# Tag Info

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It could be the base itself. If it's made of plastic or aluminum, there's a possibility that ends of the legs where the wheels mount, are slightly bent or warped. I had the same problem with my computer chair and it would always spin counter clockwise. I've even tried positioning the wheels in the opposite direction to see if it would turn clockwise which ...

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A good way to model the thermodynamics of a system like this (that involves sliding friction) is to treat the interface between the bodies (the sandpaper and the table) as a separate thermodynamic sub-system. The interface has no mass, so its change in internal energy is always zero. The sandpaper is exerting a frictional force in the positive x-direction ...

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As your hand pushes the sandpaper there is negative mechanical work done on your hand and positive mechanical work done on the sandpaper. These are (in the ideal case) equal in magnitude so that there is no mechanical energy lost between the sandpaper and the hand. As the sandpaper pushes on the wall there is negative mechanical work done on the sandpaper ...

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Heat that is generated by mechanical motion arises because of friction, as noted in the other answers given above. Note that it can also be generated on the molecular level by forcing adjacent molecules to "rub against" one another inside a chunk of solid material. Materials scientists call this internal friction and is the reason why a chunk of ...

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In short, heating due to rubbing surfaces has the same roots as Joule heating, which induces a temperature increase in a conductor when drifting electrons interact with solid lattice ions, producing phonons, i.e., quantized sound waves. Thus, in principle, everything that generates sound waves in a body makes it hotter as new vibrational degrees of freedom ...

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The negative work done by kinetic friction takes the macroscopic kinetic energy of the object it does work on and converts into the microscopic kinetic energy of the molecules of the sandpaper and wall materials, as reflected by an increase in the temperature of the surface of the materials. In effect, the rubbing action between materials increases molecular ...

