in which direction frictional force acts in a bicycle?
540441
Depends on what friction we are taking into account:\n\n1)Friction from the road: The frictional force acts in the same direction as the diection of the movement of the bicycle, because it acts at the point of contact of the tyres and the road and at that point the movement of tyres is tangentially opposite to the movement of the bicycle, and thus the friction force always acts opposite henceforth it acts in the same direction of the movement of the bicycle.\n\n2)Friction from air: Air friction will act opposite to the motion of the bicycle.
236862
it should be the pedal on the far left...leave your foot on it to switch gears and release when not needed.
1184958
Quite simple.\nYou pedal the cycle to rotate a gear . The gear is connected to a chain which rotates another gear connected to the wheel and when this wheel rotates, it pushes you forward.\nAs for the balance, it depends upon the rider.
27382
advantage:\n\nfriction, resistance offered to the movement of one body past another body with which it is in contact. In certain situations friction is desired. Without friction the wheels of a locomotive could not “grip” the rails nor could power be transmitted by belts. \n\ndisadvantage:\n\nOn the other hand, in the moving parts of machines a minimum of friction is desired; an excess of friction produces heat, which in turn causes expansion, the locking of the moving parts, and a consequent breakdown of the machinery. Lubrication is important in minimizing friction as are also such devices as ball and roller bearings.
816881
Work is commonly defined as the dot product of force and displacement.  Essentially this means that you are moving in the same or exact opposite direction as your field lines for your given force.  Example: Friction does work to stop a moving box as the force is directed anti-paralell to your direction of travel.\n\nWhen you move in a circle, centripetal force is always directed toward the inside of the circle, which is exactly perpendicular to your direction of motion in the circle.  Since the force is perpendicular to the direction of motion, there is no work done.
789212
Gyroscopics.\nThe wheels act as gyroscopes.  Precession causes any tipping to appear to the rider as a change in steering, which the rider finds easier to control using the handle bars.  Motorcycle riders use this effect by leaning over to one side to help change directions.\n\nWhether there is or isn't such a thing as "centrifugal" force depends on whether you are in a linear or a rotational frame of reference.
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
863293
In the opposite direction of the applied force or motion
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.
992106
It is opposite the direction the bicycle is moving
177252
A bicycle is a wonderful example of an object that's unstable at rest, but stable in motion. My other favorite example is a broom balancing upside-down on the palm of your hand. Let's take a moment to look at the broom because it will help to understand the bicycle. \n\nWith the broom exactly centered above your motionless hand, nothing pushes it to one side and it balances. It's in equilibrium-it experiences no overall forces. In principle, it can stay that way indefinitely, but in practice it never does. The slightest shift of the broom's center of gravity to one side and over it goes. Any disturbance that shifts the broom away from equilibrium gives rise to forces that push the broom still farther from equilibrium, so the equilibrium is unstable. Any object that has no base of support-the polygon formed by its contact points with the ground-has an unstable equilibrium and tips over when disturbed. That instability stems from the fact that its center of gravity always descends when it's tipped and it releases gravitational potential energy as a result. Potential energy-energy stored in forces-is so intimately related to the forces themselves that whenever an object can release potential energy by heading in a certain direction, it naturally accelerates in that direction. \n\nSo the broom is unstable in your motionless hand. But if you move your hand, you can stabilize the broom. You do this by endlessly moving your hand under the broom's center of gravity. If the broom starts to tip to the left, you move its handle to the left to place that handle under the broom's shifted center of gravity. In that manner, you keep returning the broom to its equilibrium. Even though the equilibrium is naturally unstable, you keep helping it out and make it dynamically stable-stable in motion. \n\nNow let's shift to the bicycle. Like the broom, an upright but motionless bicycle is in an unstable equilibrium. If you're really skilled, you can balance it by shifting your weight around to keep the overall center of gravity above the wheels. However, there's a much easier way to balance the bicycle: drive forward! Once the bicycle is in forward motion, it steers automatically and places its wheels under the overall center of gravity. Unlike the broom example, where you have to think in order to push the handle under the center of gravity, the bicycle does this for you automatically. If the bicycle starts to fall over to the left, the front wheel steers left and the bicycle drives its wheels under the overall center of gravity. The bicycle returns to upright all by itself, even when you have no hands on the handlebars. \n\nThe automatic steering occurs for two reasons. First, there is gyroscopic precession that occurs when the spinning front wheel experiences torques from the ground. When the bicycle tips to the left, forces from the ground twist the spinning wheel so that its angular momentum shifts from leftward to rearward. This complicated precession effect causes the wheel to steer left toward the left. \n\nBut the more important source of automatic steering is that the tipped bicycle naturally flexes about its steering axis and turns its front wheel in the same direction as the tip. If you tip the bicycle left, the wheel pivots so as to steer the bicycle left. The fork and frame design are important to this effect. The tipped bicycle can reduce its gravitational potential energy by flexing and, as I mentioned above, this capacity to lower potential energy leads to forces that cause the tipped bicycle to flex. When you tip the bicycle to the left, it steers left and thus manages to reduce its gravitational potential energy-the bicycle's center of gravity descends slightly. Together with gyroscopic precession, this automatic steering makes the bicycle's wheels drive under the overall center of gravity so that the bicycle keeps returning to upright. So the forward-moving bicycle is like a broom w
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.
874727
You need to sum the components. Let Fr, Fl, Fu, and Fd be to right, left, upward, and downward directed forces, respectively. Then\n\nFx = Fr - Fl\n\nand\n\nFy = Fu - Fd.\n\nThe right and upward directions are taken as the positive direction so that the minus sign indicates that the forces (Fl and Fd) are directed to the left and downward.\n\nThe magnitude of the resulting force is then computed as\n\nF = sqrt(Fx^2 + Fy^2)\n\nand the direction can be found from\n\ncos(q) = Fx/F\n\nUse some reasoning to be sure that the direction you compute is consistent with the orientation of the force components.
1024834
compression and friction
848443
when u will draw ur inclined plane and mass on it there r different forces acting in it \nfirst force will be directly perpendicular to the ground level i.e. Fg=mg\nsecond force acting is mgcos(theta) which is perpendicular to hypotenuse of the inclined and makes up Fn=mgcos(theta)\nthen another component parallel to the hyp. of the inclined \nwhich equals mgsin(theta) which goes down the ramp\nand  then u can\nequal the force going down with the frictional force\nmgsin(theta)-(mew)mgcos(theta)=ma\n(as we know frictional force is (mew)Fn=(mew)mgcos(theta) )\nsolve by putting values\nmew=0.146    ans
30757
just draw a free body diagram. you're solving for F parallel. that's just FgsinX which is just mgsin(30_ then you take this and divide it by mass to get the acceleration. so actually, you can cancel out mass and its just (2.15)sin(30) which is 1.075\n\nfor the frictional force, either you need more info, or it's zero.
212857
Harmful:\n-means that the rider must exert extra energy to accelerate and thre friction acting on the wheels slow the bicycle down\n-air friction/resistance also acts in a way that requires more energy input to maintain a constant velocity or accelerate.\n\nBenificial:\n-Without friction, the wheels would just spin in place when you pedel, you could not accelerate at all.\n-I dont know about you, but when I ride a bicycle down a big hill, I like the fact that my brakes work...this is due to friction.