L X is going to be greater than L. So um okay, trying to solve for X. So M G l minus X is going to equal MG um L minus L over two. Um but there should be a negative in here. So I'm gonna write this as X minus L divide by MG and then add L on both sides. So X is going to be mm over em L -L over two times. Uh and I am I trying to figure out X. Yeah um mm Over M x minus L over two plus L. Okay, think about this again, I am over M L minus L. Over two. Um Okay, so let's see what this gives me mm Over Em L -L over two plus L. That gives me a positive number. I didn't want a positive number. Oh yes I did. I said that X has to be greater than now. Okay, so it's going to be giving me five 64 and what are my units meters. That makes sense. Okay um Probleble 12 and made it the whole way down to E check your answer by doing the torque around .1. Okay, So now we've got MG at a distance X. And then we've got capital M. G. At a distance L over two. And then we've got- and two at a distance L. But end too is uh well, we already know him too. So X is going to be and to l minus MG. L over two over MG. Okay, putting that in a calculator and to L minus capital M. G. Hell over to for M. G. Gives me the exact same answer. Thank you for watching.

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.

Hi for this question we have been supported by two supports here and want to find the magnitude of F. one and F. two. Which about being given all the forces on it. So first of all and think the clockwise clockwise talk, indians and counterclockwise spoke and the counter clockwise talk might be causing clockwise stop for the systems we balanced. So we contact look quite stop efforts in this direction. It should be this F. One for us, this For for 15 years and force which wants it in this direction. And then for the counter clockwise of this direction which is just up to for one talk and that's it. So that's right an equation. The club where is talk. Uh huh. And Turkey is going to force them to the distance from the center of attention. So this element too. So Christ have F one times all about you plus 4 15 times L. Before and for the counter clockwise, You know, F two times available to loss mm And that's the only contact request talk. Well once you quit this too, We have F one over T. Was 4 15. Yes. Over four is equal to F. two. L. 2. This is one equation. Now we also have that the forces F one plus F two must be called to the total downward forces which would be 200 plus 4 15 to F one was able to Called to 300 was 4 15. She's a gold to 6:15. And so we can simplify from the first equation. We have F one over to you plus 4 15 before is a call to F. Positive. You can system will fight this for the What's playing two x 2 have F. One. Yes. Yeah. So 450 over to they called to have to 450 or two would be to 25, which you can substitute this into this equation. so solving simultaneous, let me get F one is go to 2 1, 2.5. Newton's F t is equal to 4 37.5 millions.


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