What is the algebraic formula to solve ultimate strength on a simple steel beam and will you explain it?
444257
The ultimate strength of a beam is found by US = max load/CSA.  There's not a lot of math to it, but you do need to measure pretty carefully.\n\nThere comes a problem, however, where some materials "stretch" before actually failing, thus making the measurement of the max load difficult.  As an example, if you've ever pulled plastic wrap apart, it gets easier at a point right before it breaks into two.  The point where after you pull on it and it gets easier is called the yield strength.  After you pass the yield strength, you don't need as much force to cause failure in the material.\n\nNow in steel, the yield and ultimate strengths are pretty close to each other that the above discussion really doesn't affect it if you are trying to break a beam on your own.  But the ultimate strength of steel is probably something you can look up in a table somewhere.  Then you can use the above formula to see how much your beam can take based on the US and the cross sectional area.
987606
usually they'd use it in engineering\n\naddition:\nwell, they use in the alogarithm form.i don't know any 'simple' analogy but the easiest way to explain to a child might be, they use it in engineering to calculate the strength of structures and stuffs like that.\ntry http://en.wikipedia.org/wiki/Exponentiation
820893
that depends on what kind of steel.  Not every steel is made the same way or with the same density so it varies.  I can tell you that a W14X56 is a 14" wide flange beam and weighs 56 lb/ft
797314
this guy explains it as easily as can be explained\n\nhttp://www.newton.dep.anl.gov/askasci/chem99/chem99297.htm
860005
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
361455
This question is both complex and easy; easy because the way to illustrate or model the shear stresses in a beam requires a simple diagram that cannot be drawn on here, and easy because once you see the procedure you will realize it is actually quite straightfoward. \n\nThe best thing for you to do is grab a Materials' Engineering book -- most will have, at some point, an accurate answer to this question.  Try your mechanics and dynamics books too, as this bordelines on many subjects.
1187223
Well, a clear question would help. \n\nAre you trying to calculate the power that a certain material can withstand?   Or are you trying to calculate strength?
901912
the moment is found by force times distance\n\nI_____v______\nI-------------------V\n\nlets say that the diagram above is the beam, the little v in the center represents the mass of the rod and the big V and the end represents the extra mass.\n\nthe moment required to resist the other forces is given by the mass of the rod times the distance from the wall (half the length of the rod) times gravity plus the extra mass on the end times the distance from the wall (full length of the rod) times gravity:\n\n(1.5 kg) * (9.81 m/s^2) * (.4 m) + (.5 kg) * (9.81 m/s^2) * (.8 m)\n= 5.886 + 3.924 Newton meters\n= 9.81 N m
456850
Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
895385
"Give a student an answer, and their grade on that assignment will go up a point, teach them how to find their answer and their score on the final exam will improve!"\n\nRadioactive decay is an exponential thing; for every 24000 years that elapses in this problem, half the remaining material vanishes.  If you express this as an algebraic equation you can figure this stuff out easily.  The hard part is figuring out the equation.\n\nLets say you start out with X_0 grams of material.  The amount of material remaining is X: \n\nX= X_0*(1/2)^(t/24000)\n\nWhere t is given in years.  See how this works?  if t = 0, then X=X_0*1/2^0 power.  Anything raised to the 0 is always 1, so X=X_0.  At t=24000 years, X=X_0*1/2^(1), so X=X_0*1/2; half as much is left.\n\nIn this problem you are given t, and X and are asked to find X_0 so  you use a calculator to turn 1/2^(21500/24000) into just a number and then you have\n\n4 grams=X_0*number\n\nPretty easy to solve that for X_0.  The point is, once you write down the equation that describes the system, you can manipulate it to solve for any of the terms in it, as long as you know the other two terms.\n\nIncidentally, I get a different answer than those other two folks who agree with one another.  They are wrong, but not by a lot.  Work it out for yourself.\n\n[Edit:  In fact I get the same result IMI does, so my method does work. Her answer was not posted when I started working on mine.]
489536
By strength I suppose you mean elastic calculation when the body is subjected to deforming forces. Actually, the centroid of a shape having an unifrom mass distirbution coincides with its center of mass. This point is central to the analysis carried out in rigid body statics and dynamics.
170255
It is much easier to just give you a link instead of explaining it:\n\nhttp://www.shef.ac.uk/physics/teaching/phy107/solvegray.html
560218
instead of counting the number of flux to determine the strength of the force, i suggest u hang a piece of metal on a string, then investigate the angle of the string when u put a magnet at different distance. Because with this experiment, u can calculate the force much more easily through the angel and length of the string with the knowledge of gravitational force.\n\nThe flux suggestion is good, but how do u determine the force? There r much more complicated theory behind it and they r less intuitive than gravitational force.
816694
hi \nbasic formula to be used here is f=ma\nhere we have 2 scenarios elevator going up(against gravity of earth denoted by 'g' which is constant eual to 9.8 or lets say 10m/s ^2.\nnow when elevator is going up means it has to defy gravity and go up which means there has to be more force to accomplish\nhence we use f=70 x (g+a)\nwhich is f= 70 x(10+3)=910N\nwhile the elevator is moving down it doesnt need to defy gravity and hence the force required to accomplish this will be much less\nhence we use f=70x (g-a)\nwhich is f=70x(10-3)=490N\n\nis the elevator is standing still then \nf=m x a\nwhich is f= 70 x 3 = 210n
880047
I can find this on a TI-84+, is that how you want to get it or do you need an algebraic solution?
144487
the answer is exponents.
860576
Let the tension of string attached to 8.7 kg be T1 and that attached to 2.1 be T2. Let common acceleration be a. Cos 23 = 0.921\n\nequations are:\n\n8.7a = 8.7g - T1\n1.7a = T1 - T2 - 1.7gx0.14xCos 23 = T1 - T2 - 0.219g\n2.1 a= T2 - 2.1 g\n\nSolve thees equations first for a then for T1 and T2\n\nI f you have doubt about the validity of any one of the above equations  you may ask your specific doubts.