calculation of crankshaft's torque?
604852
Here are a couple of web sites that approach this in different ways. \n\nhttp://www.circletrack.com/techarticles/84199/\n\nhttp://www.diracdelta.co.uk/science/source/e/n/engine%20excitation%20mechanisms/source.html\n\nHope these help.
31430
The amount of torque applied is force * length.  At 1 cm from the hinge, a 5 N force provides 5 Ncm of torque.  At 85cm, you would have 85 times as much torque, or 425Ncm.
860448
22-d, 28-b, 30-c\n\nA baseball has rotation and translational motion.  Torque is a measurement of the ability to produce angular acceleration, while force is related to the translational acceleration (hence 28-d is false)  28-a and 28-c are equivalent, which should have been a clue that 28-b is the right answer.
358957
Volume = pi * r^2 * h\n\nr is 1/2 the radius of the drilled hole = (D'/2)\nh is equivalent to the diameter of the original cylinder = D
785816
h = 4tanθ\nVh = dh/dt = (dh/dθ)(dθ/dt)\ndh/dθ = 4/cos^2θ\ncos^2θ = 4^2/(4^2 + 19^2)\nVh = (16 + 361)/4)(5/361)\nVh = (5/4)(16/361 + 1)\nVh = 1.3054 km/sec
514815
OK, the barrel´s mass is 40Kg (notice: its mass is 40 Kg, not its weight). \n\na) So, your formula is Weight = mass(m)*gravity´s acceleration(g)= 40Kg*10m/s^2 = 400 KG*m/s^2 (=400 Newton)\n\nb) Work = force * distance\nSince work is calculated only for the vertical, and the force equals the weight, you have:\nWork = 400N*5m = 2000 Joule\n\nc) Eff = 250N*5m/2000J*100 = 0.625*100 = 62.5%\n(You got 12.5% because you forgot to multiply 250N for 5 meters. Remember that efficiency is a relation of works, and work is force * DISTANCE).\n;o)
378184
When the cylinder rolls, there is a force F=15N at the top and a static frictional force f preventing slipping.\neverything in usual notation,\nacceleration A=R*a (a=angular acceleration)\nNewton's second law implies'\nMA=F-f         (resultant force)\nalso for torques,\nIa=FR+fR         (for the axis passing through the centre)\nFRom the two equatins,\n2F=mA+Ia/R\ni.e.  2F=maR+Ia/r\n   a=2F/(mR+I/R)\nI=0.5MR^2=0.05\nplugging hte values,\na=16 \ntherefore, A=aR\n            =16*0.1\n           A=1.6
964672
The answer is  - (0.635 - 0.324)* r * g = 2.9 Nm.\n\n\nYour equation for torque and other things are correct.\n\nFor the system to be in equilibrium the NET torque must be zero.\n\nSuppose that 0.635 kg mass rotates the pulley in clockwise direction.\n\nThen the other mass 0.324kg will rotate the pulley in the anti-clockwise direction.\n\nTorque produced by 0.635 kg about the axis is F x r = 0.635*r*g. (Let us take clockwise rotation  as positive)\n\nTorque produced by 0.324 kg about the axis is F x r = - 0.324*r*g. (anticlockwise direction)\n\nThe net torque about the axis is (0.635 - 0.324)* r * g = 0.311*r*g. \n\nIf there is no friction, the pulley will rotate in the clockwise direction by the action of this torque\n\nIf the frictional torque acts opposite to this torque, then the net torque will be zero and the system will be in equilibrium.\n\nTherefore the frictional torque is - 0.311*r*g. (anticlockwise direction)\n\n\nSubstitute the value for r = 0.95 m and g = 9.8m/s^2\n\nThe frictional torque is 2.9 Nm.
980357
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
1029495
It is impossible to tell from the information given.  YOu need to know the power factor and efficiency too.  \n\nAt full load of a motor that big it is likely to be 95% efficiency and have power factor of probably 0.8.  \n\nIF (!!) that is true, then the current is (75 hp * 746 W/hp)/(480  V *sqrt(3))/0.95/0.8 Amps (rms)\n\nNotice I've considered a wye-connected motor too - that's where the sqrt(3) comes from.  I also get 90 A (like the previous guy)!
