Its a question I have read in a book, that,     When two surfaces are polished, the frictional coefficient decreases. But if the surfaces are made highly smooth, the frictional coefficient increases and becomes very large. Explain why?   My thinking says that with excessive smoothness there is more ease of bond formation between the two surfaces and since friction (modern theory) is because of temporary bond formation, friction increases and hence its coefficient. Is this thinking right?
123321
I read that:  If you take a rough surface and make it smooth, the coefficient of friction decreases. But if you make it super smooth, then the coefficient of friction increases. How come?
37450
For a physics engine I am working on, I need to know two object's friction coefficient (for bouncing, collision detection, friction in general etc.). Since this physics engine will have lots of different materials, it would be inefficient to have a list of every pair of materials, with their friction coefficient. So, I would like to know if there is any value I can assign to each material, which would mean I could work out the friction coefficient between the two objects, using the two material's values.
77230
What is the 'molecular' origin of the viscosity? The molecular origin of elasticity is almost clear for me: at the very bottom the 'elasticity' comes from the attraction and repulsion between atoms when they are reasonably far from each other or very close (like Lennard-Jones potential). Is there a similar description for the viscosity as well? Where does it come from?
82375
For me, intuitively, integral $2\pi y~dx$ make more sense. I know intuition can not be proof, but by far, most part of math I've learned does match with my intuition. So, I think this one should 'make sense' as well. Probably I didn't understand the way surface area is measured. It will be great if any one could tell me how 'integral $2\pi y~dx$' is wrong. (By the way, how to use mathematical symbols in texts?)
23272
The general equation for the force of friction (static or kinetic) is $F_f = \mu * F_N$, where $F_f$ is the force of friction and $\mu$ is the coefficient of friction (its value is dependent upon the surfaces interacting on each other).  Why is it that this equation is so simple and does not contain any other variables that account for the force of friction on an object? Does anyone know how this equation was developed?
27801
Let's say that we have a car travelling at a constant velocity (V) on a flat plane , let's assume that it's applied force or the force applied from its engine ( Fapp ) balances the frictional force ( Ffrict ) , hence the power due to Fapp is equal to Fapp×V , now let's assume that the same car now moves up an incline with the same engine power as previously stated and at a constant velocity ...but this time it will travel at a much lower constant velocity since the force which the engine exerts is much higher now in order to balance Fparalell and Ffrict , but what I don't understand is what caused this reduction in velocity , if the engine applies an equal force to that of the counteracting forces shouldn't that mean that there is no net force hence no change in velocity ,also if p=FV and if we now apply a greater force for the same velocity V why does the power increase , I know the answer can be simply explained using mathematics but my intuition says that power is the rate at which kinetic energy is transferred to a particle , so why would a greater force at the same constant velocity mean that you are transferring a greater amount of KE per unit time ?
70741
Is  really independent of area? The friction force, $f_s = \mu_s N$. The equation says that friction only depends on the normal force, which is $ N = W = mg$, and nature of sliding surface, due to $\mu_S$.   Now, less inflated tires experiences more friction compared to well inflated tire. Can someone give clear explanation, why friction does not depend on area, as the textbooks says?
87701
The mirrors we use everyday have a highly reflective smooth surface, so that we can see a clear image.  But I found the same happens on a rough (not so smooth) surface, though only at shallow angles.  For example    this is the logo of physics stackexchange, reflected from the back of an ipad.  Considering that aluminum may contributed to reflection, I tried a different material:    this time a piece of printer paper.  Though not as clear, there is still some visible pattern resembling the logo.  The same effect can be observed on highways, where distant vehicles have a mirror image off the road; I do not have a picture at hand however.  So, why is this happening, and why only on shallow angles?
93775
I am facing a problem in Physics.   Problem: What will be the work done by the frictional force over a polynomial curve if a body is sliding on this polynomial($a+bx+cx^2+dx^3+\ldots$) curve from rest from the height $h_1$ to height $h_2$ (where $h_1 &gt; h_2$).  I tried to solve this as follows:  frictional force $F = k mg \cos\theta$, where $mg \cos\theta$ is  normal force at that point. $k$ is coefficient of friction  Total work done=Line Integration over the polynomial(dot product of F and displacement).  But to go ahead from this point,i do not know.
75171
I am confused by rolling friction.  Suppose you have a cylinder rolling which starts at rest at the top of an incline plane and begins to roll down the plane without slipping. Is work done by the incline on the cylinder?   I know from doing some problems that the total kinetic energy (translational and rotational) is $mgh$, which is true only if the only work done on the cylinder is by gravity. But also, the cylinder must have nonzero angular acceleration, so friction must be exerting torque on the cylinder, so work must be done by friction. One of these statements is wrong.
