I have a 4' beam with a support 3' from one end, with a fixed load on that end. How do I calculate the needed weight to counterbalance that load?
133805
If you need an exact balance, you will need too weights (combined).  You will need 3/4 the weight of the beam at the end of the short end to balance the beam (unless you can neglect that).  You then need to balance the weight (w) at the end of the long end with a counter weight CW) at the end of the short end.  A formula that woorks should look like this:\n\nW x 3ft = CW x 1ft\n\nCW = (W x 3ft)/1ft\n\nCW = 3W (plus 3/4 beam weight)
528670
Neat. You need 1, 3, 9, 27\n\nPut weights on each side:\n\n1: 1 on left, nothing on right\n2: 3 on left, 1 on right\n3: 3 on left\n4: (3,1)()\n5: (9)(3,1)\n6: (9)(3)\n7: (9,1)(3)\n8: (9)(1)\n9: (9)()\n...\n13: (9,3,1)\n14: (27)(9,3,1)\n...\n40: (27,9,3,1)\n\nThe general rule is you need powers of 3, and the reason is that each weight can be counted on the left 0 times, 1 time, or -1 times.\n\nNeat.
156693
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.
460956
The previous answerer has the right idea but the wrong details.\n\nThe four parts should weigh 1, 3, 9 and 27 pounds.  The logic is that the desired weight is written in "balanced ternary" notation, where the values of the digit positions are 1, 3, 9, 27, 81, . . . (that's "ternary"), but the allowed digit values instead of being 0, 1, 2 (that's regular ternary) are -1, 0, 1 (that's balanced ternary).  Then the digit values for any desired total weight written down in balanced ternary tell you whether the weight for that column is to go in one pan, go in the other pan, or stay on the floor.
501728
T = mg.\n\nThe length of the cable is irrelevant in this example. The beam is hanging off the cable.\n\nThe tension in the cable therefore is m*g, which is 2450 N.
68855
If it is a fixed pulley, you need the same force as the weight of the load which is 9.8*0.2 Newtons (9.8 is the acc. due to gravity and 0.2kg is the mass of the load) or simply 200gf (where gf is gram force). Now you can tell one half of this number.\n\nAlternately, if you have a movable pulley, the force required is half the weight of the load, i.e., 100gf.  You can tell one half of this number too.\n\nIf you need to find the difference between fixed and movable pulleys, look up a physics text (or science text), I am not totally familiar with the schooling system here.   Otherwise, Google should show you pictures of fixed and movable pulleys.
510432
Use a floating point, or gimbaled hydralic plates to adjust as needed. this is also a known Earthquake Resistant design.
146542
you have to be more specific then that. To get weight one would use triplebeam balance, electronic scale, spring scale, or any other type of scale I guess.
503761
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
469263
Don't you just need to balance forces in orthogonal directions and then use these components to get the forces along the cable?\n\nSince the sign is stationary, vector sum of forces must be zero.  Sketch a free body diagram and label known forces.
199508
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.
117013
that's a real vague question, try to be more specific or at least say what king of structure do u want? (steel, conc, ..?)\nanyways this is an exremely simple example.\n\nlets' say u want to design a tension member -carrying tension force = 100 tons, -between 2 supports and this member is a steel member of yield stress= 1.4\nso to get the area u need :\n100/1.4= 71.43 cm square let's say again u want it to be I-shaped steel then u'll need  I number 400 i.e IPE 400
441372
Three weighings.\nPut 4 on each side and weigh\nTake the 4 from the side that is heavier and \nPut 2 on each side and weigh.\nTake the 2 from the side that is heavier and \nPut 1 on each side and weigh.\nTake the one that is heaver.  It is the heaviest one.
282043
I doubt that a "planning board" would likely follow or be much swayed by your calculations of barycenter, especially since they use the generalization "most" which is anything over 50%.  Perhaps you can concentrate each geometric building at its vertical centerline (center of square(s) and rectangle(s), circle(s), etc.  Then assign a "mass" to each "vertical point" on a ground plan.  Does "most" refer to building volume or mass?  Likely volume which you can scale from 3-D?\n\nDraw two lines (north-south and east-west) through approx. center oof plan.  Fix one end of the first line (north-south?) and assume the buildings rotate about the fixed point (with the plan held in a vertical plane and the N-S line horizontal).  Multiply each building "mass" times the distance to the fixed point (in feet) to obtain all the "moments" about the point.  Find the balancing moment in the opposite direction (a single force for the combined mass of buildings times a distance "x").  Repeat the process with the E-W line horizontal to find "y".  The barycenter should be at the intersection of the balancing moments ("x" and "y").\n\nRepeat the process with the new buildings included.  Compare the pre- and post-building project barycenters.  Good luck.
517714
before the hiker boarded the bridge was balanced with 4055/2 on each end.\n\nWhen the hiker boarded, his weight moves the center of mass to the enar end.  Walking 1/5 the length, the mass (weight) is distributed as follows:\n\nnear end:  4055/2 + 915*4/5\nfar end:  4055/2 +915/5\n\nj
540821
(tension in crossbar) * (its height) =(half of ladder weight) * (length of ladder * sin(30/2) / 2)
646682
Cosidering that the load (2P) has been applied to the midpoint of the bar, the load which will be carried by each cantilever beam will be 'P'.\nIn that case, the maximum deflection at the free edge of the system will be  \n\n(P*L*L*L )/ (3EI)\nwhere L is the span of each beam \nEI is the flexural rigidity