What is the frictional force equation?
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.)
792677
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.
118752
The fricional force between an object and a surface is dependent on two things,\nthe coefficient of friction between the two surfaces and the normal force exerted by the surface on the object.\nIf you increase either of these, the frictional force is increased, likewise, if you decrease (or make zero) any of these, the frictional force is less (or zero).
48671
Static Friction\nStatic friction is some thing that works in th plane between the object and the surface when the object is not moving .\n\nStatic friction has values from zero to the value of limiting friction force \n\nExample if it is needed 40 N force to over come frictional force .\n\nif your pulling the object at 10 N force then the object will not move as it is smaller then the force needed to move it \n\nsince your your pulling it at 10 N force and the object is not moving , then Static friction force has a value of 10 N \n\nif you increase the force form 10N to some 35N , the object will not move at this time since it is smaller then 40 N . at this point the value of static friction will be 35N \n\nStatic frictional force is equal to the force at which your pulling the object and is developed in the surface of contact and opposote to the pulling force \n\nif you pull it at 40 N then the object will have just the motivation to move and just enough to over come the limit of static friction which is called the limiting friction .\n\nthank you .. i hope it helps you
740778
Well, according to me F= (force of friction)(normal force) is the equation to be used.\n\nSo the force exerted in the x- direction is 30N and the normal force is (12kg * 9.8)N\n\nThus you get when you solve for the force of friction\n\n(30N) / (117.6N) = .2551\n\nSo the friction between the crate and the floor is .2551. \n\nThere are no units because the newtons cancel out when you divide.
1062743
D. The road.\n\nIf you draw a free-body diagram you'll see a frictional force vector pointing in the direction of motion.
1026758
friction is equal to the normal force (such as weight) times the coefficient of friction. f = c of friction X normal. if you try to push a cardboard box on the ground, its kinda easy. now imagine putting a hundred pounds in that box. its harder to push becuase there's more friction holding the box back.
897942
You should only ask one question at a time.\n\nI will answer the first:\n\nConsider the forces on the box:\n\nGravity vertically downward\nfriction pulling backwards\nThe force of gravity parallel to the ramp \nminus the force of friction\n=m*a=m*1.22\nso\nm*sin(34)*g-f=m*1.22\nfriction can be expressed as\nf=m*u*cos(34)*g\nnote that mass divides out\nsin(34)*g-u*cos(34)*g=1.22\n\nu=.52\n\nj
707407
A body in motion stays at that same velocity unless acted on by an outside force (Newton's laws of motion). So the sum of the forces acting on the book is zero since the velocity is constant. Since you are applying a one newton force in the horizontal direction, and friction is the only other force acting on it in the horizontal direction, the force due to friction must be one newton in a direction opposite to the force you are applying.
1179191
Work done by the force 2.34 N is, ‘force x distance through which the force is moved’.\n\nTherefore, it is = 2.34 N x 1.41m = 3.2994 joule.\n\nWork done by the frictional force 0.55 N is, ‘force x distance through which the force is moved’.\n\nTherefore, it is = 0.55 N x 1.41m = 0.7755 joule.\n\nThe last question is vague. \n\nThe total work is the work done by the force 2.34 N and it is equal to 3.2994 joule. \n\nOut of this total work a part (3.2994 -0.7755 = 2.5239) joule is used to accelerate the book and 0.7755 joule is dissipated as heat energy.
556094
Friction is not a force... It just is... It only exists when two objects, or things are in contact with each other.\nIf a force is acting on an object to move it, friction acts in the opposite direction.. It's simple as that. Obviously if the force is not greater than the friction, the object will not move.\nSo the direction of friction is opposite to the direction of the bicycle's movement, so in the same "Direction" for both wheels.
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
981256
Divide the Frictional Force by the frictional constant (if they're given) and you would have your Normal Force.
