Explain the force-distance trade-off that occurs when a machineis used to make work easier?
368874
think pully or wrench. \nwhy do we use pully to lift things, we replace distance for force.\nbut the total work = force *distace is fixed.\nanother example when you use a pipe to make a wrench handel longer fo you can turn the screw. \nIt is technically torque but similar concept.
946504
Intensity varies as the square of the distance.\nThat should help
110280
Load Arm - Distance between fulcrum and load\nEffort Arm - Distance between fulcrum and effort (effort - force applied by you)
753429
Presuming you are talking about the work equation J = Fd; where J is work in Joules, F is force in Newtons, and d is distance moved in meters, then the answer is...yes you can do too much work or not enough.  Why?\n\nBecause J/F = d; which means you work to move a body a distance d.  If you were to push a mass with a fixed force F, but put in less work (J) than necessary, your desired distance moved would not be achieved.  Conversely, if you work more than required, you'd overshoot your desired distance.  Note, too much or too little work requires that a given d is specified as a required distance to move the mass.\n\nBecause you failed to describe your experiment in detail, I can only presume you are pushing various masses of various material along some surface for some specified distance (d).  Then you are calculating the amount of work done J = Fd for each mass of different material.  F in these experiments would be sliding friction force = kN; where k is the coefficient of sliding friction and N is the weight of each mass perpendicular to the surface.\n\nWhat you should discover is that the F's for each material will be different; so that, even if your d is the same for each trial, work will vary with the force exerted to slide the body.  That is, more F means more work and vice versa.  In each case, the amount of work performed is just sufficient to move the mass d meters...no more, no less.  But if you push too far, you will have put in too much work...too short, not enough work.\n\nThe F's will differ because of different k's and different N's.  In this experiment, if d is not specified, then the work done in each case is just sufficient (no more or less) than what was needed to move each different d distance.
809675
Force times distance is not kinetic energy. Force times distance is work. Work can appear in different forms. If you expend 100 Newton-meters of work pushing a dog very slowly up a frictionless inclined plane, the dog will have 100 Newton-meters of potential energy when you are done, and no kinetic energy, for it will stop moving when you stop pushing.\n\nOn the other hand, if you expend 100 Newton-meters of work pushing the dog along a horizontal frictionless slippy-slide, the dog will continue sliding after you stop. It will have no potential energy and 100 Newton-meters of kinetic energy.\n\nDo you want to know how fast the dog will be going when you stop pushing? Use\n\nE = (1/2) * m *  v^2
93773
well force is measured in kg*m/s^2 so, force will cause a function of time and distance to increase exponentially over the interval of the force applied.\n\nFor example, an object dropped from 100 meters will go 5 meters in the first second, 20 meters in the 2nd second, and  25 meters after 3 seconds, etc.
239996
Work done = force x distance moved by object i thinkin a straight line ,so ithink some conversion maybe needed.
572487
Distribution of force over area.
3541
all you have to do is calculate the work that the engine should have done.  W=Fx meaning work equals force times distance.  The force is mass times acceleration, so it would be 10 x 9.8 (9.8 meters per second squared is the acceleration due to gravity) giving you 98 (lets round to 100 so i dont have to get a calculator or think too much because im tired, and your other data only has one significant figure anyway, if you think you need it more precise you can go get a calculator and do the work im going to show you with 98 on a calculator...)\n\nanyway, so to find the work (W=Fx) you multiply the force we just found (100N) by the distance so you get 100N x 4m = 400J.\n\nNow you know that the engine should have done 400J of work, but instead it did 500J, meaning that it did extra work, so it's not 100% efficient, so to find the percentage of its efficiency, just divide 400/500 and multiply that by 100 to get the percentage.  So its .8 x 100 so 80%.\n\nAnd if i may ask, why do you keep asking very basic questions on completely random physics topics that you would clearly not be doing all at once, and if you already did all of it, you should already know these answers?
