What is the formula/equation for finding friction force?
544891
The formula is actually divided up into two cases:\nForce friction = Force Applied when Force Applied < mu Static x Force Normal (i.e. nothing'll move below this threshold)\nForce friction = mu dynamic x Force Normal otherwise.\n\nThere is no formula for the two mus; they are measured experimentally.
707332
The forces on the person are\n\nSFx = 0 = N - Fc\nSFy = 0 = Fs - m*g\n\nwhere SFx and SFy are the sums of the forces in the x- and y-directions, respectively, N is the normal (to the wall) force, Fc = m*v^2/r is the centripetal force, Fs = u*N is the friction force, m is the mass, and g is the gravitational acceleration.\n\nThese equations say that\n\nN = Fc = m*v^2/r\nu*N = u*m*v^2/r = m*g\n\nSolving for the coefficient of static friction\n\nu = g*r/v^2\n\nRemember to convert the frequency (rev/s) into speed (v)\n\nv = w*r = 2*pi*f*r\n\nwhere w is the angular speed and f is the frequency.
74421
centripetal force is mv^2/r.\nso it comes around 1,777,777.778 N\n\nI need one more parameter that is coeff of friction.\nif that is available u can use\n\nv(max)= sqrt [(rg (u + tan o)/(1 - u * tan o) ]\n\nassuming frictionless\n\nV= sqrt (rg tan o)\nso \no = 66.2 degrees if g=9.8\nand 65.7 degrees if g= 10.
900588
from the other guys F fric = 26.4\n\nF fric = Coeficiant x  F normal = 26.4\n\nF normal = mg\n\nso divide 26.4 by (3.2)(9.8) and you got the coefficient of kinetic friction
935603
The net force is F2 - F1 = 25 - 19 = 6N.\n\nFrom F = ma, a = F/m = 6/5 = 1.2 m/s^2, toward the west because F2 > F1.  Also, we're ignoring friction.
937470
Getting started, have to overcome static friction.  Therefore, force applied = static friction force just before movement\n\nF = umg\nF = 75N\nm = 20kg\ng = gravity\nfind u, plug and play\n\nsame problem for sliding friction
704245
the constant velocity indicates that the 30.0 N force is just enough to overcome the force of friction\n\nSo use 30 N in the friction formula:\n\nmu=Force of Friction / Normal Force , Normal Force = Weight of Crate\n\nmu = 30.0/(9.81 x 12 ) = 0.2548419979612640163098878695209
528746
Average Kinetic coefficient of friction (m) cannot be calculated as such, \n      since  m can lie anywhere between 0 and 1.\n   \n      Thus, its value is probably more near 0.5 than many values. That's all that can be infered.
747947
Weight of bundle=4*g\nWeight component normal to the incline=\n4gcos30.\n\nForce of friction,f =uR where R is the force normal to the incline\nor 4gcos30.\n\nSubstitute given values:\nf=0.30*4gcos30\n =10.2N\n\nKE is given as equal to 128J. \nWork=f*s, where s is our unknown \n\nBased on principle of conservation of energy:\n\nKE=Work\n128=10.2s\ns=128/10.2\n  =12.5m
205046
the kinetic energy of the sled and person is equal to the work done by the friction to stop the sled.  Since work is force times distance in this case, the force can be calculated.  From the force you can find the coefficient from the definition of friction by dividing out the normal force.  \n\nor literally, 0.5*m*v^2 = F*d = mu*m*g. solving for mu\nmu= v^2 / (2*d*g) = 0.016\n\ni think....
39453
Static friction is the frictional force needed to get an object at rest to move on a give surface.  Kinetic friction is the friction force needed to keep an object moving on a give surface.  If you know the object, and the surface, you may be able to find the Coefficient of friction and calculate the simple frictional forces involved in movement.  Static friction is always greater than kinetic friction for the same object and surface.  Also, increasing the Normal force will effect the total friction.
878213
f = ma = kN = kmg; so a = kg = .15*9.81; where a is the deceleration, k = .15 the friction coefficient and g = 9.81 m/sec^2 the acceleration due to gravity at Earth's surface.\n\nWork = fd = kNd = 1/2 m v^2 = kmgd; where friction force f = kN times distance (d) is work (energy in Joules), which in this case comes from kinetic energy 1/2 m v^2 where v = 5 m/sec and m is the mass of my damn physics book.  Upon rearranging factors to find d, we have:\n\nd = 1/2 v^2/(kg) = v^2/(2kg); where d is the stopping distance, v = 5 m/sec, k = .15 the friction coefficient, and g = 9.81 m/sec^2 from gravity.\n\nYou can do the arithmetic.  The important physics lesson learned here is that neither answer (a or d) depends on the mass of that textbook...and both depend in large measure on k in that larger k means greater deceleration and smaller k means longer stop distance.
742117
work = f*d\nso we just need the force since we know d=5\n\nthe vertical force on the surface is:\nmg-sin(15)*250\n=90*9.8-sin(15)*250\nto compute the frictional force, multiply by the coefficient of friction\nand by the distance for the work\nF=5*.2*(90*9.8-sin(15)*250)\n=817.3\n\n\n\nj
772638
try the formulas:\n\nforce = mass * aceleration\n\nwork = force * distance\n\n\n\n\nWell i dont speak english very well, but i hope i help u.\n\nKisses
857685
forces on M : grav. force Mg (vertically down) tension 2T(two strings horizontal, one on pulley) tension T (vertically down due to string on pulley) Normal reaction on M at lower surface N (vertically up) Normal reaction on M due to m R (to the left) Friction due to m on M f1=u1 times R (vertically down)(sliding is going on) friction in M at lower surface f2=u2 times N (to the left) (sliding is going on)\nnet force on M vertically up :ZERO\nto the right M times A\nforces on m  : grav force mg (vertically down) reaction due to M R (to the right) tension T (vertically up) friction due to M on m f1=u1 times R(vertically up)\nnet force on m to the right m times A\nnet force on m vertically down m times a\nunknowns N,R,a,A,T,f1,f2,\nwrite a constraint eqn for relating accns a and A. a=2A \nsolve them!!!!
818860
I'm not going to give you the complete answer because you'll learn more by working through it yourself, but I'll give a few pointers.\n\nAny object travelling in a circle at a constant speed has a centripital acceleration that can be calculated from the velocity and the radius of the circle.\n\nNewton tells us that F[net] = ma, so the object must be experiencing a net centripetal force equal to its mass multiplied by its centripetal acceleration.\n\nWhere does this force come from?  In your problem, it sounds as though the only force that could be pulling the box toward the center of the circle (the centripetal force) is the friction between the box and the bed of the truck.  So, when you add up all the (horizontal) forces to get the net force, the olny one is going to be friction.  F[net] = F[friction] \n\nNow all you need to do set F[friction]  equal to the F[net] that you calculated above (remember that F[net] was equal to your centripetal force), and solve for the coefficient of friction (your textbook shold have the formula for static friction, which will include the coefficient of static friction).\n\nGood luck!
1147561
Net force on object = Force applied - Friction force\n\nSince velocity is constant, the acceleration of the object is zero\n\nSo, Net force = mass * acceleration = zero\n\n0 = Force applied - Friction force\nor\nForce applied = friction force \n= coefficient of friction * Normal force\n=.250 * 11.9 kg * 9.8 m/s^2\n=29.2 kg-m/s^2\n\nBox is decelerating at .250*9.8m/s^2\n\nand\n\ndistance traveled = (final velocity^2 - initial velocity^2)/2a\n=(0 - 3.2 m/s)  /  2*(-.250*9.8m/s^2)\n=.65 m