1054562
The important laws of friction also known as Coulomb's laws of friction are:\n\n1. The frictional force is directly proportional to the normal force between the surface at which it acts.\n(Force required for pulling a block, initially at rest, over a surface is double of that required for pulling another block, of mass twice as the former and initially at rest, over the same surface is double.)\n\n2. Frictional force always acts in a direction opposite to the direction of motion or tendency of motion.\n(If a body is moving relative to another body in contact, then the frictional force acts on both the bodies. It acts on each body in the direction opposite to that of its motion. As the two bodies are in relative motion so the direction of motion of the first with respect to the second and that of the second with respect to the first are always opposite. So the frictional force acts on the two bodies in opposite directions.)\n\n3. The frictional force on a surface is always perpendicular to the normal force. \n(A box can be held over a vertical wall by exerting a force in a direction perpendicular to the wall because the frictional force, perpendicular to normal force, acts in the vertically upward direction thus resisting the downward motion of the box.)\n\n4. Fluid friction is less than solid friction. \n(So liquid lubrication is done in bearings to minimize friction and hence to reduce wear.)\n\n5. In solid friction, rolling friction is less than sliding friction. \n(So it is dificult to pull a cycle with brakes on than without it as when the brakes are on the type of relative motion between the wheel and the road is of sliding type.)\n\n6. The magnitude of frictional force is independant of the area of contact between the surfaces in relative motion. \n(The force required to move two blocks, initially at rest, of different shapes and sizes but with same mass over a particular surface is same. But  for two blocks, initially at rest, with different masses but same area of contact with the surface on which they move the force required to move the one with greater mass is more.)\n\n7. Coefficient of solid friction is independent of temperature.\n(The force required to pull a block resting on a surface does not change with temperature.)\n\n8. Static friction is always greater than kinetic friction.\n(The force required to move a body, initially at rest, over a surface is more than that to keep the same body, initially in motion over the same surface, in moving condition.)
885634
using conservation of energy the kinetic energy at the bottom is \n1/*m*v^2\nThe frictional force times the distance is the work done by friction\nand the displacement in the vertical direction is the potential energy that gets converted to kinetic and friction\n\n1/2*m*v^2+f*L=m*g*h\nwhere h /L= sin(15.9)\n\nPlug in the numbers and solve using algebra\n(.5*6500*3.8*3.8)+4300*L=(6500*9.8*sin(15.9))*L\n46930+4300L=17451*L\n46930/(17451-4300)=L\n=3.6m\n\nj
753429
Presuming you are talking about the work equation J = Fd; where J is work in Joules, F is force in Newtons, and d is distance moved in meters, then the answer is...yes you can do too much work or not enough.  Why?\n\nBecause J/F = d; which means you work to move a body a distance d.  If you were to push a mass with a fixed force F, but put in less work (J) than necessary, your desired distance moved would not be achieved.  Conversely, if you work more than required, you'd overshoot your desired distance.  Note, too much or too little work requires that a given d is specified as a required distance to move the mass.\n\nBecause you failed to describe your experiment in detail, I can only presume you are pushing various masses of various material along some surface for some specified distance (d).  Then you are calculating the amount of work done J = Fd for each mass of different material.  F in these experiments would be sliding friction force = kN; where k is the coefficient of sliding friction and N is the weight of each mass perpendicular to the surface.\n\nWhat you should discover is that the F's for each material will be different; so that, even if your d is the same for each trial, work will vary with the force exerted to slide the body.  That is, more F means more work and vice versa.  In each case, the amount of work performed is just sufficient to move the mass d meters...no more, no less.  But if you push too far, you will have put in too much work...too short, not enough work.\n\nThe F's will differ because of different k's and different N's.  In this experiment, if d is not specified, then the work done in each case is just sufficient (no more or less) than what was needed to move each different d distance.
955845
The object has kinetic energy (1/2)*m*v^2 when first at x=0. The work friction does consumes all that energy as the spring compresses and pushes the object back to x=0. Let the distance the spring is compressed be x1. Then \n(1/2)*m*v^2 = Ff*2*x1 = mu*m*g*2*x1\n\nWhen the spring is compressed to x1, the friction has decreased the energy in the system by half and the remaining half is all stored in the spring compression because the object has come to a stop - about to start the rebound. So\nFspring*x1 = k*x1^2 = (1/4)m*v^2\nIf you solve the 1st equation for x1 and substitute the result into the 2nd, you should get the required result.