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The fastest possible way to do a lane change is to fully steer one way on the traction limit and then steer on the opposite way again on the traction limit. The traction limit is $\mu g = \frac{v^2}{r}$ where $g=9.81\;{\rm m/s^2}$ is gravity, $\mu=0.8\ldots0.9$ is the coefficient of friction (half it in the rain), $v$ is the speed in meters per second and ...

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The following might help: $H = \frac{1}{2}(mv^2 + kx^2) + \gamma mkvx$ decays exponentially with time along the solution of the damped system. Check by integrating $H$ with respect to $t$ and using the equations of the system. So the "energy" $H$ decays exponentially instead of remaining constant.

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As a side note skidding is not a yes or no state with tires. See this answer for more details. The sum of the normal force and friction force that act on the car is the reaction force to the sum of the weight and centrifugal force of the car on the road. We can equate them in the coordinate frame parallel and perpendicular to the road. $$N=m\,g\, ... 0 The is a complex question and to cover it we need to look in depth at how tires generate tangential force on the road (as opposed to normal force that is perpendicular to the road). The answer is of course friction F=\mu N but in this case the answer is complicated by the fact that the rubber surface of the tire deforms, and thus rather than simply taking ... 0 I used to be really confused by friction and the fact that frictional force = (mu) * normal force. However, this is wrong. Frictional force is actually less than mu times normal force, and below that limit only opposed applied forces parallel to the surface. If there are two blocks on the floor of the same mass except I'm pushing one down, and my friend ... 1 What is meant by this is the force normal to the surface of contact between an object and what it is resting on. In most problems like this, this object boundary is taken to be flat. The question is slightly mischievous: it's asking you to be precise to test your understanding of friction: to make the statement precise, you need to say that if the normal ... 2 Imagine a box on a horizontal surface. If you just push it down should there be a friction force? If yes to which direction? Now imagine a box on an inclined surface. Static friction balances the component of weight that is along the surface. If you apply more force normal to the surface, does that component change? 0 In reviewing the answer supplied by troy, The explanation of his increased water pressure on the larger pipe is as follows: The pressure from the tank is based on the height of the tank. A tank on a 25' tower will supply at least 12.5 pounds per square inch. (we don't know the height of the surface of the water.) The 3/4 inch pipe has an area of .44 sq in. ... 0 It appears that you are only considering the system consisting of the vehicle, i.e. car+wheels+brakes, as @user31782 has pointed out. Examining only this system, there is no way for the car to stop: it is traveling through space at a certain speed, and that speed is independent of the spinning wheels; if you apply the brakes the wheels stop spinning (and ... 0 First and foremost, here is the free body diagram of the wheel. The wheel is moving to the left and the friction is holding it back in the opposite direction. The c here is damping from the bearing of the wheel, but I am not going to explain that further because we are focusing on the friction between the wheel and the ground. From the free body ... 0 Friction at the axles would work in favor of the 4 wheeled vehicle (more friction per axle) while wind resistance would work in favor of the 3 wheeled vehicle (more surface area). I'm not sure that friction with the ground would play a factor. 0 The truck has to call the tow company as part of their breakdown policy. Even if it means you have both vehicles stuck. I've observed that stupidity first hand. There's no magic in the tow truck, but likely the driver will have experience of getting out of the situation. 1 Yes. The manner of which two surfaces in contact interact is highly investigated by the Tribology community.In particular, the field exploring the mechanics of the interaction is called contact mechanics. Tackling problems of contact mechanics analytically/numerically is often done by solving the elasticity equations. By predicting quantitatively the forces ... 0 If force F is horizontal and the blocks do not move, the friction force is the only force acting against gravity, so it should be equal to the sum of the weights, and its value should not depend on F or friction coefficients. 4 Fundamentally, this is no different from computing the friction in a fluid (shear viscosity). The theory of viscosity goes back to Maxwell and Boltzmann, and microscopic calculations are possible for many fluids. Solid friction is more complicated, because the exact preparation of the surface obviously matters. First principles theories therefore concentrate ... 