BramVandenbo

There are 3 things to know: 1) The air pressure is about 100.000 Pa underpressure. (The pressure of the water inside the balloon is not worth mentioning because it is so small compared to the pressure outside the balloon.) 2) The elasticitymodulus of latex is about 10.000.000 Pa if I am not mistaken(?) (This should of course be pretty small compared to other materials.) 3) Latex can stand about 15 000 000 Pa. (according to wikipedia) Some calculations and conclusions: 1) E = 10.000.000 Pa = stress/strain = 100.000 Pa/(extention/originalsize) extention/originalsize = 100.000/10.000.000 = 0.01 As you can see the balloon only gets 10% bigger. 2) The latex can stand a lot more pressure so it will not explode. Ok, so the most important thing is that the balloon gets 10% bigger. (Actually it will not even be that much as the pressure inside the balloon will get lower as the balloon expands, but not much). Now we can calculate what the influence will be on the water! Actually (at 25C°) water starts boiling at 0.01 of normal pressure. ( I found a nice drawing of the curve at the following website. http://van.hep.uiuc.edu/van/qa/section/States_of_Matter_and_Energy/Boiling_Evaporating_and_Condensing/976072539.htm ) Conclusion is: the water will not boil. Please correct my mistakes I'm only a student industrial ingenieer. Not a prof

Lol, actually that last drawing is a mistake. I made some mistakes in the perspective... nevermind That 4m is placed incorrectly. Looks more like an Esscher Drawing.

Statica and Dynamica: I had 3 question: Question 1: a boat is floating in the sea. (v = speed, a = acceleration, t = time, s = distance) The boat has a speed of v =0,8t. And it is floating in a direction shaped like a circle with a radius of 120m. What is its speed and acceleration after floating a certain distance, lets say it was 20meters? Answer 1: I solved this in 4 steps. a) Because the speed was declared in function of time, I started by calculating how long it would take to reach the 20 meters. v = ds/dt v dt = ds int(v dt) = 20 int(0,8t dt) = 20 (0,4t²) = 20 t² = 20/0,4 = 50 t = sqrt(50) or (50)^(1/2) = 7,07 seconds according to my calculator b) now it's easy to find the speed namely: v = 0,8*7,07 = 5,66 m/s c) now the acceleration: We have 2 components. There's a centropetal acceleration (an) and a straight acceleration (at) touching the curved shape direction of the boat. Both cross in an angle of 90°. For the force in the direction of the boat: at = v/dt = 0,8 m/s² For the centropetal acceleration (aka middlepoint-seeking acceleration): an = v²/r = 5,66/20 = 0,28 m/s² luckely we know the speed (5,66m/s) allready! d) Now all what's left is bringing the 2 forces together that's easy ofcourse: sqrt(0,8² + 0,28²) = 0,85 m/s² Question 2: Here we go again. Jup a giant framework . As you can see there are 2 external vertical forces. At the right the framework has a little wheel. And at the left it is fixed to the ground in all directions. Oh yeah, and those 4 rods at the bottum each have a length of 1,5m. And the height of those 2 small vertical rods is 2m each. The question is: what is the pressure or tension inside the 2 red rods. And give all rods that have no tension or pressure at all? Answer 2: a) I allready marked all rods without forces, this can be done on sight: the pink ones. b) Next I calculated the responses in the left and right point. Because as we know the sum of the external vertical forces has to be 0. The question is how much at the left, and how much at the right? c) There's an easy trick to do this. Namely: the sum of the moments around a point has to be 0. So I calculated the moments around the bottum right point. 7,5*(2*1,5) + 7,5*1,5 + external_force_bot_left*6 = 0 external_force_bot_left = -5,63kN The right one can be calculated the same way. Or a faster way now: Sum of all vertical powers = 0. 7,5 + 7,5 - 5,63 + X = 0 X = -9,37kN d) Now that we know the vertical power in bottum right. We can easely calculate the right red rod tension. Because it's the only rod with a vertical force-component. For this we need to calculate it's length first though: sqrt(1,5² + 2²) = sqrt(6,25) = 2,5m (not that it matters, but that's a 3-4-5 ratio). The vertical force component is: Tension_Or_Pressure = -9,37*(2,5/2)= -11,71kN. So far so good. That's the first rod. e)Now we are at it it is easy to calculate the horizontal rod in the right-side corner. Because the sum of the horizontal forces has to be 0. This makes: horizontal_rod = (11,71/2,5)*1,5 = 7,03 kN. f) The 2nd rod is harder... We have to calculate a moment again. For these I am cutting out some part of the diagram. What a nice drawing ! So I cut some part off, and there's a moment around the green point in the top. I took the point at the top, because that makes it easy to calculate: rod_bot_left* 2m + 7,03*2m = 0 rod_bot_left = -7,03N Yup the same size, but other direction. Question 2 solved! Question 3: This one is actually the easiest. There's a metal construction attached to a wall with a cable. The cable is stretched in the diagonal 3D direction. And the metal construction is also connected to a smooth vertical tube. But as you can see the tube has a square shape, this means it will give horizontal reactions, but no vertical-reactions. The Question is: calculate the tension in the cable and the horizontal responses from the connection to the vertical tube. This picture isn't correct though. That 8m is 6m not 8! Sorry, don't feel like redrawing it . Well actually the question was a bit more complex, because I first had to calculate those 3 sizes. They were not given directly. Answer 3: Well ofcourse the vertical tension of the cable equals the weight of 45N. So that's where it all starts... cableZ = 45N cableX = 4*45/6 = 30N cableY = 12*45/6 = 90N Now we need to calculate the size of that force: Tension = sqrt(45²+ 30² + 90²) = 105N The horizontal responses from the connection are: -30N and -90N ofcourse. ==> the opposit direction of the cable.