55453
The horsepower of a motor is dependent of two parameters: speed (RPM) and torque (lb-in or lb-ft). \n\nPower = speed*torque (carefull with the units!!)\n\nAs you see, the horsepower not only depends on the motor's speed, but also on the torque applied.\n\nYou cannot measure the horsepower unless you can meassure these two parameters.
171644
It stands for Raw horsepower and Raw torque. Did you need to know what horsepower and torque were too? Wasn't sure.......maybe I should add that too eh??\n\nHorsepower: Awhile back, a gentleman by the name of Watt (the same gent who did all that neat stuff with steam engines) made some observations, and concluded that the average horse of the time could lift a 550 pound weight one foot in one second, thereby performing work at the rate of 550 foot pounds per second, or 33,000 foot pounds per minute, for an eight hour shift, more or less. He then published those observations, and stated that 33,000 foot pounds per minute of work was equivalent to the power of one horse, or, one horsepower. \n\nTorque: A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum. \n\n\nOK. Remember Watt? He said that 33,000 foot pounds of work per minute was equivalent to one horsepower. If we divide the 6.2832 foot pounds of work we've done per revolution of that weight into 33,000 foot pounds, we come up with the fact that one foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower. If we only move that weight at the rate of 2626 rpm, it's the equivalent of 1/2 horsepower (16,500 foot pounds per minute), and so on. Therefore, the following formula applies for calculating horsepower from a torque measurement: \n\n\n                                Torque * RPM\n\n        Horsepower      =       ------------\n\n                                    5252\n\nLol...besides the math....here is an article that might help you understand what you want to look for in your car. Good luck :) (but pay off the house first!!)
400555
for automobile\n\nFormulas for brake horsepower\nhorsepower = rpm x torque / 5252\ntorque = 5252 x horsepower / rpm\nbrake specific fuel consumption = fuel pounds per hour / brake horsepower\nbhp loss = elevation in feet / 1000 x 0.03 x bhp at sea level\n\nFormulas for indicated horsepower & torque\nhorsepower = mep x displcement x rpm / 792,00\ntorque = mep x displacement / 150.8\nmep = hp x 792,000 / displacement x rpm\nmep = hp x 792,000 / displacement x rpm\nmechanical efficiency = brake output / indocated output x 100\nfriction output = indicated output - brake output\ntaxable horsepower = bore2 x cylinders / 2.5 \n\nElectric Motor Horsepower Calculations, see http://www.bacharach-training.com/norm/electric.htm\n\nPrimer for understanding Horsepower, http://vettenet.org/torquehp.html
746380
Torque = change in angular momentum over change in time\n\nT=Iw/t      T=mrrw/t     T=1.1(.09)(93.6)/1.7  T=5.45 N/m
290214
Its the standard measurement for torque like ft in leght/ it is how many pounds of pressure it can produce  /the chevy duramax diesal can produce 650 Lb-Ft of torque the best in it's calss/ these is one of the things that allows it to pull more easier
11536
There are actually several formulas I know of to calculate power.\n1. Work (Joules) divided by time (seconds)\n2. Force (Newtons) times velocity (meters per second)\n3. Current (Amperes) times voltage (Volts)\n4. Current squared times resistance (ohms)\n5. Voltage squared divided by resistance\n\nIn your case, applying the first formula gives you no power in the system since no work is being done unless the blades accelerate tangentially (centripetal force is not parallel to the direction of motion). If there is definitely a non-zero answer to this question, then use the second formula.
834512
Torque is not the right question to ask in this problem!\nFirst define the generator output in terms of watts.\nAssume an efficiency of 0.8 for the generator and define the Input of the generator (In = Out / 0.8).\nThis is the output of your motor (assuming some diesel/petrol).\nTorque is expressed as N/m (force x radius) and is part of the electric generator specs.  See there to get the equivalence.\nUsually, the motor power is also 0.8 efficient, so the motor power should be W / (0.8*0.8) and that will give you the right ratio.