Within a shaft/shaft coupling system, what effect does rate of torque application have on the frictional coefficient of the surfaces involved?
198911
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.
406856
Torque is a force which is applied about an axis of rotation at a distance.\n\nToque = Force * distance\nIn SI units,\nTorue = Newtons * meters = N m\n\n* It is important to note that the force here is the force which acts perpendicular to the radial distance and axis of rotation.  If the applied force is not perpendicular, you must use its perpendicular component.\n\nYou can see the implication of this formula...\na small force applied at a very long radial distance can amount to a very large toque force.  Example, using a long wrench to loosen a tight bolt requires less force than if one used a short wrench.\n\nJust like with a normal force which causes an acceleration as per Newton's 2nd law,\nF= ma\nA torque also causes an acceleration, an "angular acceleration" since the object is moving about an axis,\nT = I * alpha\nWhere I is the ojects rotational inertia, and alpha it the resulting angular acceleration (units of radians per second squared, radians measured the angular displacement)\n\nA common type of example if the torque force involved in opening a door.  A door rotates about the hinges and a force is applied on the door (on the door knob) at some distance away from its piviot point.\n\nOne might have a problem such as this,\nIf a 10 Newton force were applied to a door (with a rotational inertia of 5 kg m^2) at 1 meter away from its hinges, how long would it take to open fully (move a distance of pi / 2 radians)?\nIn this case, one has a force which acts as a torque to cause an angular acceleration and thus a displacement.\nT = F * d = 10 N * 1 m = 10 N m\nT = I * alpha = 5 kg m^2 * alpha = 10 N m\n--> alpha = 2 rad/s^2\ndelta theta = 1/2 alpha * t^2\npi / 2 radians = 1/2 * 2 rad/s^2 * t^2\nt = sqrt (pi / 2) seconds
557961
Depends, if the attraction force of the magnet is bigger then the force of friction of the disc with the oil combined with the force necessary to break the surface tension. The weight and size of the disk are also very important.
505057
They all involve friction, but I think you mean which has the most force due to friction, right? :-&#92;  If so, I believe the answer is that they are all the same.\n\nIn simple situations, the frictional force created by sliding two surfaces against eachother is given by:\n\nf = u * N\n\nWhere u (mu) is the frictional coefficient (an experimentally-determined constant) and N is the normal force, which is mg (mass times accelleration of gravity) if the surface isn't tilted.\n\nSo, assuming the frictional coefficient is the same for all sides of the book, there is no reason why changing the sides would make a difference.
377700
For your solid cylinder  I=m*r^2/2, m=10.1kg, r=0.499m,\nw=w’*t1, w angular speed 5.88 rad/s, w’ angular acceleration, t1=0.531s\nu=w’*t1^2/2 angular path to achieve w, u=w*t1/2. \nThus the total work of motor A=A1+A2, A1 useful work, A2 to overcome friction resistance,\nA1= I*w^2/2, A2=T*u=T*w*t1/2, T friction torque or torque to overcome friction.    \nNow motor off. \ny=y’*t2^2/2, y=26.8*2*pi rad, y’ friction deceleration, t2 time of deceleration\nwhile w=y’*t2, thus t2=w/y’ & y=y’*(w/y’)^2/2, y=w^2/(2*y’) or y’=w^2/(2y)\non the other hand T=I*y’=I*w^2/(2y) or I=T*2*y/w^2\nnow p=A2/(A1+A2)=\n=(T*w*t1/2)/((2y*T/w^2)*w^2/2+T*w*t1/2)=\n=(w*t1/2)/(y+w*t1/2)=1/(1+2y/(w*t1))\nthat means: to find percentage p three values – y, w & t1 – are essential, others do not matter,\ncheck it & delete everything unnecessary I’ve written here!
509015
a) 1. friction coefficient\n2. whether the bead is tilted as a whole or some portion of it. if a portion then you need to know mass distribution of patient and where the bent in the bead is.\n\nb) assuming the bead is tilted as whole then F_friction=mu*mg*cos(32deg) ,ie a projection of force due to gravity onto the normal axis.\nthe friction force is balanced by the projection of force of gravity to surface: mg*sin(32deg),\nequate the two, then you get mu=tan(32deg)=0.625
290575
It doesn't actually come in contact with anything to cause friction.  When a "machine" makes sound, it is made from the moving parts as they hit or rub against one another.  Since a moving pendulum is relying on gravity and its own weight for the arc movements, it doesn't have anything to rub against.
