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Beam of mass Mand length supported from the ceiling two strings, shown:If the beam uniform, show that Ifthe beam has its center of mass ofthe way from the left end,...

Question

Beam of mass Mand length supported from the ceiling two strings, shown:If the beam uniform, show that Ifthe beam has its center of mass ofthe way from the left end, showthat tane 3tand_

beam of mass Mand length supported from the ceiling two strings, shown: If the beam uniform, show that Ifthe beam has its center of mass ofthe way from the left end, showthat tane 3tand_



Answers

A uniform beam of mass $M$ and length $L$ is held in static equilibrium, and so the magnitude of the net torque about its center of mass is zero. The magnitude of the net torque on this beam about one of its ends, a distance of $L / 2$ from the center of mass, is a) $M g L$ c) zero b) $M g L / 2$ d) $2 M g L$

Here we have the RARD that is suspended at an angle on one and on a man be on the other end. Let's say this is the one on this is to now the way this drive off centered weight on That's a This distance is X and this just Dantes one minus x on the total Dantes, Let's say and so that's extend on one minus extent. And this is the center of mass. Talk about the center of mass equals zero. So we have. Do you on signed fee dimes and one months X, It was Do you do? Sign it off Times X now forward X equals half or the wedge to be at the acting at the center. We apart these two terms, cancel out and we will have the one sign feed. It was Do you do scientist now something off F X equals zero with Give us two on design fee. Equal student. We'll sign. Okay. Now. So from these three questions, if we divide one by the other, we have fancy equals stand. So we have gone be equal status, which would imply Do you want because you do now In the second case. We have the central gravity's 3/4 from the left end, which means we have exit was 14 So we will have 34 and dines one minus X is before and dance. Do you wan? Signed feed equals 1/4 l dance Studio sign, kiddo. Odd three d one sign fee. It was to sign Terra are ever This equation is still valid, so we would have damn theta equals three. Yeah.

Okay. What we're going to do is we are starting with two thin rods of constant density that are actually welded together at a right angle. Okay. Um this rod is going in the X. Direction, this rod is going in the right direction. Um If we let this right angle kind of be in a coordinate plane, um the length of the rod and the extraction is just L. And the length of the rod in the Y. Direction is two times that length. Okay? And what we want to do is find the frames center of mass. Okay. And so what we're going to do is um since it has a constant density, Mhm. Um We're gonna actually look at um each separately first and then try to pull everything together. So because each rod has a constant density, it is safe to assume that the rod in the X. Direction, his center of mass is actually um halfway in the middle, in the middle. So he's exactly in the middle, right, and in the Y direction. It is also exactly in the middle of the rod. Okay, so now what we're trying to do is we're going to try to look at the frame. Now we're going to look at the frame altogether. Okay. And so what we know is the center of mass. The center of mass is given by the some of the masses times their distance away from the origin or what is called a pivot point divided by the sum of the masses. Okay, so um the frame center mass is going to be the some of the masses um in the X. Direction. And so that is going to be the mass of and there are equal matter, there have equal constant density, they are equal masses. So it's going to be this thin rods mass times his Centre point, which is L over two Because that's the distance in the X. direction away from the zero mark. Plus the vertical rods Mass times his distance away from in the X. Direction away from the zero point which is zero. And then this would be to em. So this is going to be L over four now in the Y direction, it's the same thing. So we take the in the Y direction, the horizontal bar, the distance away from in the Y direction, the horizontal bar from the center 0.0 plus the vertical bars center a mass away from That zero Mark, which is L. All over the some of the masses. So this is all over to. So that is telling me that the frame center of mass is L over for comma Eleanor too. I hope this helped.

Hello and welcome to this video solution of numerous. This question is based on principles of static equilibrium. Here we are given a uniform beam of mars M. B. That has a length of L. And supports two blocks and one of them to two points. As shown in the figure. Also this game rests on two points. Now we have to calculate the value of X. For which the B. The beam will be balanced at P. So that the normal force it too easy. So we are given that first, let us brother forces directing on the screen. So let's say the masses are shown. The which will act downward, the which will act down works. And also the reaction force due to this resting points are also shown. But there is something even which is our one equal to zero is Cuban. Right now you have to calculate the value of X. For which the system is in equilibrium. So for that first we need to calculate the submission of forces along the Vertical direction and equated to zero. Here we have the masses, So we do to the masses M1G plus the mass of the beam, Mbg plus daughter muslim to g minus R. One, miners are too Will be equal to zero. Okay, Unfortunately our one is already given a zero. So we can say that are too is equal to we can consider G. S. Common, then we have M one plus and two plus M M B. Right? Yes, we have the value reaction force at the second fixed point. We can say now we have to calculate uh in six we can take a moment about this is to explain itself about this fixed point. That's a B in the unit. It's a P. This point. So the talk about B Is equal to zero. Right submission of would be. So, first of all, we can say that the talk due to the mass M. two which is a clock first clockwise one so minus M. Two G. The way due to Muslim. two times X. Right exists. What? And there's a clock face talk. And uh now we have the weight due to the beam that is plus because it will produce anti clockwise to so M. B. G. Times of the small distance D. Now this will be uh Now that will plus plus we have another force you to another mosque which is in one G. G. To himself elevator plus and the red D. Now this will be equal to zero. Now we haven't considered Erdogan because everyone is still to do so we can neglect that. Now here from here you will get X equal to uh it's pretty simple. So we have one by I am too. Now we can actually cancel G from every time out here. So actually won by M2 times M. B. D plus in one L by two plus in one B. Right? So at this value of X, we can say the system will be at equilibrium and there will be no reaction force that the first a fixed point. I hope this is clear to you and have a very good question.

Yeah, with the masa being one that is M one is 10 kg and Marcelo being to tourists for the length of being to his 2 m length of Yemenis. Mm. Like a The expedition of the center of gravity for being one is excellent. Thanks to you know, 35 ft and x two for B two is one plus one meeting our X two equals to 2 m. The center of gravity X C t is Sigma m I X I right. Take my and my r x C d. S 10 and 2015 plus What field to to by 10. Press 40. Uh huh. Sieges 1.7 from the left. Now the second part. So the position of the center of gravity is that 1.7 1.7 m from the left part, and the whole way is passing through the center of gravity. Should that sister distance between the center of gravity? And the weight is zero that are approved by the way, That is doubt. W yes, R w I doubt that you is see So the next hour polluted away to the access our quotation passing through the central priority is you see? Thanks


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