53833
When I got more into physics, I started asking myself if just like acceleration represents the growth of speed, something else could also represent the growth of acceleration itself. And it came that is exists and is called jerk. Before I thought about this, I thought that acceleration was the highest derivative that made sense on physics.  My question is, why can we perform accurate calculations without account for jerk? I never needed it in a single physics class! I've also read that it is seldom used.  Also, while position, speed and acceleration seem familiar to me due to everyday experience, why doesn't jerk seem familiar to me? I never tell someone when driving a car "you're jerking too much", whereas I could say "you are accelerating too much".  Is it that jerk has little influence because it is almost constant? Even if it is not, what about higher order derivatives? Why don't derivatives higher than acceleration have much influence on calculations? We perform a lot of reasoning in everyday life without resorting to it!
78631
Hi Guys, this is my first time to make a fluid simulation. However, I got some problem. The surface looks so weird. I am trying to make an animation so I do not decide to use sculpt mode to fix it. Is it any other method to fix this problem
87454
Is the coefficient of friction always less than unity? What are the cases in which coefficient of friction is greater than unity?
116307
Two discs mounted on different thin, lightweight rods oriented through their centres are made to rotate about their axes seperately such that the angular momentum of the two about their respective axes is same in both magnitude and direction. When both are brought in contact, they stop due to force of friction. Why is the angular momentum not conserved about their axes in this case?(Was positive before contact, but zero after the process completes, no external force present either)
119592
When we say a function is smooth? Is there any difference between smooth function and continuous function? If they are the same, why sometimes we say f is smooth and sometimes f is continuous? Please help me. Thanks.
100574
As an introduction this is what's written in my textbook:  $Power=\frac{work\ done\ by\ driving\ force}{time}$ $=\frac{Driving\ force×distance}{time}$ (as distance over time = speed) $= Driving\ force×speed$ Note: If the power is constant and the body is accelerating then the speed is changing thus the driving force must also change to keep the power constant.  And in most of the questions we solved in class we assume that the power (of a car) is constant and the &quot;note&quot; in the text I mentioned explains how it will be constant and I understand that. What I don't understand is why will it be constant in the first place? Shouldn't the car exert the same force all the time— so the speed will increase increasing the distance— and thus the work done will increase as time passes and the power will keep increasing? My teacher informed us that we &quot;assume&quot; that the power is constant and that it variates between a certain range. • Yet why does the power have to be limited to a certain range? Why doesn't it increase forever? Or at least increase until it comes to its maximum speed due to air resistance? • Why does speed decrease when the driving force gets bigger even when there's no friction to prevent the car from speeding up forever?
82212
Consider two blocks, one on top of the other on a frictionless table, with masses $m_1$ and $m_2$ respectively. There is appreciable friction between the blocks, with coefficients $\mu_s$ and $\mu_k$ for static and kinetic respectively. I'm considering the fairly routine problem of determining the maximum horizontal force $F$ (say, to the right) that can be applied to the top block so that the two blocks accelerate together.  The problem is not hard to solve symbolically. If the two blocks move together, their accelerations are the same, and the top block doesn't move with respect to the bottom block, so only static friction is in play. In a standard coordinate system (with $x$ oriented to the right), the sum of horizontal forces for the top block is  $$F-F_{sf}=m_1a$$  and for the bottom block  $$F_{sf}=m_2a$$  where $F_{sf}$ is the force of static friction. Solving for $a$ in these two expressions, and then equating them, gives  $$F=\frac{(m_1+m_2)F_{sf}}{m_2}$$  The maximum such force will therefore be achieved when $F_{sf}$ is maxed out at $\mu_s m_1g$, so  $$F_{max}=\frac{m_1}{m_2}\mu_s(m_1+m_2)g$$  I understand this solution, but conceptually I don't have a response to the following nagging question: $F_{max}$ is clearly larger than the max static friction force $\mu_sm_1g$ (because $\frac{m_1+m_2}{m_2}&gt;1$), so why doesn't the application of a force of magnitude $F_{max}$ to the top block cause kinetic friction to take over? This line of reasoning would suggest that applying a force $F$ of magnitude greater than $\mu_sm_1g$ would cause the top block to start moving with respect to the bottom block (in which case the blocks no longer accelerate together, as in the above solution). I'm at a loss, conceptually, to say what's wrong here. I suspect it has something to do with being careful about reference frames, but a clear explanation would be much appreciated.