484248
In the absence of friction, the acceleration of an object down a 12-degree slide would be\n\n9.81 * sin 12, or around 2.04\n\nInstead, your obejct is accelerating at 1.22.\n\nThe loss of acceleration due to friction is\n\n(9.81 * sin 12) - 1.22\n\nand the coefficient of friction is thus\n\n[(9.81 * sin 12) - 1.22] / (9.81 * sin 12)
515512
You cannot get away from friction.  Material properties are also not going to be exact.
449637
the correct answer is not listed ( 32 would be correct if box 2 did not move ) \n\nsince the friction of the surface is not given the correct answer is unknown - it would be the amount of force required to overcome the friction OF THE SECOND BOX ONLY ( you said BETWEEN the boxes and when in motion force is proportional to resistance )\n\nor you did not phrase the ? properly ( likely )
482386
For most fluids, you would expect to see a negative gradient if you would plot the friction factor vs. 64 / Re ...top number is relative based on units or something.\n\nAre you sure you're plotting it correctly?  The friction factor should be decreasing as the velocity of the fluid in the pipe is increased.\n\nI'm rusty w/ my chen material so this is all I have for answer; for now.
512276
The mechanical advantage is the factor that the device multiplies the input force (so it's not subtraction, as chattycatty implied).\n\nMA = output force / input force \nMA = force on steering column / force on steering wheel\nMA = 132 / 55 = 2.4 (solution!)
128489
The Output Speed is generally the main difference. To increase the torque (strength)of the drill, the motor is geared down. The motor in the high speed drill and the high-torque drill may be turning at the same speed but the chucks or bit-holders will be very different.\nMost drills come with two speeds. If the speed is controlled by a gear assembly, then the torque is much higher at the lower speed.
20910
The last poster needs to have his dynamic physics retaught, or stop trying to sound like an engineer.\n\nThe first mistake, is identifying newtons = force x distance.  This is actually a measure of work or energy and measured in joules.  The second mistake identifying force as being measured in joules.  Force is measured in Newtons.  Since it was Isaac Newton in the second law of motion that defined Force.\n\nYour question cannot really be answered without more information.  What is the direction of motion, horizontally or vertically?  \n\nVerically?  Then 150 lbs of force.  Pounds were once used to measure force.  If you want to use metric, convert the 150 lbs to kilograms, multiply by 9.8 and then you must exert at least this much force in Newtons to overcome the force of gravity.  150 lbs is 68.04 kg.\n\nHorizontally, what type of surface?  Determine the coefficient of friction, and multiply it by the mass in kilograms.  Then this is the amount of force you need to exert to overcome the force of friction.  There are two types of friction, static and dynamic.  In other words it takes more force to move an object at rest, but once the object is in motion it takes less force to keep it move (opposing the dynamic force of friction).\n\nWatts by the way is a measurement of power, which is energy derived by time, Power = dE/dt.  Joules per second.
535045
All of the power from the motor goes to friction because they just let the wheel coast and burn up all the energy.  But, I'm sure that wasn't the intended answer for the problem\n\nYou need to make a few preliminary calculations:\nFirst you need the Inertia of the flywheel which you can find given the mass and radius of the wheel.\n\nThen you need to calculate the angular acceleration required to bring an object at rest up to 5.88 rad/s in 0.531 sec.  Using this, you can get the torque of the motor - friction.\n\nFinally, calculate the angular acceleration required to bring an object spinning at 5.88 rad/s down to rest in 26.8 revolutions.  Using this, you can get the torque of the friction by itself.\n\nAdd the second result from the first to get the torque of the motor.  Take the friction torque and divide by the motor torque times 100 to get percentage of friction.
402821
Look at flexions at the point of supprt:\n1 is the support solid so its flexure can be ignored\n2 is the steel rod held by a grub screw. If it extends any distance to lock the rod then some rotation will occur due to it bending\n3 if held in a chuck check slippage.\n\nIn motor cars torque transmission is usually accomplished using splined shafts, but for small oscillations the backlish in these will be large.
534132
your coefficient of friction would be .2608 and i'll send u an email on how i didn't it.