for most of the daily life situations and calculations, it is taken to be less than 1.it is also said that when two interacting surfaces are made highly smooth,friction coeffficient becomes very large.can someone please explain this fact?
209287
When the calculations were set up for determining coefficient of friction 1 was set as the upper limit. There can be no coefficient larger than 1. Highly smooth surfaces of the same shape (for example, two rectangular blocks) are in almost perfect contact with each other so that almost every molecule of one is in direct contact with every molecule of the other (that's a bit of an exaggeration unless the surfaces are micropolished, but it will serve). The extremely large surface area in contact produces enormous frictional resistance to sliding movement. It helps to understand this if you can make the distinction between smooth and slick. Think for example about trying to slide one heavy glass plate over another without any lubrication between them. They "stick" to each other like glue.
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.
547181
is the ques complete is there any friction between the boxes or any other force. can't understand the ques itself do we have a figure somewhere
358600
Shear stress is a parallel-acting force, where two surfaces (or, in your case, two layers inside the fluid) slide against each other.  This is in contrast to compression (perpendicular-acting force), tension (stretching force), and torsion (twisting force).\n\nAn everyday example might be to place your hands together with the palms flat against each other.  Now rub them, and notice that there is some resistance to the rubbing due to the features of your skin.  Try the same thing holding a piece of sandpaper in each hand, the resistance (or shear stress) will be much greater.\n\nThe same thing happens in a fluid.  If you put some fluid in between two plates, then slide one plate by pushing on the edge, you will encounter resistance (shear stress) due to the layers of fluid rubbing against each other.  A fluid like water will endure much less shear stress than a fluid like molasses.\n\nIn the hand-rubbing example, the property of the materials that quantifies this phenomenon is the coefficient of friction.  In the fluid/plate example, it is called viscosity.\n\nShear rate is simply the rate at which this shear deformation occurs.\n\nThis is a very crude explantion, but it should suffice for your purposes.
56740
I think "µs" and "Ms" are both the coefficient of static friction; they're just in different notations. "~" means "is similar to". The first equation is just saying that the coefficient of static friction and coefficient of kinetic friction are similar quantities: static is for non-moving contact surfaces, and kinetic is for moving contact surfaces.\n\n"Mk = f (v,m)" just means that the coefficient of kinetic friction is a function of both mass and velocity. This means that the coefficient is dependent of both mass and velocity (if either of them change, the coefficient will change also). It's important to note that mass and velocity are not dependent on each other.
403865
What I said about inclined surfaces works just the same on a horizontal surface; you will compute sin(theta)=0 and cos(theta)=1 so the normal gravity force=g*m and there is no gravity force parallel to the surface.\nWhen KE or PE decreases due to friction energy loss, the decrease equals the friction energy loss which is friction force * distance. I assumed you already knew that. If my answer helped I'm happy. I'd appreciate it if you go back to that question and select my answer as best; this is the normal procedure even though it would eventually be selected by public vote.\nP.S. A tip on yahoo answers. I only found this question by accident. You can add followup questions to your original question; just click on the original question and there should be a link that lets you add more words, probably called Additional Details. That's what I was referring to when I said I'd be checking for followup questions.
411876
i got 100N to start the container moving and 80N to keep it moving at constant speed, but i'm not sure if i got it right. What is the answer on the back of your book? if it is the same as mine, i will help you.\n\nstatic friction is how "rough" the surface is between the base of the container and the ground.\n\nnot sure about this one but i think kinetic friction is how rough between a moving object and the ground.\n\nJust think of you are stand on the hill. The static friction keep you still but once you start to slide down, the friction is less.
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.
369966
avg v = (start v - end v) / 2\navg v = d/t\n4 m/s = 80 m / t\nt = 80 m / 4 m/s\nt = 20 sec\n\na = change in v / t\na = 8 m/s / 20 sec\na = .4 m/sec^2\n\nF = ma\nF = (.5 kg) * .4 (m / (sec^2)) = 0.2 newtons\n\nThe coefficient of friction (also known as the frictional coefficient) is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. The coefficient of friction depends on the materials used -- for example, ice on metal has a low coefficient of friction (they slide past each other easily), while rubber on pavement has a high coefficient of friction (they do not slide past each other easily). Coefficients of friction need not be less than 1 - under good conditions, a tire on concrete may have a coefficient of friction of 1.7. Magnetically attractive surfaces can have very large friction coefficients, and, theoretically, glued or welded together surfaces have infinite friction coefficients.\n\n(.2 newtons) / (.5 newtons) = 0.4
265551
FRICTION .... causes a Polarized( + or - ) energy buildup .... which is released in the form of lighting....\n\nAs when you bush your hair and the hair is attracted to the Brush....  it's the same thing.. ONLY millions of times stronger...
