Rate of Torque Application?
443097
A flexible coupling connects two shafts, end-to-end in the same line, for two main purposes. The first is to transmit power (torque) from one shaft to the other, causing both to rotate in unison, at the same RPM. The second is to compensate for minor amounts of misalignment and random movement between the two shafts. Belt, chain, gear and clutch drives also transmit power from one shaft to another, but not necessarily at the same RPM and not with the shafts in approximately the same line. \n\nSuch compensation is vital because perfect alignment of two shafts is extremely difficult and rarely attained. The coupling will, to varying degrees, minimize the effect of misaligned shafts. Even with very good initial shaft alignment there is often a tendency for the coupled equipment to "drift' from its initial position, thereby causing further misalignment of the shafts. If not properly compensated, minor shaft misalignment can result in unnecessary wear and premature replacement of other system components. \n\nIn certain cases, flexible couplings are selected for other protective functions as well. One is to provide a break point between driving and driven shafts that will act as a fuse if a severe torque overload occurs. This assures that the coupling will fail before something more costly breaks elsewhere along the drive train. Another is to dampen torsional (rotational) vibration that occurs naturally in the driving and/or driven equipment. \n\nEach type of coupling has some advantage over another type. There is not one coupling type that can "do it all". There is a trade-off associated with each, not the least of which can be purchase costs. Each design has strengths and weaknesses that must be taken into consideration because they can dramatically impact how well the coupling performs in the application.
658877
I have tried this free program but I have to pay for the full program if I want to work correctly... If you can help me out then you are the the "bomb"...\n\n\nRegistry Mechanic 5.2
269311
assuming that force is applied in the same direction that displacement....\nwork = force * displacement * cos(angle between force and displacement)\nwork = (10N)*(5.0m)*cos(0)\nwork = 50 Joule
311499
Work = force * distance\n= 1000 N * 10 m\n= 10000 Nm\n\nYou wouldn't take into consideration the mass of the car because the force you are applying is horizontal (assuming that you are pushing straight on the car and not down at an angle).  The only time you would use the 3000 N is if you took friction into consideration.  In that case, you would have a friction force opposing your 1000 N, thus reducing it.  (Force of friction = coefficient of friction * normal force)\n\nPower = work / time\n= 10000 Nm / 100 s\n= 100 Nm/s
859502
What you are likely looking is W=F*D \n\nWork = A Force * A Distance\n\nCheck out wikipedia for additional help:\nhttp://en.wikipedia.org/wiki/Mechanical_work
989843
net force IN THE DIRECTION OF DISPLACEMENTis (1.7*10^6 * cos 14)*2 \n= 3.3* 10 ^ 6 N\nwork done = 3.3*10^6*860=2.838 * 10^9 Joules
456275
9.81 m/s² or 32.2 ft/s²
94996
Torque is the rotational force applied at the diferential--put simply, it's how much force is applied at the drive wheels
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
829645
This is half a question! Would you be able to understand this question without more info? Are answerers supposed to be clairvoyant? But... read on.\nHow was the calculated MI calculated? Experimentally?\nYou show a 3.6% error. This could be considered reasonable experimental/measurement error. Could you have ignored (perhaps you were told to ignore) some contributors to error, such as: mass of a connecting string; error in pulley diameter (exactly what length of string per revolution can be difficult to determine because of string thickness); error in effective radius of mass of pulley (if it's the pulley MI you're dealing with) or wheel. Add more information to this Q via the "Additional Details" option and you may get a better answer.
205454
well i have no clue what your units are. but torque is force x distance. and if the drum is 12 inches in diameter, its 6 in radius. so the distance from pivot point is 6 inches. and the force would be um 200lbs? i guess. if that could be considered a force. and then multiply. \n\ni can do it in SI units for you though. 1 in = 2.54 cm = 0.0254m. and 1lb = 2.2kg. \nso 6 in = 15.24cm = 0.1524m. and 200lb = 440kg. F = 9.8*440 which is around 4180N. so 4180N * 0.1524m
963784
tau = I*(alpha) where tau is the torque\nand alpha is the angular accelleration, \nI is the moment of Inertia = 0.5M*R^2 (for a cylinder)\nTau = r*F = 0.16m*40N \n6.4 = M*(0.0256)*90rad/s\nHence M = [6.4]/(0.0256*90)\nM = 2.78kg
979659
Torque is defined as\n\nT = r x F = |r|*|F|*sin(q)\n\nwhere r is the distance vector, F is the force vector and q is the angle between them. In equilibrium the torque from the two forces balance each other out.\n\nT1 = r1*F1*sin(q1) = r2*F2*sin(q2) = T2\n\nSince F1 = 2 N and is applied at the end of the meterstick, r1 = 1m. Also, it is applied perpendicular to the stick so q1 = 90 deg. Thus T1 = 2 N-m. F2 = 6 N and the angle q2 = 30 deg. Use the above equation to solve for the distance, r2.
414568
.75 seconds, and Acceleration
512039
I hope this is what you were looking for: \n\nLet A = your acceleration(RPS^2)\nLet D = your deceleration(RPS^2) (use positive value)\nLet RPS = your target velocity in RPS\nLet RPM = velocity in RPM\nLet N = the number of revolutions the motor needs to turn\nLet t (sec) = time needed\nSqrt = square root. Make sure you pay attention to the brackets.\n\nThen RPS = ( -t + Sqrt(t^2 + 2N(A+D)/AD))x AD/(A+D) \n\nAnd RPM = RPS/60\n\nI believe this formula is correct since I checked that both sides of the equation have the units of velocity.\nHowever, I have noted that many people on this site give out incorrect info. It is always a good idea to get a second source confirmation. Perhaps a servo motor manufacturer or distributor could provide the info and give helpful tips for a specific application. Also physics or math faculty at most colleges and universities are pretty co-operative when it comes to requests like this. If you fax the details they'll usually respond. Best wishes!
131445
Torque is force*distance.\n\nFind the weight of the person (M*g) = 89kg *9.81m/s^2 \nM*g= 873.09 Newtons\n\nthere is a 1:3 lever arm there--> 4cm/12cm\nThe muscle is having to exert a force of 3*873.09N or 2619N\n\nThe torque around the pivot is 2619N @ 4cm (or .04m)\n2619N * .04m = 104.77N*m\n\nYour question says you want to "find the torque" yet you ask for the "forces pushing up".  Make sure you know which one you need.
984947
W=F*L – work done by F=145N on distance L=7.2m,\nU=k*m*g*cos(a)*L, work wasted on friction, where k*m*g*cos(a) is force of friction, while\nk=0.31, a=11.7deg, g=9.8m/s^2, m=10.3kg.\nY=m*g*L*sin(a) work done to hoist m to height of L*sin(a) meters.\nE0=m*v0^2/2 initial kinetic energy of m, v0=1.3m/s \nNow energy balance: E0+W=U+Y+E1, E1 is final kinetic energy of m E1=m*V^2/2, V is final speed of m. Thus kinetic energy change dE=E1-E0=W-U-Y.\nAs E1=E0+W-U-Y=m*V^2/2, then V=sqrt[2*(E0+W-U-Y)/m], \nI hope you cope with numeric calculations, all are SI units