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(II) Consider again Example 11 but this time assume theroadway is supported uniformly so that $\frac{1}{2}$ its mass $M$$\left(=7.0 \times 10^{5} \mathrm{kg}\right)$ acts at the center and $\frac{1}{4} \mathrm{M}$ at each end support(think of the bridge as two spans, $\mathrm{AC}$ and $\mathrm{CE},$ so the center pin supports two span ends). Calculate the magnitude of theforce in each truss member and compare to Example $11 .$

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$\frac{M g}{4 \sqrt{3}},$ in tension

Physics 101 Mechanics

Chapter 12

Static Equilibrium; Elasticity and Fracture

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

Hope College

University of Sheffield

University of Winnipeg

Lectures

04:12

In physics, potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. The unit for energy in the International System of Units is the joule (J). One joule can be defined as the work required to produce one newton of force, or one newton times one metre. Potential energy is the energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is associated with restoring forces such as a spring or the force of gravity. The action of stretching the spring or lifting the mass is performed by a force which works against the force field of the potential. The potential energy of an object is the energy it possesses due to its position relative to other objects. It is said to be stored in the field. For example, a book lying on a table has a large amount of potential energy (it is said to be at a high potential energy) relative to the ground, which has a much lower potential energy. The book will gain potential energy if it is lifted off the table and held above the ground. The same book has less potential energy when on the ground than it did while on the table. If the book is dropped from a height, it gains kinetic energy, but loses a larger amount of potential energy, as it is now at a lower potential energy than before it was dropped.

04:17

05:56

(II) Figure 71 shows a sim…

02:49

(II) A heavy load $M g=66.…

16:23

(III) The truss shown in F…

24:49

A bridge truss extends 100…

03:11

The two steel channels ar…

13:05

A bridge truss extends 200…

so first set the three body daijaan from figure 12.29 on, Modified to indicate the changes in the roadway mass distribution. You might also need to draw the three body diagrams fall point A and B So I have drawn them separately over here. Now that example, if the road bay moss is one point four times stand to the bar six kg. So this was the old Ramos in that example. Then for one dress, we should use em. Tow the equal toe seven times 10 to the par five Katie mullets right The conditions for equilibrium for the entire trust by considering vertical forces and talks about the point, eh? It's about this point on here I am taking clockwise stocks to be positive. So whilst Lex tried the dark in question on this Gibbs Half N g Dying's this distance Easy which is given Toby 32 meters less, 1/4 off and Eve. It is this force over here times the distance e and that is 32 plus 30 does since both are equal A C and C segments are equal. So we have talked to bless that to do with 64 meter. Then we have a counterclockwise talked youto f toe Sylvie take after dimes the same distance. And that gives Edo 1/4 off. Indeed. The force over here and f one don't give any talks because the past through point a so and be a calculating the talks about point is that they don't give any talk about that point. So if you simplify this, you give you get f toe equal, do half off mt. Now let's like Newton's law for the vertical direction. So we have f one going up. If you're going down on Dhe, then three off days. So let me like that together and all these forces go down and that is equal to zero. So this gives f one less f too minus and James of these fact those somewhat bong. And from here you can find F one which comes up to the M G minus after. So you like this value off md over here and you get a friend to the equal dough half a Fendi. So now we have another force No more that the problem Miss Stern symmetric about this vertical line through C Let me draw that line. So the problem is the magic about this point about this line. So the forces at the ends at the end of each beer half off the week off that side off the structure, So it will be clear after Ah, I right when I started working on the on those spots now, the second thing you need to do is analyze the force on a lease. All the forces on the bin at point B so have drawn the free body diagram over here. So if we analyze that on Light Newton's second Law for vertical and horizontal direction and solve them after so for the vertical direction we have half N G minus one foot off engine minus F A B science extra dignity to be equal to zero and for horizontal direction, we have f A C minus. If a bee costs extra degree to be equal to zero, so basically here and taking the components off F A B in horizontal and vertical direction. So as you see it, this angle is 60 degree. So if a bee cost 60 degree will be the horizontal component towards right and if it be signed 60 cripple the vertical component towards up, up aunt here and taking go No left side. Toby, was it too? And down Could be sorry. Okay. Oh, I guess I Yeah. Okay, So f a b cost 60 degree east towards left for the pin A for the pain. That point is, if it because 60 degrees towards left. So therefore there is a negative sign here on if it be signed 60 degrees downwards. So here there is a negative saying on DDE we're going with the usual assumption that is Ah, positive xx is is two words right on DDE Was it? The Y axis is upwards now solving this we get if they see Toby equal toe f A B 16 inkley cause extensively. Now, if you solve this a question, let me just move this. We're here now. If we solve the first a question you find it's a b to the equal do mg All right. Two times through Dover off three. Then you can find if a sea by plugging this value off its baby over here. So doing that, you get if a sea to the equal toe mg Oh, my for your tree then go find F B C and F Baby. We will analyze the forces at the for the pin at going to be on DDE. We go with the same Matar Sylvie right? Newton's second law at point B for the horizontal and vertical direction. So for the vertical direction we get if baby signed 60 degree minus if BC signed 60 degree to be equal Placido. So this gifts F B c do the equal do f a C f A b on. We'll call it the calculated F A B over here, so just plug that value and you get f B C. The magnitude off. FBC No, But if edge for the horizontal forces we have if baby go 60 degree plus if BC cost 60 degree minus. If deep on this gives f D B to the equal dough F A B less F B C times cause ecstatically on using the values off F A, B and F B C, which basically are equal on equal to M g over to victory. So we like this value over here and we get if debate to be equal, do twice off mg forward to tree time Scott 16. It is one over two, and therefore this value is mg or maturity. Now we have four off these forces. Basically, we have all the forces for one side off the drills. So we have all these forces off the left side of the crush on By symmetry off the geometry began to tell mine other forces so f d e which is here so f d e will be equal to f A B. This will be equal to M g over duty on DDE BC will be equal to it. ABC and this was the equal do and you to room three on F C will be equal toe f k c. This will be quite omg over for tea, as we calculated, depending on the direction off this forces and which of these segments are getting stretched on dhe getting compressed. So basically the direction off the forces that be found if they see if a B and F B, C and F D B oh, you can find which of these are in tension or compression. So if the e and we are getting compressed, as you can see over here, so they are in compression. This is so These are in competition. Stands for C. This is intention on C E C R. Intention. Again. Now, if you compare this solution Toa example 11 you'll find that each force is reduced by a factor off.

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