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K6. A 2.0 m long beam with negligible mass is attached to a vertical wall on its left end and supported horizontally by & wire on its right end: The angle fy d ...

Question

K6. A 2.0 m long beam with negligible mass is attached to a vertical wall on its left end and supported horizontally by & wire on its right end: The angle fy d between the wire and beam is 359 as shown: A 17 kg basket of apples is then placed on the middle of the beam (1.0 m from the wall) If the contact point between the wall and the beam doesn't cause a torque on the beam; what is the tension in the wire? UA (Abd= (,3 .9,8.6 2 8 Us-a)= 8 1).74d = 7 - Sd 'Jd 330 20m

k6. A 2.0 m long beam with negligible mass is attached to a vertical wall on its left end and supported horizontally by & wire on its right end: The angle fy d between the wire and beam is 359 as shown: A 17 kg basket of apples is then placed on the middle of the beam (1.0 m from the wall) If the contact point between the wall and the beam doesn't cause a torque on the beam; what is the tension in the wire? UA (Abd= (,3 .9,8.6 2 8 Us-a)= 8 1).74d = 7 - Sd 'Jd 330 20m



Answers

A beam is attached to a vertical wall with a hinge. The mass of the beam is $1000 \mathrm{~kg}$, and it is $4 \mathrm{~m}$ long. A steel support wire is tied from the end of the beam to the wall, making an angle of $30^{\circ}$ with the beam (Figure 9-24). (a) By summing the torque about the axis passing through the hinge, calculate the tension in the support wire. Assume the beam is uniform so that the weight acts at its exact center. (b) What is the minimum cross-sectional area of the steel wire so that it is not permanently stretched? The yield strength (elastic limit) for steel is $290 \times 106 \mathrm{~N} / \mathrm{m} 2$, $290 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}$, and the ultimate breaking strength is $400 \times 106 \mathrm{~N} / \mathrm{m} 2^{400 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}}$. Example $9-8$

Notice first role it was first draw the Kanye T diagram of the system. So this is yeah center of mass and this is point this point and this is going to be so it is T. A. It is T. V. It is M. B. And this is center of mass and this is rotating with alpha center of mass. The next lesson, linear X lesson is a center of mass so we can write, we can right excellence in at A. Is equal to ALF. I multiplied where are A. So let's put the value. Alfa is four multiplied by are a 7.5. The It will be 30 in per second square. No we will calculate excellence and linear excellence and maybe we will be called to Alfei be multiplied where RB. It will come out to be six multiplied where 1.5 that is 45 in. But 2ndly Square. no now the NATO excellence and appoint a no tax lesson, appoint a center of most will be called to a center of my mask minus alpha, Multiplied Way 18 is equals two. No tax dollars. And or two. Let us put the values it will be a center of mass minus 18. Alfa Center of Mass is equal to Excellent Sanity. Is 13. Similarly or a similarly for being we can right a center of mass. Yes. Alfa Center of mass multiplied by 18 is equals to a B. If we So it will be equal to 145. So on solving these two equations we will get on solving from solving However, the question a poet person we will get we will get a center of mass Is equals to 37 25 inch per second square. And alfa centre of Morris is equals to zero 4 1 double six radiant. The 2nd is good. No, let us right. The net force. The net force will be equals to mass, multiplied by excellence. And so the force will be e less E b minus MG is equals two. Management played by a center of mass. Let us support the values here. Uh yes, he'll be minus masses To 18 divided by 32.2. Multiplied by G is 32.2. This will be equals two. Waiting Very well. 32.2 multiple way a center of mass. The center of must we know that we have calculated as 37 15 inch per second square. So let us convert it into eight per second squared. It will be equals two. Yeah, less he be equals two. 307 point 17 But from here Figure PB is equals two. He entered 7.17 E. Let the secret ambiguous in one. No, let us right Nate moment about center of mass. So it will it will be I. G. Alpa, ideal for El Fassi. M The next moment will be the b minus. Yeah, Multiplied by in the distance 1.5 It will be close to eyes. Family squared divided by two my deployed way alpha C. Um Let's put the values here. It will be T v minus E. Multiplied by 1.5 is equals two masses. 280 divided by 32.2. Multiplied where early square early's nine square divided by to help multiplied by alpha alpha C. And we have already calculated as zero point 0.41 double 641 double six. So from here also we get ub minus the is equals to 16 Please zero. Let this situation because the number two upon solving upon solving reason one and two. we get a bigot. We get to be a muse equals two. 100 45 points 43 on L. B. Similarly, E b is equal to 100 61 point 73 Thank you, baby. So this is the required value of 10s and 10s and in the string. Tencent in a string at point B intense and instrument point.

