I have a conceptual question about friction

Will give you a scenario to understand.

Suppose a body is moving in a straight line and there is frictional force acting on it in opposite direction.

We have applied a force on the body equal to the frictional force in such a way that the body moves with constant velocity.

Now in one other perpendicular direction, I apply a force on the body.

Since friction is there, any motion in the perpendicular direction will be opposed to some extent.

So by this way the net frictional force value is vector sum of coefficient of friction X Normal force + force in perpendicular direction

That means net frictional force is greater than coefficient of friction x normal force

But isn't it violation of the law of friction??


First, there is no "law of friction". The friction force is actually very complicated and can vary based on many things such as speed and contact area.

Now, as far as the situation you described; the perpendicular component to your movement would change the direction, but not the magnitude, of the friction force.

The object would begin to accelerate in the direction of the new net force vector; and friction will oppose the direction of motion.

  • 1
    $\begingroup$ Yes I agree friction is a complected force having base in coulukbs attractive and repulsive forces...and it a doesn't depends a upon speed as much if the speed is less than 10m/s...So mean to say that in the perpendicular direction, their won't be any resistance in motion by frictilm $\endgroup$ – lancer Jul 6 '17 at 9:32
  • $\begingroup$ That's coulumb force...my bad ...and so you mean in the perpendicular direction even a small amount of force will lead to change in the direction of motion? That means no resistance against the force by friction in that direction ??? $\endgroup$ – lancer Jul 6 '17 at 9:35
  • $\begingroup$ Yes, even a small amount of force will change the motion (the smaller it is the less it changes it, but because $F_{net} = ma $, any change in net force direction changes the acceleration. It doesn't mean friction doesn't resist in that direction. It means the direction of friction moves to match the new movement. The amount of friction doesn't change. $\endgroup$ – JMac Jul 6 '17 at 9:39
  • $\begingroup$ +1 Except for 1st paragraph, which I think is wrong and not relevant. Empirical "laws of friction" do exist and one of them is being violated in lancer's explanation. $\endgroup$ – sammy gerbil Jul 8 '17 at 1:30

If you wish to use the "model" of friction that you are envisioning, then the friction force has to be oriented in the direction opposite to that of relative motion(i.e., the velocity vector). Suppose that initially, the force is in the x direction, and the velocity is V. The force balance is $$F_x=\mu_k N$$. Now, you push in the y direction with constant force $F_y$. The force balances now become: $$m\frac{dv_x}{dt}=F_x-\frac{\mu_kNv_x}{\sqrt{v_x^2+v_y^2}}$$ $$m\frac{dv_y}{dt}=F_y-\frac{\mu_k Nv_y}{\sqrt{v_x^2+v_y^2}}$$subject to the initial conditions $v_x=V$ and $v_y=0$. You solve these two coupled ODEs simultaneously.

At very long times, the solution to these equations approaches: $$v_x\rightarrow \frac{\left[1-\frac{\mu_k N}{\sqrt{F_x^2+F_y^2}}\right]}{m}F_xt$$ $$v_y\rightarrow \frac{\left[1-\frac{\mu_k N}{\sqrt{F_x^2+F_y^2}}\right]}{m}F_yt$$

  • $\begingroup$ Thanks for your patience to answer @chester ....however in my model of scenario, the block does not possess any relative motion in the perpendicular direction. The block only possess a tendency to move in that direction because of the applied force, which in turn has been opposed by the friction applied in the perpendicular direction. $\endgroup$ – lancer Jul 10 '17 at 11:33
  • $\begingroup$ Once the block is sliding in the original direction, static friction has already been overcome, and any (even slight) force in the perpendicular direction will cause the block to move in that direction (as well as continuing to move in the original direction). $\endgroup$ – Chet Miller Jul 10 '17 at 12:39

I will try to answer your question within your chosen model of friction. Remember friction is a complex play between different surfaces that has to do with the intrinsic properties of those surfaces, or their interfaces. So your model of friction, I presume, is one basic model of mechanics of dry surfaces ("Coulomb's dry friction"). Remember also that this model implies ideal bodies, since some properties like roughness or their fragility (and how much of them is "peeled off" during the interaction) are synthesized in only 1 constant, and ideal forces (where the interplay is only in the macroscopic scale, while in reality the real process is happening between both surfaces, and hence is microscopic in essence).