5 When you inhale you create an area of low pressure immediately in front of your mouth, like the Venturi of a carburetor. You would be drawn toward the low pressure area as the incoming stream of air accelerates down your throat, maintaining the low pressure in front of your mouth. Until your lungs are full. When you turn 180 degrees and exhale, you reverse ... 3 Looking very close at the surfaces that touch at friction, this is an illustration Both surfaces are rough. They have ticks, holes, gabs, pits, spikes, and edges on the microscopic level. The smoother, the lower the coefficient of friction \mu. This constant is thus to be considered as a combined "roughness" between these two surfaces. Intuitively and ... 0 When two surfaces are at rest, the only friction between them is static friction. Static friction is a force, and takes a value between 0 and F_{max} = \mu_\mathrm{s} F_{n} When two surfaces are in relative motion, the only friction between them is kinetic friction. Kinetic friction is a force and has a constant value of F_{k} = \mu_\mathrm{k} F_{n} ... 3 This is the simplest analogy I could think of. Imagine a long narrow carpet sliding across a huge ice rink at 1kph. On the rear end of the carpet stands a very fat (200kg) man wearing roller skates. You want to bring him to a standstill. You could grab the man and dig your ice skates into the ice until he eventually stops. Alternatively, you could grab the ... 5 There are two parts to this question. Part 1: will the card slide?if I have a card at an angle, is there a limiting vertical force that will make it slide sideways? The force diagram looks something like this: This is a bit like the "climbing a sliding ladder" problem, in which case there is going to be a limiting force F - once you exceed that force, ... 2 ...a truck in motion and it has stack of hay (lets suppose) on the back. Now if the truck comes to a sudden stop will it stop faster if the force exerted by the truck on hay had overcome the friction force (another wording: will it be faster if the hay slips forward) or will it stop faster if the hay remains constant. I tried to find a braking ... 14 Not to detract from Floris' answer, but I think this is an instance where it is nice to think in terms of limits. If the hay is tied down, you're stopping an object with mass (truck + hay). If the hay isn't tied down, but on a sufficiently sticky surface such that it doesn't move, it should be the same as stopping it if it were fixed, since the outcome is ... 14 On the whole, static friction is higher than dynamic friction. This means that if you can brake without your wheels skidding, you will come to a halt more quickly. So let's assume that the truck brakes without skidding, and see where that gets us. Let's assume that your truck has weight W = Mg with a haystack with additional weight w = mg on top. ... 4 The essence of static friction is that it acts to prevent motion even in the presence of some outside force. The desk I'm sitting at while I type this is homemade and thus almost certainly not perfectly level. Yet the items on the desk are all fixed in place, not sliding down the slight slope. (OK, pencils tend to migrate by rolling, but..). It's the ... 1 All the other forces except friction acting on the ball have their line of action pass through the center of mass. So, friction is the only force which can provide torque in the above example. Therefore, in order for the ball to roll, you need friction. 2 No the point of contact is not at rest. It moves with the block. You are probably confusing with rolling motion in which the point of contact is always at rest. There the point of contact is rest because the lowest point on the disk has two contributions, one due to forward motion of disk as a whole (v) and one in the backward direction due to rotation ... 2 According to Newtonian mechanics, it is true that the table exerts an equal but opposite force against gravity that results in \Delta y = 0, where y is the up/down dimension. However, sliding block is clearly moving in the x dimension (i.e. horizontally across the table). And it is also acted on by a force, namely friction. The block does not generate ... 4 Let's figure it out. First let's figure out the force you would need to slide the object. In order to slide it you'd need a force that overcomes friction.$$ F_{\text{slide}} > \mu m g  where $\mu$ is the coefficient of friction. In order to tip it, you'd need to cause a torque that will cause the can to rotate about its far bottom corner. If you ...

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You've stumbled on an interesting idea: how do classical systems that dissipate heat or energy via frictiom arise from quantum systems that perfectly conserve energy in their interactions? Particles in the collision kind of scenario you described don't really exhibit friction. One convenient point is that temperature and heat transfer in quantum physics is ...

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Whatever you do it produces heat. This has to do with electromagnetic interactions between the two objects which interact. Moving one surface on an other, the surface electrons from both surfaces will be disturbed, they couldn't be at the same place and will be moved slightly. Every displacement has elements of acceleration and any acceleration is escorted ...

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