What r 2 factors that affect the friction force between 2 surfaces.?
343753
the type of surface\nthe surface area
858739
Use Bernouillis equation and you will have your answer. You can use the equivalent lengths method to find the friction.
362474
It's almost impossible to calculate that, for air friction there is a physical constant that is mentioned in mechanical physics. but for a table, there is still a massive amount of friction being applied from the table. so i don't think that it can be calculated since the air friction changes in regards to speed. it could probably be approximated in a lab thou.
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
587906
Whether friction force depends on velocity or not depends (further) on what sort of friction you are considering.\n      Friction between a solid-solid interface is seldom affected by the relative velocity of the bodies. There, in fact, the force merely depends on the NORMAL reaction on the surface. For fluids, however, friction is substantially affected by the relative velocity. You may check out "NEWTON'S HYPOTHESIS" & "STOKE'S LAW" in this context.
1171207
Let us write coeff. of friction = k     \n                    acceleration due to gravity = g     m/ sq. s\n                    uniform velocity of motion = u      m/s\n                 \n                    Since k=F/R where R = weight of the box acting downward and F = Frictional force acting against the direction of motion.\n                   Hence, F = kR. Again F= ma, where a is the retardation on the box and R=mg.\n                    Therefore, a=kg. (m cancels out on both sides).\n\n                     By using the formula:\n                            v^2=u^2 - 2aS  ( here a is retardation) we have\n                      S= u^2/(2a) = u^2/(2kg).\n\n            Substituting the values:\n                         u=8 m/s\n                         k= 0.16\n                         g=9.8 m/sq. s\n                        The result is S= (8^2)/(2*0.16*9.8) = 20.41 m (approx.).
109063
Strictly speaking, friction is force that is the product of a constant for the friction producing device (coefficient of friction) and the perpendicular force between two rubbing surfaces. When sliding occurs, friction does not change with speed, but stopped (static) friction may exceed sliding friction.  Traction typically refers to the force that prevents a tire from skidding and may include friction, adhesion, hydrodynamic, aerodynamic and aerostatic (suction) effects.\nEDIT:\nAnd contextually, traction is almost always beneficial while friction may be.
543978
Frictional force, f = uR\n\nwhere u - frictional constant\n(it is actually "miu", but I can't typr it out.)\n\nand R - the normal/reaction force (which is the reaction caused by the action of the object's weight.)
609398
FRICTION is the energy generated when two bodies have some physical interaction with each other. \n\nGRAVITY is the force of a mass which pulls an object towards its center.
757648
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
751590
1. Changing roughness - make the surface smoother\n\n2. Instead of sliding surfaces together, you can add rollers between the surfaces, thus changing over to rolling friction.\n\n3. By adding a thin layer of oil or even water between two objects, you convert the sliding friction to fluid friction. This lubrication is used in many applications to reduce wear and to make the sliding much easier.
1161373
Air resistance is fluid friction. Air resistance occurs between the surface of a falling object and the air that surrounds it. Rolling friction occurs when a rounded surface moves over a solid and sliding friction occurs when a solid moves over another solid. Static friction occurs when a solid touches another solid but there is no movement.
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
345166
Friction, Heat, Acid.
65267
It's me again, Dhinesh,your classmate.\nrolling frictin is the friction caused due to rolling. it opposes rolling & it acts on a single point as there is a single pt. which has contact with surface
856335
Here are the key things you need to know:\n\nForces acting on the small box:\n- gravity attracting its mass (pulling it downward)\n- large box (pushing it upward)\n- friction force between the two boxes (pushing it forward)\n\nForces acting on the large box:\n- gravity attracting its mass (pulling it downward)\n- the weight of the small box (= the force of gravity on the small box) (pushing it downward)\n- friction force between the two boxes (pushing it backward)\n- friction force between the large box and the underlying surface (pushing it backward)\n- the horizontal force (pushing it forward)\n\nFirst, treat the two boxes as a single mass, resting on the surface and being accelerated by the horizontal force. Calculate the force of friction between the large box and the surface (remembering that the weight of both boxes is acting to push it against the underlying surface). Calculate the force needed to both overcome this friction and to accelerate the mass of the two boxes at the specified rate (the second part of this requires the formula F = ma). That is the horizontal force.\n\nNow calculate the amount of force required to accelerate the small box at this rate. (Use F = ma again.) Divide this force by the weight of the small box to find the minimum coefficient of friction that will allow the large box to apply this force to the small box.\n\nIf you understand all this, you should be able to draw the two diagrams and answer the questions.\nGood luck!
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.