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$a$ is the acceleration of block A with respect to the reference frame $S$ in which the ground is at rest. We do not need to introduce a pseudo force because $S$ is an inertial frame. However, $a$ is not necessarily the same as the acceleration $\frac F {M+2m}$, which is the acceleration of the centre of mass of the whole system. We can only say that $a = \... 0 The normal force on each small segment of area may vary, depending on how the mass is distributed, but the friction force is always proportional to the normal force, so the total friction is proportional to the total normal. 0 I don't know what you mean by "size" but the angle$\theta$where both blocks begin to slide, that is, where the maximum possible static friction force occurs, only depends on the coefficient of static friction, and the coefficient of static friction equals the tangent of$\theta$. So if the coefficient of static friction is the same for both ... 0 TL;DR Whether or not the block will start moving (sliding) depends only on the inclination angle$\theta$and the static friction coefficient$\mu_s$of the inclined surface. The block will move when the following condition is satisfied $$\boxed{\theta > \arctan \mu_s}$$ Below I give detailed derivation for this condition. The magnitude of static ... 0 Nothing of the above mentioned thing matter, only the coefficient of friction between the two surfaces. 1 Every book says that friction is independent of surface area in contact. It is pretty obvious that equation for our friction doesn't have any "area term" in it. The "area term" is built into the value of the normal (perpendicular) force,$N$, between the contacting surfaces. Think about$N$being the pressure (force per unit area) on the ... 0 Friction does depend on the normal force. As the area gets bigger, the normal force per unit area and the friction per unit area get smaller: f = (f/A)A = [μ(N/A)]A = μN. 0 The reason is that the weight of the cube is now spread over a larger area of contact, so each part of every plate is now pressed more lightly in contact with the surface. 1 You have assumed that the coefficient of friction for the car is the same as that give for the trailer (0.15) so the frictional force acting on the car and the trailer is$1630 \times 9.8 \times 0.15 = 2396.1$Newtons You have also assumed that friction acts in a backwards direction on the car. But you are told that the force exerted by the car on the ground ... 1 The free-body diagram at exact collision time should look like: If I assume there is no mass loss the first equation cannot be used, otherwise, J would always be zero. I will define some parameters: e: coefficient of restitution u: coefficient of friction t_c : time collapse during collision vi: initial velocity before collision vf: final velocity after ... 0 So during a typical collision calculation with no friction, there is a single impulse$J$along the contact normal that is applied in equal and opposite measure on the colliding bodies. This impulse is calculated by $$J = (1+\epsilon)\, m^\star\, v_{\rm imp}$$ where$\epsilon$is the coefficient of restitution,$m^\star$is the reduced mass of the system ... 5 First, consider the hand: The force of friction on the hand is acting in the opposite direction of the hand’s velocity. (Specifically the velocity of the material of the hand where the force is applied) So the mechanical work is negative meaning that mechanical energy is leaving the hand. Next, consider the table: the force of friction on the table is equal ... 2 To begin with, I don't care much for the diagram you obtained as it implies a gradual transition over time between the maximum static friction friction force and the onset of the constant kinetic friction. In my view a better diagram of the friction plot is one relating friction to applied force as found here: http://hyperphysics.phy-astr.gsu.edu/hbase/... 1 I misunderstood the question with my first answer. Here's a better one. The two blocks will, when the surface underneath them starts accelerating, stay stationary, influenced by static friction, until the surface accelerates so much that they individually break free from static friction and start sliding and lagging behind, influenced by kinetic friction ... 0 The main reason why ABS shortens the braking distance, is because the static friction between the tyres and the road surface is larger than the sliding friction. By making sure the tyres are rolling, the tyres will have the larger static friction with the road surface than if they are sliding, resulting in a larger maximum deceleration of the car. A reason ... 0 ... friction is suppose[d] to be present only when there's a force pulling an object forwards ... Incorrect. Friction always acts on objects to oppose relative motion between two surfaces, regardless of whether a force is or is not applied to one or both objects. If two touching objects are in motion relative to one another then kinetic friction will always ... 0 I have always, always hated how friction is explained at the school level. The issue you are facing is that all these shenanigans about static friction, kinetic friction and even the misnomered monstrosity the rolling friction is are highly empirical. That is, given Newton's laws of motion we try to describe how objects around us behave, almost completely ... 0 Friction is not about other forces. It is not even just about speed. It is about sliding. Technically, that means that it is about relative speed. Solely and purely. Friction comes into existence in order to stop sliding. And that is why a moving object that slides over the ground comes to rest. A moving object in space wouldn't come to rest as it wouldn't ... 3 I think you are confused about differences between the static friction and kinetic friction. OBJECT IS AT REST When an object is at rest, there is a static friction on the contact surface $$F_{fs} = \mu_s n$$ where$n$is magnitude of the normal force, and$\mu_s$is coefficient of static friction. In order to move the object, the resultant force from other ... 1 The object will come to rest. Friction is "always" present. There is no frictional force if the object isn't currently moving, but when you exert a force or if the object is moving, there is a frictional force. The value of that friction is$\mu R$where$R$is the reaction force. 0 There are two main types of friction, static and dynamic friction. Static friction determines the amount of force required for a non-moving object to accelerate, and dynamic friction determines the deceleration of an object that is already in motion. Shaking or vibrating a surface simply has the effect of adding additional forces that aren't biased in any ... 2 Imagine rain falling on a landscape in which there are hills and valleys. A small lake may form in a hollow half way up the side of a hill. The water doesn't know that it's half way up the hillside, so it stays where it is. If a storm blows, or if there's an earthquake, the water in the hollow may be shaken out, and descend the hillside to the valley below. ... 0 from the free body diagram you see that the car move due to the constraint force$~F~$between the wheel axel and the car suspension, for a static case this force is equal to the force between the wheel and the road$F_T~\tau~$Engine torque (transmitted to the wheel)$~F~$constraint force$~F_T~$Tire force or constraint force (in case of no slip ... 