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Molecules are undergoing elastic collisions. For any given substance, the faster its molecules are colliding, the hotter it is. When an object slides along another object, the total kinetic energy plus gravitational potential of its molecules plus gravitational potential energy must be conserved. It can be proven that in Newtonian physics, for any system ...


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What comes first? Pressure or flow? Without potential, a pressure differential, there is no flow. It's pressure that needs to be there first. So rather than ask what pressure drop you get, it's better to ask what flow you get from applied pressure. We can never measure pressure drop across an infinite pipe because we can never reach the infinite point. ...


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As drawn the disc's angular speed $\omega$ is too fast as related to the velocity of the centre of mass of the disc $v$ for the no slipping condition ($v = R \omega$, with $R$ the radius of the disc) to be satisfied. You can think if the frictional force as trying to accelerate the centre of mass whilst at the same time the frictional force applies a torque ...


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The motion (perpendicular to the wind) is called Ekman Transport. It results from a combination of the wind and the Coriolis Force (due to the rotation of the Earth), which together produce an underwater motion called the Ekman Spiral. Icebergs, which extend large distances below the surface, experience forces from the whole depth of the spiral. The ...


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I thought about this in the molecule level, but then if molecules are at constant velocity then there is no "pushing" force that they apply on each other. To clarify my misunderstanding - I imagined the system at constant velocity at molecule level as if it was accelerating so molecules of A push molecules of B and also the opposite. If we are speaking about ...


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If the blocks initial velocities are zero (i.e. the blocks start to move from rest), then it is impossible for block B to move with constant velocity. Because the only horizontal force acting on it (if there was) is friction force due to block A. We have: $$\Sigma F_B=m_Ba_B$$ If block B moves with constant velocity, then $a_B=0$ Thus, friction force acting ...


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When you move the pen at constant velocity, you are correct that you apply $F=\mu_kmg$. This is true regardless of what constant velocity you choose. The difference is you have to supply more force initially to accelerate the pen to a faster speed, but once it is up to speed you only need $\mu_kmg$ to keep it at that speed. So it really isn't the force ...


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What do people actually mean by "rolling without slipping"? This question and the answer should give you better insight into your question. Think about the definition of rolling without slipping given and how friction is created and works for rolling objects maybe make a free-body diagram.


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power = force × velocity so at 10 km/h , power is 4000 N × 10 km/h while at 100 km/h, power is 4000 N × 100 km/h this way fuel will b consumed approx 10 times faster at 100 km/h than at 10 km/h


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Because water "squishes out", allowing the sole of your show (which has treads for that reason) to contact the ground directly, and experience friction. Shoes with smooth flat soles are quite slippery on water, if the floor is also smooth and flat. Oil, because it is more viscous, does not "squish out".


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Friction is an adhesive process. The friction isn't caused by the load, what the load does is flatten asperities on the surfaces and increase the area where adhesion can take place. For a discussion of this see How is frictional force dependent on normal reaction?. So you can have friction without a normal load if the surfaces are already smooth enough for ...


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Are you suggesting that friction could come about as a result of the Normal Force (Fn) between the two interacting objects? If so, Normal Forces act only perpendicular to an object's surface (opposite to the force of Gravity. The Normal Force serves basically to prevent solid objects from moving thorough each other. Friction implies some horizontal ...


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You have found the critical angle $\theta_c$ at which the block begins to slide. That gives you the coefficient of static friction $\mu_s = \tan\theta_c$. Kinetic friction comes into consideration when the block is actually moving. Before the block can move, the force $mg\sin\theta$ acting down the incline must be at least equal to the maximum possible ...


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First off, you should draw free body diagram for each block. Then, you should recognize if relative motion between blocks occurs or they move simultaneously. For this purpose, you can consider to the maximum friction force between the blocks. If second block ($m_2$) moves with respect to the other blocks, then friction forces $f_1$ and $f_2$ must reach to ...


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Friction is a result of asperity-tops adhering together (as if they melted together) through chemical bonding. The more asperities that stick to the other surface, the larger the friction. To make them slide across each other, friction must be overcome, meaning that these "glued-together" tops must be broken. I gave this explanation to another question not ...


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If the two surfaces are real and perfectly smooth at an atomic level, and made of the same material, there will be chemical bonding between them. They will fuse together as though they had always been the same object [1]. Even different metals can fuse together if left in contact for sufficient time, or if forced together, due to mutual diffusion of atoms ...


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The Laws of Dry/Solid Friction - such as that the friction force is independent of area of contact, proportional to the normal force between surfaces, and is independent of their relative speed - are empirical laws based on observation, like Hooke's Law and Ohm's Law. They are generally and approximately true in most situations for relatively small forces ...


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In general field theories Lagrangian are not derived: they are instead assigned (postulated) and proven against the equations of motion. Every Lagrangian that gives rise to the correct equations of motion is in principle a good Lagrangian for the system. One can prove that for mechanical systems described by conservative forces $\textbf{F} = - \textrm{grad}\...


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As you only asked for a hint, not an answer: Your answer is coming just by considering the force F to be horizontal. But what if it's not horizontal or what if it have a vertical component too? This vertical component will decrease the normal on the right block, so it will also decrease friction. For minimum force, assume the force to be at some angle, ...


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Static if there is no relative motion between the ground and the tyres at the point of contact. If it was a block then as there was relative movement between the block and the ground then it would be kinetic friction. You need to produce a centripetal acceleration and so need to provide a force towards the centre of the circular trajectory. Although the ...


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The original problem wants to find minimum force to move horizontally and this doesn't mean that force $F$ should necessarily be horizontal. We have: $$N_1=Mg-F\sin\theta\;\tag 1$$ $$F\cos\theta-T-f_1=0\;\tag 2$$ $$T-f_2=0\;\tag 3$$ $$N_2=Mg\;\tag 4$$ Because of the condition of verge of motion, we have: $$f_1=\mu N_1\;\tag 5$$ $$f_2=\mu N_2\;\tag 6$$ Then, ...


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This effect is significant in debris flows, where large numbers of huge boulders move like a fluid. For a technical discussion see: https://profile.usgs.gov/myscience/upload_folder/ci2013Mar07174849246641997.Iverson.PhysicsDebrisFlows.Rev.Geophys.pdf For a spectacular demonstration: https://www.youtube.com/watch?v=51C7vEAVbxk


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One thing everyone seems to have forgotten is the "tar" in tarmac. There is more than just friction on a race track there is also adhesion. drag strips will typically have fresh tar at the beginning and be very tacky. think of it like the tires being glued to the track both by the tar and the melted "rubber" material of the tires. the more surface area for ...


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The claim that frictional force does not depend on area of contact is not, in general, true. It is a good enough approximation for many cases that it can be used for many real world calculation and even more back-of-the-envelope estimates, but it fails under several circumstances. Conditions for failure of the not-area-dependent approximation include (but ...


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Having tested it, using maximum force screwing it on, I always managed to open it again. The test was repeated with my other hand btw, to make sure it wasn't due to a difference in strength between the muscle groups involved in both actions. Still not perfect, because grip is the limiting factor (strain-sensitive receptors in the skin detect the amount of ...


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You have a few questions here. They mostly concern the energy loss due to friction. In this problem there is no energy loss because of friction. Friction is present between the rope and the pulley. This is static friction. It makes the pulley rotate. Energy is lost due to friction (dissipated as heat) only when there is relative motion between the ...



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