What is the principle behind the Chinese steelyard balance?
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The basic principle behind a Chinese steelyard balance is torque—or leverage, if you prefer.\n\nA symmetrical balance has the two weights equidistant from the point at which they are suspended (let’s call it the “fulcrum,” for simplicity’s sake):  when the weights are equal, the balance levels out.  But let’s say you’re trying to measure the weight of a two-pound object.  Then the balance will level out if you place a two pound weight equidistant from the fulcrum, or a one-pound weight half as far from the fulcrum, or a half-pound weight a quarter of the way from the fulcrum.\n\nThe steelyard balance puts the object to be wieghed at a specific distance from the fulcrum, then moves the known weight along the length of the opposite arm.  The ratio of the distances between the fulcrum gives the ratio of the two weights.\n\nThe link shows the best picture of one that I could find.\n\n\nRyan Wyatt\nRose Center for Earth & Space\nNew York, New York
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There is a weight mounted off-balance on a shaft so that it hangs down. It has a one-way ratchet connected to a reduction gear so that motion in one direction winds the spring. In the other direction the shaft just turns freely.\n\nAs you move about, walking, running, gesturing, whatever, your wrist moves back and forth. As your wrist changes position the weight causes the shaft to turn and the ratchet captures the motion to wind the main spring.
505706
the striker transfers some of its momentum to the coins. if it hits a coin straight (in line with the COM) and if the collision is elastic, the striker should com to a stop and the coin wil aquire its momentum.
297895
We have to insure that the coin is fairly weighted. (if one side is heavier then the other, it could land wrong.)  The horizontal surface must be completely flat.  The coin should be dropped on edge, in a repeatable fashion. (So that whatver drops it isn't angling it or affecting it.
1184958
Quite simple.\nYou pedal the cycle to rotate a gear . The gear is connected to a chain which rotates another gear connected to the wheel and when this wheel rotates, it pushes you forward.\nAs for the balance, it depends upon the rider.
973095
Unless your leaving out some necessary charge, it is already balanced. On the left side if the arrow: 1 Zn, 1 S, and 1 O2. On the right side: 1 Zn, 1 S, and 1 O2. Now if there are charges or subscripts that are being left out, then it may not be balanced. The total charge on each side must be equal. Just remember that to balance, make sure that there is the same number of atoms for each element on each side of the arrow.
230825
Teach them how to demonstrate centrifugal force by taking a metal coathanger, holding on to the hook and pulling the center of the straight part so it looks like a skinny oval with the hook on the bottom, put your finger in the end of the skinny loop away from the hook and hang the hanger on your finger so the hook is down, then balance a penny on the point of the hook and carefully spin the hanger around in a vertical circle.  The penny will stay balanced on the end of the hook.  Believe it or not, this works and if you practice it it's easy - you can even stop the coat hanger without losing the penny.  Try it!
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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,
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When the train is levitated, magnetic force of repulsion between the rail and the train balances the weight of the train and makes it suspended above the track.\n\nAccording to Newton's third law of motion, every action (the upward force acting on the train, here) has equal and opposite reaction (downward force on the train).\n\nThat is why the weight of track plus the train will load the scale and the scale will read 50 T.
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try http://www.wfu.edu/~ylwong/balanceeq/balanceq.html which is an interactive site OR try http://members.aol.com/profchm/balance.html if you want a lot of examples.
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(a) The particle is held stationary between the plates by balancing the weight of the particle with the force due to the electric field, which is expressed as the charge times the electric field (F = q*E). Since the top plate is negative and the bottom plate is positive, the sphere must be positively charged (deficient of electrons) in order for the electric field to balance the particle's weight.\n\n(b) The balanced position means that the net force on the sphere is zero so that the force due to the applied field balances the weight of the sphere\n\nq*E = m*g\n\nFor the parallel plate geomtery, the electric field is expressed as\n\nE = V/d\n\nwhere V is the applied potential and d is the separation. Solving for q in terms of V and d yields\n\nq = m*g*d/V\n\n(c) The charge deficit, in this case, is then\n\nq/e\n\nwhere e = 1.602 x 10^-19 C is the value of the elementary charge.
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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.
664775
if you're talking shaft out of balance you need an accelerometer\nother two look for steps in something, as in something is there or not there.\nThe accelerometer measures force
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When you first learn a chair that you can push and lean against can help you keep your balance while you learn.  At some rinks they have special metal bars that do this as well.
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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
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* spring balance\n    * pivot balance\n    * load cell
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Hi Either one of two answers. If the string is pulled and releases from the flywheel it allows more turns (faster wheel motion). If the string stays attached then you want the weight as small as possible.