The total pressure or tension is 0, or undefined actually. When the pieces are perfectly horizontal, then suddnely both powers (tension and pull) are present: the huge positive one and the huge negative one. I guess this gives a total different answer. As you can see the first side goes to +infinit, and the other side goes to -infinit. If you look at the huge numbers I get, just before I get 90. Īctually what I didn't show last time, are the numbers below. But feelings and emotions have of course nothing to do with Statica. On the other hand, I must agree that 10N would make sense, It's the first reply I would have given as well. If alpha = 90 then it's a vertical rod and (assuming there is only 1) tension would be equal to the weight of the object (assuming the system is in equilibrium).Īctually according to what my book and prof have told me, the pressure and pull always go in the direction of the rod. (internal_pulling_A * sin(89.99) ) - (internal_pulling_B * sin(89.99) ) = 0īecause the system is in equilibrium the 2 horizontal components are identical, so one subtract the other is 0. The whole of the 10 going into one thing results in 10! (That's 10 exclamation mark, not 10 factorial!)

Oh, as there is NO vertical component the force must have all gone into the horizontal component. This will act as a second proof for our big 57296 Newton 'internal pulling'. As each iron thing is at the same angle and because I am assuming this model assumes the two iron things are identical the vertical force supplied by each will be 20/2 = 10N The VERTICAL = 10N because we know the the sum of the vertical component is 20N because the system is in equilibrium. represents more decimal places)Īnd if you use the exact (unrounded answers) then use Pythagoras and it all works out.ĥ7295.7798 squared - 57295.77893 (with more decimal places) squared = 100 (which is 10 squared). Point about 3) You are rounding, they are not identical, actual value (using unrounded figures) is: 57295.77893. The way I would go about it is by doing what you did in #2 and I get the same answer: 57295.7798 (exactly) Does that mean the internal force becomes infinit? This can't be correct can it? The result of the last equation is: 57 296 Newtonģ) Point A and B have both a vertical power of 57296 Newtonįirst of all is my way of thinking correct? As you can see the weight of 20N created an internal force of 57.296 Newton!!!Īnd what if alpha would have been 0. The_internal_pulling_or_pressure_inside_one_of_the_cables = 10N/cos(89,99°) This makes that in point A and B there is an vertical power-component of 10N.Ģ) 2*(cos(89,99°) * the_internal_pulling_or_pressure_inside_one_of_the_cables) = 20N The angle of the cable (alpha) as seen in the picture = 89,99° (or almost PI/2)ġ) first of all the sum of the external vertical powers should be 0. The cable is connected to a weight of 20 Newton. Next there's a cable hanging in the connection point. Let's say there is a framework of 2 equal horizontal iron pieces both are attached to opposing walls at the same height. Something is probably wrong in my way of thinking. I am stuck however on a problem I made up myself.

And I can say with proud I am finally starting to get it. I have been studying Statica for the past month.

Almost impossible question 106.7 today