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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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i want to increase my water pressure in my house which has a header tank ,if i raise it 400mm will it make much difference?


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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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Friction force always opposes relative motion between two surfaces. In most cases (like yours), one of those surfaces is fixed. So, you should recognize that how does the other surface tend to move. For this purpose, you should consider that what force or torque want to move the body. Then, you can determine the correct direction of friction. While ...


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Remember that static friction is a constraint force: it enforces the rule "no motion between these surfaces" as long as the force needed to do so does not exceed the maximum. The force of static friction will have the value needed to prevent relative motion unless that value exceeds the maximum. So there are two possibilities to what happens here. ...


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When $F=2N$, the a friction of $2N$ would also act in the opposite direction on $A$ This is wrong. Friction force isn't equal to force $F$. It is equal to $F_f=m_Ba_B$ Free body diagrams of blocks are as below: When $F=2\;\mathrm N$ then we have: $$F-F_f=m_Aa\Longrightarrow2-F_f=2a$$ $$F_f=m_Ba\Longrightarrow F_f=4a$$ Hence $$a=\frac 13\;\mathrm{m/s^2}...


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You say "rolling without slipping." But what do you think will happen in this situation? Do you think the large disk will "roll without slipping"? Can you do an experiment at home to find out? You can use whatever method gets you to the correct answer. You are given forces (F, weight) so try using forces. Try solving the equations lucas gives you. ...


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Transmit the force $F$ to the center of mass and add its torque. Consider to the relative motion between disc and ground. Then you can recognize the correct direction of friction force (friction force opposes relative motion). Free body diagram of disc is as below. (See this answer to understand better.) Equations of motion are: $$F-F_f=ma$$ $$N=mg$$ $$...


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Potential energy change of mass $m$ is equal to kinetic energy change of system ($m$ and $M$) plus wasted energy due to friction between spool and axle. $$mgy=\frac 12mv^2+\frac 12I\omega^2+W_f\tag 1$$ $\omega=\large{\frac{v^2}R}$ As the tension force of the string is constant, then the net force acting on mass $m$ will be constant ($F_{\textrm{net}}=mg-T$)....


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The first thing first -The concept of friction Friction come in play when there is any tendency of relative motion between 2 surfaces and it is in opposite direction of relative motion (Note- I used word relative motion not simply motion). In simple word friction try to reduce relative motion. in 2 block problem and this problem the difference is T (tension)....


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I draw one for you,hope it helps!


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You can easily find direction of friction force by drawing free body diagram for the cylinder. FBD of the cylinder is shown below (Note that I have drawn diagrams for counter clockwise rotation) As you see in figure above, friction force $f$ opposes to rotation of the cylinder. Equations of cylinder motion are as below $$f=ma_G\;\tag 1$$ $$N=mg+F\;\tag 2$...


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If the cylinder is not accelerating and the rolling no slip condition $v_{cm} = R \omega$ is satisfied then the frictional force between the cylinder and the ground is zero. The horizontal force on the cylinder is zero as it is not accelerating in that direction. The weight of the cylinder is equal an opposite to the normal reaction pn the cylinder due to ...


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If you allow a cylinder with a piston where there is friction present between the walls of the cylinder and the piston to expand, some of the work done is used in overcoming the frictional force (released as heat). Now if you supply the same amount of heat which was released as work and try to compress the piston back to its original state, some of the work ...


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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 ...


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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.


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(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 ...


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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 ...


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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: $a=\sin(\theta)-\mu \cos(\theta)$,...


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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 ...


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First off, I think it is better that we draw free body diagrams and write motion’s equations related to a motorcycle (for example). Consider that motorcycle is moving with acceleration $a$ on a rigid road. Assume that motorcycle’s wheel are rigid too. Rear wheel is driver. ($T_e$ is engine torque and $R$ is radius of the wheels) Free body diagram for rear ...


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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 ...


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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})=\...


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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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The 'formula' is the Principle of Conservation of Energy. Whatever energy (work) is put in via gear wheel 'a' is either dissipated as friction in the gears or bearings, or increases the rotational KE of the gearshaft, or appears as work done on the output from gear wheel 'b'. I assume the gearshaft is balanced so there is no change in its gravitational PE. ...


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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 ...


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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 ...


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Energy lost is due to sliding friction when the gear teeth engage and disengage. Most gears are designed with involute profiles which allow rolling about the contact point when the contact force peaks (at the pitch circle). So the actual energy loss comes from a scrub calculation where the frictional force is multiplied with the scrub velocity to determine ...


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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\...


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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.


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@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 ...


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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 ...


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One way of looking at this arrangement is to consider the external force $F_{\text{ext}}$ which must be used to prevent an object of mass $m$ moving. When the object is on level ground $F_{\text{ext}} = 0$ When the object is on an inclined plane, angle $\theta $ with the horizontal, the magnitude of the external force on the object has to be $mg \sin \...


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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 ...


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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 ...


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Frictional force doesn't depend on normal reaction. Frictional force depends on contact surface properties and from a exact view point, frictional force depends on electromagnetic forces between molecules of two bodies that are in contact. But, maximum frictional force can be measured by normal reaction and friction coefficient. It doesn't depend on normal ...


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When we speak of the horizontal force affecting only horizontal motion and so on we mean the total horizontal force. It just means that once you (somehow) know the component of net force along any one direction (as a function of time) you can easily find out how the particle changes its position along that direction - without the information of any other ...


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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 ...


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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 ...


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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 ...


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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 ...


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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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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 ...


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The resonant frequency is equal to the natural frequency when no damping and no external force at all is applied to the system. When damping is applied so that now the decay time (decay of amplitude) is in effect, the resonant frequency decreases a little below depending on magnitude of damping.



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