I just had my exam this morning. Big success! Don't know my score yet, but probably pretty high. Thank you for helping me in my studies.

According to my prof and my book there is a difference between a vertical rod and 2 vertical joining rods. And they should be regarded as 2 different subjects. The difference is that there is a turning point present. This changes the way pressure is handled and always makes the forces point in the direction of the pieces. Actually according to what my book and prof have told me, the pressure and pull always go in the direction of the rod. On the other hand, I must agree that 10N would make sense, It's the first reply I would have given as well. But feelings and emotions have of course nothing to do with Statica . Actually what I didn't show last time, are the numbers below. If you look at the huge numbers I get, just before I get 90. (I copied them below). As you can see the first side goes to +infinit, and the other side goes to -infinit. I guess this gives a total different answer. When the pieces are perfectly horizontal, then suddnely both powers (tension and pull) are present: the huge positive one and the huge negative one. The total pressure or tension is 0, or undefined actually. (Then again nothing is really 100% horizontal in reality.) force angle ------- ------ 10,000 0 10,002 1 10,006 2 10,014 3 10,024 4 10,038 5 10,055 6 10,075 7 10,098 8 10,125 9 10,154 10 10,187 11 10,223 12 10,263 13 10,306 14 10,353 15 10,403 16 10,457 17 10,515 18 10,576 19 10,642 20 10,711 21 10,785 22 10,864 23 10,946 24 11,034 25 11,126 26 11,223 27 11,326 28 11,434 29 11,547 30 11,666 31 11,792 32 11,924 33 12,062 34 12,208 35 12,361 36 12,521 37 12,690 38 12,868 39 13,054 40 13,250 41 13,456 42 13,673 43 13,902 44 14,142 45 14,396 46 14,663 47 14,945 48 15,243 49 15,557 50 15,890 51 16,243 52 16,616 53 17,013 54 17,434 55 17,883 56 18,361 57 18,871 58 19,416 59 20,000 60 20,627 61 21,301 62 22,027 63 22,812 64 23,662 65 24,586 66 25,593 67 26,695 68 27,904 69 29,238 70 30,716 71 32,361 72 34,203 73 36,280 74 38,637 75 41,336 76 44,454 77 48,097 78 52,408 79 57,588 80 63,925 81 71,853 82 82,055 83 95,668 84 114,737 85 143,356 86 191,073 87 286,537 88 572,987 89 ??? 90 <<== Not Inifinit , but unknown +oo -oo = Undefined -572,987 91 -286,537 92 -191,073 93 -143,356 94 -114,737 95 -95,668 96 -82,055 97 -71,853 98 -63,925 99 -57,588 100 -52,408 101 -48,097 102 -44,454 103 -41,336 104 -38,637 105 -36,280 106 -34,203 107 -32,361 108 -30,716 109 -29,238 110 -27,904 111 -26,695 112 -25,593 113 -24,586 114 -23,662 115 -22,812 116 -22,027 117 -21,301 118 -20,627 119 -20,000 120 -19,416 121 -18,871 122 -18,361 123 -17,883 124 -17,434 125 -17,013 126 -16,616 127 -16,243 128 -15,890 129 -15,557 130 -15,243 131 -14,945 132 -14,663 133 -14,396 134 -14,142 135 -13,902 136 -13,673 137 -13,456 138 -13,250 139 -13,054 140 -12,868 141 -12,690 142 -12,521 143 -12,361 144 -12,208 145 -12,062 146 -11,924 147 -11,792 148 -11,666 149 -11,547 150 -11,434 151 -11,326 152 -11,223 153 -11,126 154 -11,034 155 -10,946 156 -10,864 157 -10,785 158 -10,711 159 -10,642 160 -10,576 161 -10,515 162 -10,457 163 -10,403 164 -10,353 165 -10,306 166 -10,263 167 -10,223 168 -10,187 169 -10,154 170 -10,125 171 -10,098 172 -10,075 173 -10,055 174 -10,038 175 -10,024 176 -10,014 177 -10,006 178 -10,002 179 -10,000 180 How come constructions like this don't break in real life? There's no material that can hold nearly Infinit pressure or tension.

Sorry, I meant "what if alpha is 90°". Not 0°, but 90°... my mistake. F = 10N/cos(alpha) Does this mean... 10N/cos(90°) = 10N/0 = Infinit?

Hi, I have been studying Statica for the past month. And I can say with proud I am finally starting to get it. I am stuck however on a problem I made up myself. Something is probably wrong in my way of thinking. Excercice: Let's say there is a framework of 2 equal horizontal iron pieces both are attached to opposing walls at the same height. In the center they are joined together. Next there's a cable hanging in the connection point. The cable is connected to a weight of 20 Newton. The angle of the cable (alpha) as seen in the picture = 89,99° (or almost PI/2) My answer: I solved it in 3 steps... 1) first of all the sum of the external vertical powers should be 0. This makes that in point A and B there is an vertical power-component of 10N. 2) 2*(cos(89,99°) * the_internal_pulling_or_pressure_inside_one_of_the_cables) = 20N or ... the_internal_pulling_or_pressure_inside_one_of_the_cables = 10N/cos(89,99°) The result of the last equation is: 57 296 Newton 3) Point A and B have both a vertical power of 57296 Newton (= sin(89,99°)*57296N) My question towards the readers of this: First of all is my way of thinking correct? As you can see the weight of 20N created an internal force of 57.296 Newton!!! And what if alpha would have been 0. Does that mean the internal force becomes infinit? This can't be correct can it? Thank you in advance