The constant is somehow related to (and simplifies) those surface properties between the contacting surfaces that I will call their "roughness", for simplicity. But it is only a model for the "roughness" in one direction (in other directions there could be other "friction constants" implied, since the properties of 2 surfaces aren't necessarily the same in both directions of a plane or even perpendicular to the plane).

If you exert a parallel force to the surface, then you use the parallel constant and the normal force to calculate friction. If you exert a normal force then you would use a "normal" constant (representing the property of "roughness" between 2 surfaces in the normal direction) and the parallel force to account for the friction in the normal direction. The problem with this approach is that, if the bodies are rigid (as it is assumed in the model, since they don't break or deform and hence there is no "waste" of force due to deformation) then both bodies cannot approach each other (move in the "y" direction) while they are moving in the parallel ("x") direction, or else you need to account for the loss of matter.

So suppose the constant does account for the loss of matter (there's always some peeling off of atoms, at least). Then there could be some movement in the "y" direction, which will increase the friction (and the peeling) but to a really low extent, since there will also be an opposing force of equal magnitude from the other surface. Hence at some point there will be either an equilibrium closeness or (more realistically) a bouncing on and off of the surfaces (due to roughness). But in any case, there's a much more complex model of the problem going on, where you need to take a more "microscopic" point of view that the model doesn't really account for. Remember, what you called the "law of friction" is just a special case (and a simplification) of Newton laws in the microscopic scale applied to a macroscopic scenario of two dry rigid bodies in one direction of motion.

So, to finish answering your question, the "law" is not broken since: 1) your case intends to apply a simple model that does not account for the problems you are introducing (in the model, there is no movement in the "y" direction since the bodies are rigid and their "roughness" is only considered as a constant that limits their movement in a specific direction); 2) in case you are considering microscopic effects (that aren't taken into account by the model but you wish to make a more complex model using the same simple constants), then there's "a friction" in each direction. You could consider a "one vector" of friction that is composed by both directions of the force, but you also need to consider the different forces implied in the calculation. In that case, the "law of friction" still holds in each direction and in the composed vector, but at some point you will also need to account for the fact that there will be an equilibrium of some sort where a) there won't be movement in the "y" direction anymore (in the ideal case); or b) there will be bouncing, where your model of friction wouldn't apply anymore since you are considering the microscopic scale and what you used to call "friction constant" now is just the "laws of Newton" and you are considering the bodies have a realistic complex surface rather than a soft surface and a magic constant that makes them rough.

  • $\begingroup$ Thank you for your patience and time lukesw... It has given me some insights ....however just to be reassured of your point of view- you mean to say that the actual value of friction may be much more than (mu* N) ...which is nothing but a simplification of friction in one direction? $\endgroup$ – lancer Jul 10 '17 at 11:30
  • $\begingroup$ The simplification in 1D accounts for the linear effects exerted by the normal force. So in a sense, it accounts for the movement in the y direction (it assumes it is null and incorporates it in nu: its an empirical model!). But the linearity of the problem only holds to some degree (if the force is not too strong and there is not much peeling). What you could account for is friction in the y direction (through some nu_y), but as I tried to explain, you cannot have movement in both x and y direction within this model. You could TRY and calculate both, but I wouldn't count on its accuracy $\endgroup$ – LukeSW Jul 10 '17 at 22:06
  • $\begingroup$ I don't know if I answered your question in a way you will be satisfied, so I will try and address it more directly. The value of friction itself accounts for the movement in y, so friction (within this model) IS NOT bigger than mu*N... at least while the model still applies. That's because it is an empirical model, so the "nu" constant just takes into account everything (I know it sounds simplistic, but that's because IT IS simplistic, since it's empirical). BUT the model doesn't apply to every situation: non-rigid bodies, strong forces, etc... they need a new model. I hope this helps better $\endgroup$ – LukeSW Jul 10 '17 at 22:15

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