In this problem, we are hanging a flowerpot using a horizontal cable and a beam angled from the horizontal cable to a wall. And we want to find the tension in the cable and the force at the hinge of the beam. Eso. What we'll do here is we will take our attention about the hinge and set it to zero, and that will allow us to solve for the T tension in the cable and then weaken some forces in the X and Y directions to find the force at the hinge. Eso talks about our hinge. First one, we've got one coming from our massive the beam which was 2 kg so minus two G and that has an arm of zero 0.75 and the vertical or the perpendicular component of that to the beam Insigne 53. And then we've got the torque due to the flower pot and a mass of 2 kg arm of 1.4. And the particular component of that would be a sign 53. And then they talk. Due to the tension on the cable, that's gonna be a key an arm of 15 and the perpendicular component of that. Be a co sign 53 degrees okay. And solving that for our tension in our cable t, we got a two g sign fifth, three times a zero point 75 plus 14 our numerator and our denominator, a 1.5 co sign 53. And if we plug all that end, we'll get our attention is a little over 37 Newton's, uh next thing we wanna find our horizontal and vertical components of the force at the Hinch eso Let's some forces in the X and y directions and some of forces in the X direction. There are hinge force, the X component and in the opposite direction on the the tension and those something, those equal to zero gives us the hinge force. Uh huh. X component is equal to achieve a little over 37. Then in the UAE direction. Yeah, we've got our hinge force vertical components and we've got the weight of the beam and the flowerpot acting in the opposite direction and setting all that to zero. We get our hinge force in the y component is equal. Thio four g, which is a little over 39 which and we also want to find the magnitude and direction of our hinge force eso the magnitude gonna be square it of the X and the why opponents squared. And if we plug those values in that works out to uh huh 54.507 millions and the angle of our hinge force we'll get from the our tangent of the Why component over the X component and that will work out to be close to 47 degrees. Alright, So collecting all of our information here we've got our if tension in our cable forcing our cable 37 Newton's The magnitude of our hinge force is 55 Newton's and are and force in. Our X direction is 37 Newton's and our hinge force in our Y direction is 39 Newtons. And so those air all to choose significant figures

Hello Problem. 47 has a uniform being that's four meters long and weighs 25 100 Newtons and I have a 3500 and wait sitting on the end of it and it's supported by a guide wire as well. So let's take a look at what we got here. We know this is a uniform board, so it's wait, located in the center and it is 25 100 Newtons and that would be two meters from the end because it's a four meter board. Now this wait here that it's supporting is 2.5 meters from the end in its 3500 news. And this is our tension wire here at a 30 degree angle and we also know it's attached here. So we've got a horizontal and vertical force that the hinge must bar exert some vertical or horizontal and vertical force. They were asked to find, um, all three of these forces tension, the horizontal and the vertical force. We know it's an equilibrium, so we have three conditions that must be meant. Arnett torques on this or zero our forces in the ex direction are zero in our net force is in the wind direction. So let's look first at our forces in the extraction, we have one that's totally horizontal. The horizontal force we know that would be equal to the horizontal component of the tension. So it's tension. Sign that. Take that back because I'm 30 now we have a few more vertical forces. We have to downward vertical forces and we have to upward are vertical and our vertical component of the tension that we know that our vertical force of the hinge plus our tensions sign 30 must support the two weights 2500 and waas the 3500 new lights who help us out here at 6000. Right over here on our torques, we have our vertical or horizontal since our pivot points right here are not providing torques because our liver arms or zero, but these two weights to provide a clockwise tour more attention in the wire is providing a counterclockwise tour. So our attention in the wire now it's love arm is the perpendicular distance to this force. So that would be our entire length of four and that would be signed 30 and it would be equal to the 2500 Newton force its level on love too. Wasthe e 3500 didn't force in its lever arm not to it's left arm is 2.5. So here we can actually solve for attention. Well, when we fall for attention, you get 68 75. Since which helps us out here we can then put in 68 75 co sign 30 to get horizontal 59 54 mittens and down here are vertical component, then 25 63. Thank you for

Now we're told that we've decided to hang another plant from a 1.5 meter two kilogram horizontal being noticed has two hands on the wall on the left cable catches the right end goes 37 degrees above the beam. To connect your point about the hinge on the wall, we hang 100 Newton pot from the beam 1.4 meters away from the wall. What is the force that the cable exerts on the beam? Okay, so we can, um, draw free by diagram of our situation here, Since they gave us the mass of the beam, we can use it include its weight in the problem. So we have forces from the hinge acting on the beam have the weight of the being macking the flowerpot here and then the tension in the cable. And so we need to weaken right balance of moments so that the sum of the moments and getting the best place to take moments about is this point so that these forces don't come into the equation. Um, that has to be zero. So we get, um Let's hear this distance here was 1.5 meters. The why component of that attention in the court is tea time. Sign of 37 degrees. Um and then we have So we're taking counterclockwise positive. And so then we have minus 100 Newton time. This distance was 1.4 meters and then this The beam had a massive two kilograms. So we have minus two kilograms times 9.8 meters per second square ano assuming that the beam is uniform, that it exit the center here and that's 1.0 point 75 meters. We can then, um Sol for tea and we know everything in here except for tea. And if we do that, we find that he is on 171 Newtons and in this case we can figure out what the forces in the forces here are s o. If you balance forces in the horizontal direction, you know that F X must be equal to t times Sign our co sign of 37 degrees and that's 137 Newton's. And this force here has toe be equal to offset this force in this force. But they're why component of this attention to contribute so that we take a minus Sign here. So we get f of why equals minus two times 9.8. I'm sorry I said my f of why it was two times 9.8 plus 100 Newtons, minus T times seven sign of 37 degrees. We plug out all the end we get that this force here is 16.5 Newton's.


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