how you calculate the mechanical advantages of  a wedge?
962609
The M/A of a wedge (disregarding friction) = cot(wedge angle)\n\nUnfortunately, there's a LOT of friction, so it's not so simple after all!
878371
The incline is a simple machine that permits less force to be applied to move an object. The work associated with moving an object on an inclined surface, in the absence of friction, is the same as raising the object of weight (m*g) through a distance H.\n\nFor the inclined plane with an inclination angle of 30 degrees, moving the mass (m) over a distance L would raise the object a distance H given by\n\nH = L*sin(30) = 0.5*L\n\nThe work then is\n\nW = m*g*H = m*g*L/2\n\nFor the values given\n\nW = (1.0 kg)*(10 m/s^2)*(0.5 m)/2\n\nW = 10/4 N-m = 2.5 N-m\n\nThe same answer can be found by looking at the net forces on the object and multiplying by the distance traveled.\n\nFu = 10 N (the upslope force)\nFd = m*g*sin(30) = (1 kg)*(10 m/s^2)/2 = 5 N (downslope force)\n\nFn = Fu - Fd = 5 N (Net force)\n\nW = Fn*L = (5 N)*(0.5 m) = 2.5 N-m
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
898267
Given: mass, m, of crate=14kg\n   therefore, weight of crate=14g\n   initial speed of crate=1.56m/s\n   applied force along the incline=150N \n   distance traversed by crate=7.99m\n   mu=0.28\n   acceleration of gravity,g=9.8m/s^2\n   angle of incline=25 degree\n\na) Initial kinetic energy, KE=1/2mv^2\n                =1/2*14*1.56^2=17 Joules\n\nforce of friction=mu*R, \nwhere R=14*9.8cos25=124.3N\n\nNet force,F=150-124.3\n                  =25.7N\n\nF=ma\n25.7=14a\na=25.7/14\n  =1.84m/s^2\n\nv^2-u^2=2as\nv^2-1.56^2=2*1.84*7.99\nv^2=29.4+2.43\n     =31.8\nv=5.64m/s\n\nFinal KE=1/2*14*5.64^2\n              =222.6 Joules\n\nChange in KE=222.6-17\n                      =205.6 Joules\n\nb) 5.64m/s as computed in a) above.
985550
consider the forces on the block:\n\nThe force of gravity\n\nThe frictional force\n\nWhen the forces are translated into the frame of reference of the surface of the slope,\nthe forces parallel to the slope are:\n\nfriction:\n0.60*cos(60)*m*g\nWeight\nsin(60)*m*g\n\nnote that in order for the block to be sliding down the slope, the parallel component of the weight must be greater than the frictional force \nor sin(60) must be greater than .6*cos(60)\nwhich it is\n\nTo compute the acceleration,\nF=m*a\nwhich is\nm*a=sin(60)*m*g-0.60*cos(60)*m*g\ndivide out the mass\na=g*(sin(60)-0.60*cos(60))\na=5.55 m/s^2\n\nb)  For this part, only the force necessary to overcome kinetic friction plus the parallel weight is necessary to keep it moving up the slope.  Since it was given a shove and we don't know how much force was applied in the shove, we can only compute the minimum force necessary to over come kinetic friction and weight\nF>g*m*(sin(60)+0.60*cos(60))\n\nj
280562
The definition of Mechanical Advantage is "The ratio of the output force produced by a machine to the applied input force."\n\nWhen you add more rope segments, you have to do less work to move an object held by a pulley, because each segment of rope is doing a piece of the work.\n\nSay you have a 100lb weight, held by 4 sections of pulley.  Each section is holding 25 lbs (including the section you're holding onto), so you're only doing 25 lbs of work.  The MA is 4.\n\nSo the answer is, "the number of rope segments".\n\nNote this is the ideal mechanical advantage.  The true MA is reduced by things like friction in the ropes.
936671
The lever gives you a 5:1 advantage, this means that what ever you get for the hydraulic advantage, you will divide the answer by 5 because of the added mechanical advantage of the lever.\n\nThink of it like this: Without the lever, the hydraulic advantage is 1.3/.3 or 4.33:1\nThe pressure to balance on hydraulic port #1 is 500/4.33 or 115.38 lbs. and the lever makes the total needed to balance 23.08 lbs. or rounded to approx. 23 lbs.
778096
The height of the ramp give potential enery in the following format\n\nE = mgh      where E is potential energy, m is the mass of the car in Kgs, g is the gravitational acceleration of the planet you're on (in meters per second per second (9.8 on earth)) and h is the height of the ramp.\n\nThe angle or slope of the ramp will not affect the potential energy, but will have an effect on the rate of change of potential energy.
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.
22465
Once you can wedge turn you want to increase the pressure to your outside (of the turn) ski while releasing pressure of the inside ski then rapidly turn across the hill and pressure both uphill edges
238117
If by "calculate" you mean process the results of experimental meaurements of speed, force, etc., then you measure the initial total energy (potential, kinetic) of the test object, measure all known energy input that occurs along the path (integral of applied external force times velocity dot product), and subtract from the total of those two the final total energy. The result is the energy lost to dissipative processes like friction and air resistance.\nIf instead you use calculate in the sense of predicting the dissipative losses, you need a model of the object (its drag coefficient and the coefficient of friction). If the path is constant-slope and speed and air density are constant, you can calculate the end state analytically because drag force is constant and so is normal force and thus friction. Otherwise you have nonlinearities which require a time-stepped simulation to solve.
473777
Wedge is a movable inclined plane; Some times two inclined plane  are joned toghther.Hence it is a simple machine,\n\nIn inclined plane, the plane is fixed and load is moved upward by effort.\n\nIn wedge load is fixed and effort and the plane are moving and hence do work.
26736
look up stuff like static friction, and your coefficient of friction...
286094
This depends on a number of factors, from the course you play on to the type of ball flight you hit, to the type of ball you play. \n\nYou should ask a good player at the club you play at for recommendations on what wedges to carry. Or better yet ask the pro at the club, he should be able to point you in the right direction.
10661
Depends on weight of the block angle of inclination, friction. Get resultant velocity by getting velocity along X (w cos theta) and velocity along y (w sin theta).... blahblahblah... check other sources but basically it goes like that.
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
376523
it depends on the angle of ur effort. lets say there u are trying to move a cart. if u push the way u will push ur mower, one component of ur force will be downwards, thus effectively pushing the cart down and thus increasing the frictional force . but if u pull the cart this force will not be downwards but upwards and hence the frictional force will be reduced. so its better to pull.