an 80cm long, 1mm diameter steel guitar string must be tightened to a tension of 2000N . how much is the string stretche????\ni don't know if im using the right equation.....\nchange in length = force * initial length/ area * Y (youngs modulus) - but i don't know the value for y so how do i work it out???
216288
looks to me like that is the correct equation to use.
47566
Don't screw the bridge to the body. Well, I guess it wouldn't be too big of a deal if it's a guitar you don't really care about and are using as a back- up or extra. I use my tremolo bar very rarely, and I usually don't even have it attached to the bridge. \n\nIf you are having trouble tuning, you should probably take a look at the bridge itself. Are the saddles in good shape? Or, how about the springs? Either of those can be fixed or replaced fairly easily. How about trying some different strings? Do you change between tunings often? One more option for you is to have the tuners checked by a pro.
544521
That'd be your B string. Take the peg out of the bridge and put the end of the string that has the little round bit in. Then place the peg back in and pull the string so that it locks in place under the peg. Pull the string up to the head (it's the bottom middle peg that it needs togo through) and put it through the peg. You'll want a bit of slack, but not too much. About enough that you can put both your hands under the string one on top of the other should do it. Then hold the string down at the nut (the little plastic bit where the neck meets the head) and start turning the tuning peg toward the body. This will make the string wrap around on the correct side so it doesn't rub the E string. While you are doing this make sure to keep pressure where the string is on the nut and don't let the peg come out of the bridge or the winding string pop off the top of the tuning peg. Just tighten it up and there you go!\n\nHope this helps! Good luck!
597119
1 ksi = 1000 psi\n\nWhat do your points represent and how did you get that answer?\n\nLater...\nE= stress * length/deflection, or force*length/(deflection*area)\nIf the tester was zeroed, your result is non-linear - .004 deflection for the first 150# and .006 for the next 150#. Could indicate yielding - or slipping. If you're sure of your work, check the data and the setup.
432909
how long is a piece of string?\n\nit depends where in the world you live, which city, whether you're going to buy it online or in a store, brand new or second hand.\n\nwithout details we can't tell you anything.
470236
Force acting down at the center:\ng*m = 9.81*60 = 589 N\n\nTwo tension forces, one to the left and one to the right at angle tan-1(1.2/25) = 2.75 degrees.\n\nForce acting up at the center equals force acting down\n2*T*sin(2.75) = 589\n\nSolve for T\nT = 589/(2*sin(2.75)) = 6130 N
257674
Take a rope ( that doesn't elongate/stretch easily), wind round both pulleys (just the way belt goes)and mark one end where it touches the other. Measure this and the size in inches is the size of the belt. As to whether 'A' or 'B' or 'C', it is based on cross-section. The smallest is 'A'. I hope I am understood.
83501
ok so for A, the first set has 2 seats. so between the 2 seats, theres 2.5feet of steel. and then from each seat to the end, theres 2 more feet (on each side) so thatz 4 feet + 2.5feet = 6.5 feet for the first set.\nfor the 2nd set, 3 seats so 2*2.5 between the seats (2.5 between seats 1 and 2 and then 2.5 more between seats 2 and 3) + 4 =9feet of steel for the 2nd set.\nfor the 3rd set, theres 4 seats so that means (3*2.5) + 4 = 11.5feet of steel.\n\nso you should get the pattern hopefully. if theres 5 seats, then 4*2.5 + 4. 6 seats is 5*2.5 + 4.\nso then the expression would be [(n-1)*2.5] + 4. i dont really know how to explain why its correct. i guess you could just say  it works and it makez sense. for n seats, theres n-1 gaps between the seats and each gap is 2.5feet long. and then theres alwayz 2 feet on both ends added to it which is the +4. \n\nanywayz hope this helps. let me know if u need mroe help or whatever. laterz
469860
a) As the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso w²= g/(L cosa)  \n\nso v = r sqrt(g/(Lcos a)) = 1.4 sin 30  sqrt( 9.8/1.4 cos30) = 1.99 m/s\n\nb) 4 = 1.4 sin a sqrt (9.8 /1.4 cos a)\n\nc) As mw²R = S sin a\n    and mg = S cos a\n\nw = sqrt (S²- m²g²) /mR
298840
Remaining length of string\n= 7m - 2.56m\n= 4.44m\n= 444 cm\n\nHow did I do it, below is the working.\n7m is also same as 7.00m.\n\n  7.00\n- 2.56\n------\n  4.44\n------\nCancel the last 0 to 10.\nCancel the middle zero to 9.\nCancel the 7 to 6.\n\n1m = 100cm\n4.44 m = 100 cm * 4.44 = 444cm
593651
Another string theory, \nG – hmm, \nwhy not then a G-string theory?\n\n\nCurios question it is. Let us as assume that you will put a geosynchronous satellite at about 300km from the center of the Earth. Since it is a geosynchronous satellite it will stay with the same point on the Earth say over your house.  The air resistance will be minimum as I‘ll assume that the planets atmosphere stays with the planet. Here are some numbers:\n\nEarth rotates at Ve= 1,700 km/hour (diameter ~13,000 km)\nOur geosynchronous satellite Vs= ~40,000 km/hour (at 300,000 km above center of the Earth)\nThis was just to show off, now let us do some serous computation.\n\nAnd the weight of the string say at 1gm/meter will weigh …. will weigh about 126.87 kg by employing the inverse square law of gravity.  (These were very tough computations).\nThat will apply a force of about 1,300 N on that satellite.  Why apply a steady pull? The weight of the string will bring the satellite down.\n\nThe daunting question is how are we are going to attach one end of that string to the satellite in the first place and then getting the other end to your house ;)? Or perhaps…\n\nThanks for all the fish
