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107

Your derivation is composed of correct statements and indeed, if something is known to act as a lubricant, we want the viscosity to be as low as possible because the friction will be reduced in this way. For example, honey is a bad lubricant because it's too viscous. However, your derivation isn't the whole story. The second condition is that the two ...

27

A good lubricant tends to effectively minimize direct contact among components of any device that need it Keeping this in mind, viscosity is not the only factor involved. Grind a graphite pencil lead, and it makes a mighty fine lubricant. It might be that in the case of water placed between two surfaces, a water droplet which was supposed to act as an ...

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The parallel plate situation that you describe is not the typical condition encountered in practical lubrication operations. In addition to facilitating the surfaces sliding over one another, the lubricated bearing must also support a normal load. To do this, the gap between the surfaces varies with location along the bearing. For example, in a journal ...

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Why Oil is Slippery Explaining why oil is slippery requires a look at its chemical properties. First, oil is non-polar, which means it does not have a positive or negative charge. Some molecules, like water, have a “charge distribution,” which means the molecule acts almost like a battery, part of it has a positive charge and part of it has a ...

9

To apply Noether's theorem, which is what you are alluding to here, one needs to look at continuous symmetries of a Lagrangian description of a system's dynamics. The damped oscillation equation you have written, although it is invariant with respect to a time translation as you rightly say, is not a Lagrangian description. If you write the Lagrangian for ...

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Conservation of energy is related to time translation invariance for systems that can be described by a Lagrangian. Dissipative systems in general are not describable by Lagrangian mechanics (without altering the formalism that is) and so Noether's theorem cannot be applied to check whether or not energy is conserved. EDIT: the dissipative system the OP ...

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Assuming rolling without slipping, this can easily be solved by means of Energy Conservation. Let's assume the incline has a height $h$. During the travel down the incline, potential energy $U$ is then converted to kinetic energy $K$: $K=U$ $K=mgh$ The kinetic energy $K$ is a combination of translational and rotational energy: $\frac12mv^2+\frac12 I\... 4 @tbf is right; lubrication, and tribology in general, is complicated. That's why there is that high effort to understand it and to design advanced materials. There are several phenomena that cause the friction force exist and the ones you have neglected causes that oils are superior to water in most industral applications. In dry sliding we can identify ... 3 For the first figure: You are misinterpreting the second law, it only says what happens to an object that is subject to a net force. In this case, if the impact force is larger than the friction, the object will accelerate. The second law doesn't say what happens to the object that produces the force. Now, the third law does, which means that the actuator ... 3 As long as the object does not slide, the static frictional force is equal to$mg\sin \theta$. Once the frictional force also becomes equal to the normal force$mg\cos \theta$times the coefficient of static friction, the object begins to slide. But the static friction law is an inequality, not an equality. $$F\leq \mu_s N$$ As long as the F is less than ... 3 The conceptual problem seems to be however there is no way to define the friction for the rope since it has two parts, with apparently distinct potential energies. Don't worry about the "distinct potential energies". You can just compute the gain in potential energy for each part separately. For the bit already hanging down, as the rope slides by a ... 2 Free body diagram of C is as below: Am I missing something important in interpreting the free body diagrams? Yes. You have assumed that$\vec N+\vec F+\large{\frac 12 m_C\vec g}=\vec 0$. But, this is not correct. The correct equation is: $$(\vec N)_y+(\vec F)_y-\large{\frac 12}m_Cg=0$$ or $$N\sin(\large{\frac \alpha 2})+F\cos(\large{\frac \alpha 2})=\... 2 Because, dissipative forces convert some of the work to heat. If we want to have a reversible process we must be able to return system and its environment back to their initial states without any change in universe. For this purpose, we must extract heat from environment and convert whole of that to work and it is impossible due to the second law of ... 2 Water can not bear normal loads as well as oil. Water is bound to escape from high pressure bearings to lower presser places in an open lubrication loop leaving bear contacts. Water can create bubbles around cavities and corners and break the laminar flow which will compromise the separation of moving parts. Water will react chemically with surfaces. There ... 2 The mass of the wheel is located farther from the axis of rotation, thus the moment of inertia is greater for the wheel. So it rotates slower with the same rotational energy and therefore the ball rolls faster down the slope than the wheel. 2 (Forgive me if I am stating the obvious.) Taken individually, any body will accelerate at a rate that is inversely proportional to the coefficient of friction between itself and the surface on which it sits, regardless of its mass. This is easily proven algebraically: the mass affects both forces that run parallel to the plane (the parallel component of ... 2 The observation that if \mu_1 < \mu_2 then a_1 > a_2 applies only when the two blocks are separated and on the same incline. If the two blocks are in contact, with block 2 lower down the incline (as in the illustration), this is obviously false, because it is impossible for the upper block to pass the lower block. Neither can we conclude that the ... 2 You need to start by considering the microscopic origin of the frictional force. In most circumstances surfaces are rough so when to touch two surfaces together they actually only make contact at the highest points on the surfaces. We call these high points asperities, and in the diagram below I've outlined in red where the asperities on the surfaces are ... 2 Because there is two types of friction, static and dynamic friction. Static friction is always greater than dynamic friction. Static friction opposes motion of the object while it is not moving. Dynamic friction is what opposes motion of the object while it is moving. In a closed lid, turning the bottle cap requires more force because the static friction ... 1 Let's prove it:$$\sum F=ma\\ f-w_x=ma\\ \mu n-mg\cos(\theta) =ma\\ \mu mg\sin(\theta) -mg\cos(\theta) =ma\\ \mu g\sin(\theta) -g\cos(\theta) =a$$So, smaller$\mu$gives smaller (more negative)$a$. So acceleration is growing downwards along the incline for smaller friction coefficient. 1 You always need to make a force diagram to be safe, however, for this particular configuration (assuming the blocks are not in contact otherwise will have a common acceleration) you get almost always that response. The reason is that if you calculate the forces along an axis parallel to the plane you get, after simplifying:$a=\sin(\theta)-\mu \cos(\theta)$,... 1 force is change in impulse over time, so saying that impulse is lower than static friction does not make sense, in the same way that saying that some speed is larger than some acceleration, they are different things. Impulse from a moving object is transferred to the one at rest through a force, that results in an exchange in momentum. During the ... 1 Why using torques? Because you have three unknowns,$\theta$,$F$and$N_2$and that requires three equations. You also have$N_1$as unknown, but by using torques you can get rid of that! I would do first the torque part (the second half of the answer), then Newton's law and then the friction model formula (the first half). Find normal force$N_2$by ... 1 You are right: the frictional force does not directly depend on a force that is normal to it. The frictional force depends only on the longitudinal force and is precisely equal and opposite to that force (hence, there is no motion). But there is a maximum beyond which the frictional force cannot go (hence, if the longitudinal force exceeds this, there is ... 1 I think It is an very critical condition of kinetic physics. When an object moves it faces friction.and the friction does not depends on the relative speed of evolved object because we see p=m×v and f=p×t then f=m×v×t so there in frictionforce if we increase the value of v(velocity) then the value of t(time) will decrease because after increasing in speed ... 1 According to the definition of the work ($\delta W=F\mathrm dx\$), it is better that we say friction doesn't do work. Because, at each point of contact area, friction force is fixed and doesn't move. Friction converts some portion of energy used to move the box, to heat.

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