410992
F - Ff = m*a\n\nThis is because the force you're applying is in one direction, while the frictional force is ALWAYS in the opposite direction. Frictional forces act to prevent the applied movement.\n\nBecause you're moving at constant velocity, you therefore have no acceleration, so:\n\nF - Ff = 0\n-or-\nF = Ff\n\nFf is the force due to friction which:\nFf = u * Fn\n\nWhere "u" (it's actually the greek letter mu, but this will do) is equal to the coefficient of friction, and Fn is equal to the perpendicular force acting on the box. On flat ground, the normal force is equal to the gravitational force, but on inclines you have to break it up into its vector components.\n\nF = u*Fg\nF = (0.20)*(25kg)*(9.8m/s^2)\nF = 49N\n\nYou see that the velocity at which you push it has no bearing on the answer because neither your applied force nor frictional forces are due to constant velocities.
441616
Kinetic friction is equal to the Frictional force divided by the Normal force  (k = Ff / Fn)\n\nStarting at the standard equation F = ma , we need to modify this because there is some friction involved.\n\nThe two forces are our Applied force (Fa) and our Friction force (Ff).  Because Ff is acting in the opposite direction, it will be negative.  Therefore we have Fa - Ff = ma.\n\nThe key to this question is to know that because there is a constant velocity, the acceleration is equal to 0.  \n\nSo Fa - Ff = 0.  However is we re-arrange this, we have Fa = Ff\n\nThis means that the Applied force (Fa) is equal to the Frictional force (Ff).\n\nAlso, the Normal force (Fn) is equal to the Force of gravity (Fg) because we are on a level floor with no vertical displacement.  Remember though that Fg = mg.  g is yor gravitational constant.  Because there is no movement in the vertical direction, it will not be negative.  \n\nSo to solve this question, \n\nk = Ff / Fn\nk = Fa / Fg  (because like I mentioned above Ff = Fa and Fn = Fg)\n\nNow work with the number that you have.  We don't have Fg but we can solve for it using the given mass (12kg) and the constant g = 9.8\n\nk = Fa / mg\nk = 30 / (12)(9.8)\nk = 0.2551020....\n\nNoting your significant digits of 3 decimal places...\n\nThe coefficient of friction is 2.55 x 10 ^ -1\n\nHope that helps you out a little bit.
47326
With a sliding mass on a level surface and subject to kinetic friction, the friction force is constant in magnitude and opposite the motion direction.  The force magnitude is the normal force (in this case = weight) times the coefficient of friction.  With constant friction force, the power required to maintain a given velocity is proportional to the velocity.  (Thus if a car were subject to friction but not to air resistance, it would use the same energy per mile regardless of speed.)  When the driving force is removed the sliding mass undergoes a constant deceleration to zero velocity.  A greater mass will have the same deceleration as a lesser one.\nSome structures that have potential vibration problems use friction dampers instead of (or as well as) the more conventional viscous dampers whose retarding force is approximately proportional to speed rather than constant.  Friction dampers have the advantage of removing a constant amplitude of vibration with each oscillation cycle, as compared wih viscous dampers that  approach zero amplitude asymptotically.  Why a constant amplitude decrement?  Energy is proportional to peak amplitude squared (X^2).  With friction damping, d(Energy) therefore d(X^2) is proportional to X, thus d(X) is constant.
478640
The force due to friction (whether static or kinetic) is dependent on the appropriate coefficient of friction (static or kinetic), between the materials, and the mass of the object (which defines the NORMAL force between the two surfaces). \n\nBy the way, the coefficient of static friction is ALWAYS greater than for kinetic friction.\n \nNow, no matter what the surface area, the normal force will remain the same, since the mass remains the same. In other words, no matter what side you put the block on, the force due to static friction and the force due to kinetic friction will not change.\n\nHope this helps!
639011
If we put two smooth glasses together (without glue them), normally what is the distance in between? \nIn an ideal situation, let's say the two glass surfaces are perfectly smooth and 100% flat, and we superpose one of them onto the other, there should be an uniform distributed air-layer between the two pieces of glasses even though it looks we put them in direct contact, so what is the thickness of this air strip? round 10nm? or more than 1 μm?\nAny guess is welcomed :)\n\n\nif ur mentioning that it is an ideal situation - glass 100% smooth etc. then it is stupid of u to ask such a question. \n\nThere wud b no air gap b/w them.\n\n(ur quest included so u may not modify it later)\n\n\nArchitectural glass is made from three principal raw materials, easily found in nature: silica, lime and sodium carbonate.\nAvg size of molecular terrain would be approximately 300 - 3000A. (neglecting secondary composition of glass). Thus logically this has to be the thickness of this strip of air.
91142
With the right number of steps forward and enough time you will eventually reach the edge and fall. The more steps you take the closer you get to the edge but never quite reaching it because your distance to the edge is merely halved with each step taken. So, the greater the number of steps forward, the closer to the edge.\nThe instant you make step number infinity you will have reached the edge. The problem is when and how many steps will that take. This is one eccentricity of infinity.
480475
Think of it like this, if you have a car on ice, there is no friction. The wheels will spin, but the car won't go anywhere. If you have a car on pavement, the wheels spin and the car moves. This is  because the friction where the wheel is touching the pavement doesn't move. Instead,the wheel to rotates and the next part of the wheel touches the pavement and 'doesn't move'.\n\nHope that makes sense. I was never all that good with physics.