521457
Young's modulus, also know as the modulus of elasticity, is a mechanical material property that defines the elastic nature of a material. For example Young's Modulus (E is used as symbol for this property) of steel; E=3x10^7 psi (pounds per square in). You would need to read up a bit on the definitions of stress and strain to understand further, but if we use the simple stiffness equation of a bar: d = (F * l ) / (A * E ) where d is the displacement, l is the bar length, A is its cross-sectional area, and F is the applied force. It is basically telling you how much the bar will stretch if you pull on it.\n\nPoisson's ratio, is also a mechanical material property. It is usually 0.3 - 0.33 for metals. Simply explained it is how much a material shrinks perpendicular to its loading direction. Imagine an elastic band if you strength it, it not only becomes longer, but the strands become thinner; the amount that the strands thin compared to the amount that they stretch is defined by Poisson's Ratio.\n\nThe notion of a datum is completely unrelated to mechanical properties. A datum refers to a reference (on an engineering drawing or 3D model). Its like the origin of a coordiniate system, often used to simplify how the dimensions are drawn on the drawing. A datum is usually given a letter designation like A , On the drawing you would see it as a straight line with the A in a box touching the line.\n\nNot sure if datum geometry is basically the same thing. The only thing I can think of for that is that there could be different types of Datum i.e. a plane, a point, a cylinder, or a sphere (but I have only come across datum planes in my work, but I could imagine the other shapes could be useful under certain situations.)
267837
1) Take the old string off.  Sometimes there is a peg that holds the string in on the body of the guitar.  If this is the case, you may need to find something to pull the peg out.  Be careful not to use something that will scratch the wood.  Unwind the rest of the string from the capstan on the neck.\n2) I assume you have a new string to replace it with.  Put the string in the slot in the body first.  Slip the other end of the string through the hole in the capstan.  Be sure to leave enough string so that the capstan will have enough to grab when you wind it.  In other words, don't pull the string tight before you wind it. Leave some slack.\n3)Hold the string in place above the capstan and start winding.  It helps to have one of those winders as it makes the job so much faster.  If you don't keep winding.\n4)Don't let the string slip over the top of the capstan.\n5)As the string tightens pull up on it to make it seat better.\n6)Don't over tighten.  I've broken a few strings this way.  As the string starts to get tight pluck it to see if it is getting close to where it needs to be.\n7)You may want to clip off the excess after you are finished.  It makes the guitar look neater.\n8)Be prepared for the string to stretch after a minute or two. You will have to keep tuning it until it gets fully stretched.  It may keep doing this to a lesser degree over a matter of days.\n\nLet me know if you need more specific details.
5140
It sounds to me like you have a problem with the key. By this I mean the itemthat you twist to get a string to get tighter. This is often a problem in cheap guitars.\n\nI am not sure what you mean about "C". The Guitar has strings on a normal tuning as follows:\n* E\n* A\n* D\n* G\n* B\n* E\n\nWhat do you mean by "C"?\n\nI have played guitar for over 25 years. So I do have a clue about this beyond personal experience in a general sense. You can contact with me directly, but I am not sure if I am allowed to post my email due to the answers rules. Therefore I do not post it
5513
I always liked my strings as tight as possible but really it depends on how you play the game. Here's the general grid:\n\nStandard: 50-60\nMidsize: 55-65\nOversize: 60-70\nSuper Oversize: 60-70
279517
You need to find the spring constant, k, which should be in units of force (e.g. meters per Newton or m/N) per unit  length (m).  In this case:\n\nforce (N) = mass, m (kg) times acceleration, a (m/s^2)\nm = 0.2 kg\na = g = 9.8 m/s^2\n\nf = ma = (0.2 kg)(9.8 m/s^2) = 1.96 N\n\nk = force, F (N) divided by elongation, delta L (m) = F / delta L\n{The actual equation has a negative in front. Don't worry about the negatives in this exercise. Since force is in the opposite direction of the extension, one of them would be negative, and therefore would cancel out the leading neagtive sign.}\n\ndelta L = 0.1048 - 0.1037 = 0.0011 m \n\nk = 1.96 N / 0.0011 m = 1781.81 N/m\n\nFn = 0.5(3.14)sqrt[1781.81 / 0.2] = 148.26 Hz
461516
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.