help me with friction?
917525
Normal force is you're weight from the force of gravity.   And you can't increase the friction force because it is directly related to your weight.  The main reason for taking smaller steps is to make sure that the force that you apply does not exceed the friction force.  By taking longer steps, you are essentially pushing off harder on the ice to propel yourself horizontally.  And if this horizontal force exceeds the frictional force of the ice, you're going for a tumble.
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.
882649
First solve for the force of friction between the block and the table using the formula,\n\nf=uR, where f is the force of friction, u the coefficient of static friction, and R is the force normal to the table. R therefore is the weight of the block equal to 31g.\n\nf=0.425*31*9.8\n =129.1N\n\nThe weight of the sand added to the bucket plus the weight of the bucket itself should be just slightly equal to 129.1N so that the system will begin to move. At that point, however, the block will begin to accelerate because the force of friction acting on it will no longer be static friction but kinetic friction, which is lower in magnitude. Thus there will be a net force equal to 129.1 minus the force of kinetic friction on the block. Thus,\n\nNet force, F=129.1-0.320*31*9.8\n                 =129.1-97.2\n                 =31.9N\n\nF=ma, where F is our net force of 31.9N, m the total mass of block plus mass of sand and bucket, and a is the acceleration of the system.\n\nMass of sand and bucket=129.1/9.8\n                                    =13.2kg\nSubstitute known values:\n\n31.9=(31+13.2)a\na=31.9/44.2\n  =0.72m/s^2
913014
F(net) = m x a\n\nF (net) = F (forward) - F(air friction)\n\nso  F(air friction)  = F (net) - F(forward)\n\nMass is given 20 kg\n \nnet acceleration given 2.0  m/s^2\n\n\nSolve for the Net Force, subtract the forward force, the difference is the force due to air friction.
613880
sorry can't help you bro.  But from my experience with these types of things in general, all I can say is DON'T SCRATCH
905447
The net force is 0 Newtons. The force of friction is 100N.\n\nThe key here is that\n1. No acceleration.\n2. Horizontal force only.\n\nThis means that the forces are balanced in the horizontal direction. The object isn't accelerating so net F = ma = 0.
910713
at the moment there are the horizontal component of weight and the friction that opposes it are in equilibrium. we must apply a force that with the already present weight compontns will equal the kinetic friction. \nmeaning that \n\nWx + Fp = Friction.\n\n500cos20 + Fp = 0.6*500sin20 =>\n\nfp =367.24N
909261
basically it is the amount of friction, caused by movement, or no movement. below is a GREAT link to explain. hope this helped!!!!\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html
734736
first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me
737967
Draw a free body diagram as shown below: \n\n                [m1]------> T<---[m2]------->89N\n              <--------            <---------\n                f1 =19.6N        f2=29.4N\n\n\nFirst solve for f1 and f2:\n\nf=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values\n\nf1=0.2*10*9.8\n   =19.6N to the left\nf2=0.1*30*9.8\n   =29.4N to the left\n\nLooking at the free body diagram:\n\nfor m1:\n\nnet force, f1=T-19.6\n                      \nFrom the formula, f=ma  where f is the net force, m the mass, and a is the acceleration.:\n\nf1=10a\n\nThus,\n\nT-19.6=10a\n         a=(T-19.6)/10\n\nfor m2:\n\nNet force, f2=89-29.4-T\n                  =59.6-T\n\nSimilarly as for m1:\n\n59.6-T=30a\na=(59.6-T)/30\n\nNow a for m1 is also the same a as for m2, therefore:\n\n(T-19.6)10=(59.6-T)/30\nThe zero on each side cancels out. \n3T-58.8=59.6-T\n4T=59.6+58.8\n4T=118.4\n  T=29.6N in the direction shown in the above free body diagram.
920572
coefficient of kinetic friction = Friction Force / Weight\n\nFriction Force = (coefficient of kinetic friction)x(55x9.81)\n                       = 188.84 N\n\nSo Net Force = 220 -188.84 = 31.16 N\n\nF=mxa,  a=F/m=31.16/55=0.5665 m/s2
882327
easy ;-)\n\nthe friction force, call it F, is equal to the normal force (the force perpendicular to a surface - in this case, the car's weight), times the friction coefficient, call it k. And the weight would be equal to the mass, m, times g.  So you'd have\n\nF = k * m * g\n\nnow the acceleration is the force divided by the mass, so:\n\na = F / m = k*g = (0.642)*(9.81)  (in metric)\n\nthe formula you suggest to use is the correct one ;-)\n\nnow i'm used to metric so i'll convert your mph to m/s, 40mph is 17.8 m/s.\n\nso you have\n\nd = (17.8)^2 / 2*(0.642)*(9.81)\nd = 25 metres\n\nif you want feet, and don't want metric in between, then you'd have to use 32 ft/s2 and not 9.81 m/s2 for g.  And you'd have to convert your 40mph in ft/sec (about 58.3 ft/s).\n\nhope this helps
939533
I would use conservation of energy.\nx is the distance traveled on the ramp\nthe forces at work are:\nGravity.  The potential energy will be\nsin(5)*m*g*x\n\nFriction:\n.48*cos(5)*m*g*x\n\nSince there are no brakes all of the kinetic energy will be lost to friction and the potential energy gain:\n\n.5*m*v^2=\n.48*cos(5)*m*g*x+sin(5)*m*g*x\nnote that m divides out\nI get 110 m\n\nj
845875
Ah, time to dig out the physics from last year.  If I remember correctly:\n\nStatic friction is when the force of friction is greater than the force applied; thus basically what amount of force it takes to move the object from rest.\n\nKinetic friction is when you apply a force for an object to keep sliding.  In this, it "breaks free" of your push and continues sliding.\n\nAnd the force needed to initially break the static friction (to get it moving) is much more than what is needed to keep it going.\n\nHope this helped a bit.  Just remember that static friction is associated with objects at rest and kinetic is objects in motion, and don't get the two mixed up. :*P\n\nHave fun with physics!
205045
obviously you have to use the least steep slope, which will cause your marble to move slower. i think you are trying to find ways to slow down the marble itslef??? you would want to increase friction, and this can be done by applying something onto the marble or onto the slope (my favourite would be honey or syrup) im not sure if that is what you were looking for?? good luck
67664
friction is F=uN where u is the coefficient and N is the normal force.  In this case 260N is the normal force. So:\nu*260N=150N* cos 40\nYou must subtract off (150 * sin 40), also because you are counteracting part of the normal force.\nSo the final eqn should be\nu = (150N* cos 40) / (260N - (150* sin 40))\nLet me know how it goes\nemail: ke7alw@yahoo.com
515562
Friction, Heat generated to name 2 reasons.