208227
It means that for each 2 turns on one gear, there are 3 turns on the other (though I assume you know that already).  The efficiency is related to friction loss-the greater the ratio difference, the greater the force transer and thus friction loss, so the lowest ratio difference is the one with the least friction loss.\nThere is no change in distance from the fulcrum on a see-saw unless you are talking about the very small bending of the board, and that is likely too little to overcome other small forces, such as friction, movement of the people on the seat, etc.
991659
the key to the tip is the "movable block".  mechanical advantage refers to the relationship between how far (how hard) you pull the rope versus how much the load moves.  if you have a rope running through a fixed block and down to a load, you can change the direction of force (you pull down, the load goes up), but you have no mechanical advantage.  for every foot you pull the rope, the load goes up one foot.  \n\nif you have a fixed block and a movable block attached to a load and the rope runs through the fixed block (F) once, down to the movable block (M), through M and back up to F, you have two parts through M (one down and one up).  if you attach a load to M and pull, the load goes up 1/2 foot for each foot of rope you pull, so you have a 2:1 mechanical advantage.  remember, the rope running through F at the top isn't giving you any mechanical advantage, its just changing the direction of your pull from up to down.  if you stood above M and pulled up, you'd get the same 1/2 foot up for each foot you pulled and the same 2:1 mechanical advantage.  \n\nso yes, the tip seems to work.  try it on paper for some other blocks, for example, double blocks at F and M.  \n\ngood luck!
619423
Pretty general question not knowing forces needed, space conisderations, and operating conditions. Lots of gears are stock and can simply be ordeed online IF you know the parameters.
42606
ADVANTAGES: Most powerful mechanical computational devices, able to solve complex problems such as playing games or recognizing patterns, can model complex mathematical equations (such as paterns in virus structures to find cures).\nDISADVANTAGES: Prohibitively expensive for home use, generate tremendous ammounts of heat and thus need to be kept in strictly controlled environment, relatively expensive to design and maintain, are ussually designed with systems to solve a specific problem or type of problem and are thus difficult to switch from one task to another, require a great deal of space.
207863
see:\nhttp://en.wikipedia.org/wiki/EngineerIng
69843
work = force times distance\nW=Fs
890476
It is often a problem in gyms with machines using pulleys.  If the pulley is a single one, then whatever weight is on the stack is what you're lifting, but many machines have some sort of mechanical advantage, so there might be 100kg on the stack, but you aren't actually exerting 100kg force (kgf).  On Earth a kilogram mass exerts a kilogram force downward, so we can ignore the "force" term.\n\nI keep a little retractable tape in my gym bag so every time I come across a new machine, I do the following (it's easier if you have someone to help you):\n\nMeasure the distance the weights actually rise during the full range of the exercise, and multiply that by the total weight.  That gives you total work in kg-metres.  So you might have 50kg on stack, and your shoulder press raises it 0.45m. So that's 22.5kg-m of work.\n\nNow measure the distance your hands move through the exercise (usually significantly further than than the weights). This might be 0.65m.  Divide your work (22.5kg-m) by this distance (0.65m) to get the force at your hands, in this case 34.6kg. So if you did a free weight shoulder press, around 35kg would give you the same effort as 50kg on the machine.\n\nA special case is the 45 degree sliding leg press machine.  Your legs may push the slide about 0.5m, but the weights have only risen vertically by 0.5 x sine 45 deg, or 71% of the slide distance i..e. 0.35m.  So pushing the 200kg on the rack is only 141 kg force at your feet, and your total work is 71kg-m.  That doesn't count the platform weight of course.\n\nI always record work as well as weight, as a heavy weight is not necessarily heavy work.  As an extreme example, anyone can lift a wheel of a 1-ton truck off the ground with one hand - using a jack!
650834
it depends on what machine you use for digging, if you want to open a tunnel like channel then you can use tbm's(tunnel boring machine) it gives you machine force, if you use another machines then the time will be increased