Law of friction violated?

Question

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??

Answer 1

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.

Answer 2

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: $$v_x\rightarrow \frac{\left[1-\frac{\mu_k N}{\sqrt{F_x^2+F_y^2}}\right]}{m}F_xt$$ $$v_y\rightarrow \frac{\left[1-\frac{\mu_k N}{\sqrt{F_x^2+F_y^2}}\right]}{m}F_yt$$

Answer 3

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.