0 Friction is a rather general word, which can mean several things here. The dissipation processes that reduce the energy of the ball when it heats the surface, by transforming a part of this energy into heat. Friction of the ball against the air, which slows it down. Sliding friction when it touches the surface: it may actually force the ball to start ... 0 Consider the system as the yo-yo and a short portion of the string attached to the yo-yo. For this system, the external forces are: gravity and the tension on the string. Friction between the yo-yo and the string is an internal force for this system. The motion of the Center of Mass (CM) of the yo-yo is determined by the net external force. (The tension ... 2 But how does the ground actually make the car move? By applying a static friction force forward on the car. The ground's static friction force acting forward on the wheel is equal and opposite to the rearward force the wheel applies to the ground, per Newton's third law. Since the static friction force acting forward on the wheel, and thus the car, is the ... 5 . . . . linear force from the wheel on the ground and the linear force from the ground on the wheel would not cancel as the two forces act on separate bodies and the wheel will spin. 0 The Newton equations of motion are $$m\,\ddot x+k\,(x-x_0)=F(t)+F_c-F_\mu\tag 1$$ $$I\,\ddot \phi+d\,\dot\phi=-F_c\,R\tag 2$$ and the kinematic equations $$x=R\,\phi\quad\Rightarrow ~,\dot x=R\,\dot\phi~,\ddot x=R\ddot\phi\tag 3$$ you have now three equations for the three unknowns$~\ddot x~,\ddot\phi~$and the constraint force$~F_c~$you obtain $$(I+m\,R^... -1 I would imagine that this problem is best solved by using a work-energy balance, but it looks like it fits a second order DE system well. Could someone provide me some guidance on how to solve for the travel function x(t) for this? Because there is friction involved kinetic energy is not conserved. So you need to set up a Newtonian Equation of Motion.$$F(t)... 57 It’s like shaking a measuring cup half full of sugar to make it level out—in both cases there’s an energetically favored configuration you’re trying to reach, but without agitation, friction prevents the grains from moving to that configuration. Each time you tap the cardboard or shake the cup, you give the grains a new opportunity to settle in a new ... 0 It's not necessary that increasing the temperature will only increase the friction on the surface it can also decrease with time which depends upon temperature difference between the heat source and surface or the mechanism which is providing heat to the surface and rising the temperature of the surface. Let's take an example that how friction can decrease ... 1 The assumption you are making that breaks everything is that static friction is required for the ball or cylinder to freely roll at constant velocity on a horizontal surface without sliding. It is not. The friction force you show would only be required to start the ball or cylinder rolling backwards without sliding from rest, for example in response to a ... 0 If i am correct, temperature cannot make frictional force increase or decrease Possible explanation: If the temperature increases then distance between the particles of the body increases due to thermal expansion which results in smoothening the surface accordingly and the frictional force between them decreases. 0 The harder material deforms and wears less than the softer one, but wears nevertheless. However the hardness relation between materials makes a lot of difference in industrial processes. For example: we can machine a part of cast iron using a tool of tungsten carbide. But after repeating the process with several other iron parts, the tool wears, and ... 2 Rolling resistance is an interesting topic. It is clear from experience that a rolling wheel does eventually slow down and stop. So clearly there is some dissipative process that removes the energy from the wheel and transfers it to the environment. However, it is more than that, momentum and angular momentum are also conserved, and you have to consider ... 3 If the energy were indeed conserved, the ball would roll without stopping. The reason why it stops is Rolling friction, which results in converting some of the mechanical energy into heat. Note that this is a type of friction that is different from more familiar static friction, which is also present in case of an object rolling, but which does not result in ... 1 In an ideal case, once rolling starts, friction stops acting as there is no relative motion at the point of contact and the ball keeps on rolling. In a real life scenario, there is some deformation at the point of contact, and the normal shifts slightly and no longer passes through the centre. On performing torque analysis at the centre of mass, you will ... 5 Yes, energy is always conserved. If not then you have to widen your definition of the system to include all relevant parts. For rolling motion, the main energy absorbers that cause slowing down is the contact with the surface and the mechanics of an axle (if its a wheel on a car e.g.). Kinetic energy might be lost as heat from friction in the axle bearings, ... 4 Energy is conserved when a real wheel or a real ball rolls along a real surface, but purely mechanical energy is not conserved. Some of the mechanical energy is converted to heat, which is considered to be another form of energy. One important mechanism of energy conversion is hysteresis. No object is truly rigid. A rolling wheel or ball is continually ... 1 Critical damping is the condition which transitions between overshoot of the static equilibrium position (under damped) and an exponential decay to the static equilibrium position (over damped). With less damping than critical damping the system will reach the static equilibrium position quicker than one which was critically damped but will overshoot and ... 0 Simplest answer is that it is just a matter of jargon. Obviously, with very little damping the system may vibrate increasingly (unstable), or the vibration may dampen out after a long time. With a lot of damping it will get to the stable position but it may take long, very long, or "forever". So somewhere in between is the amount of damping where ... 0 it begin by spinning its wheels. At first the car won't be moving, so if the wheels were to begin spinning without the car moving, there would be a relative motion between the wheels and the road. That describes a skid. It would work but be inefficient. Instead for most cars and drivers, the engine applies a torque to the wheel. The ground applies a ... 2 What the OP asks for is indeed possible but only for very specific initial conditions. There are a few things to discuss, the first being what does "returning to equilibrium faster" mean. First, we are trying to compare the behavior of two different systems (I will assume they have the same damping coefficient$\gamma\$ but different resonant ...

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I agree with the idea that the first block provides total friction in that case. Here is an imaginary experiment, assume there are "two" equal forces pushing in obverse directions on the two blocks. In the beginning, the two blocks are "separated", and they move to each other until they are in "contact". Now, as they are in ...

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