# Is friction always opposite to velocity?

Let's say an object is sliding on a slope and is the object has a velocity of $$(0,0,5)$$. The friction would be acting in the opposite direction of motion, being $$(0,0,-1)$$.

However, gravity is also affecting the object on the slope. The gravity is exerting a force of $$(-1,-1,0)$$ in the direction of the slope tangent. Should the object also experience friction in the direction of $$(1,1,0)$$? Or would it only experience friction in the direction of $$(0,0,-1)$$?

What confuses me is that if friction is only acting in the opposite direction of motion, when an object is standing still on a slope due to friction acting in a direction opposite to gravity, when you add a perpendicular impulse the object will suddenly be sliding downwards as well because friction is no longer acting in the direction opposite to gravity. Is that correct?

• What do you mean by a "perpendicular impulse"? Commented Apr 27, 2021 at 19:40
• @BobD I tried my best to illustrate i.imgur.com/hapvc9b.png Note that the impulse is actually in the Z direction and not in the XY plane but it was hard to illustrate this. Commented Apr 27, 2021 at 19:48
• can we assume the impulse force is temporary, i.e., that it ends once sliding starts? And can we also assume the magnitude of the impulse force is not great enough to separate the object from the slope? Commented Apr 28, 2021 at 10:39
• Also, I am trying to understand your use of coordinates on the png file. Apparently, the z-axis is perpendicular to the slope (the direction of the impulse force), the x- (or y-) axis is down the slope and the y- (or x-) axis is into the page. In the first paragraph you have the velocity in the z-direction, which wouldn't make sense. Commented Apr 28, 2021 at 10:57
• @BobD Sorry for my poor representation. 1. The impulse is temporary, I thought that's what an impulse was? 2. For the purpose of this problem, the object will never get separated from the slope. 3. The Z axis is tangential to the slope, I guess I've just drawn it weirdly. It's a default left handed coordinate system. and the Z axis is perpendicular to the XY plane. The XY plane is your screen. The Z axis goes through the screen. Just default. Commented Apr 28, 2021 at 12:22

As pointed out in the answer by Nuclear Hoagie: there is the distinction between static friction and dynamic friction.

Here is an example of a situation where I expect a very large difference between static friction and dynamic friction.

You have a floor that is carpeted, and a piece of carpet is lying upside down on that, and the upside-down piece of carpet is pressed down by an upside down table, and it's a very heavy table.

Your task is to drag that upside-down-carpet-and-table assembly to the other side of the room. The hairs of the two carpets will tend to interlock, so it takes a lot of force to get going, but you know that once you get it going you will be able to keep it going. Once you get the two carpet sides to go out of interlocking there will still be friction, but not as much as at the start.

In most cases the difference between the static friction and the dynamics friction will be smaller than in the above example, but there will always be some difference.

So let's say an object is stationary on a slope and there is just enough static friction to keep it from sliding down the slope.

Next you give it a push to the side, making it slide sideways with respect to the slope. Now that the object is moving the friction is dynamic friction.

Dynamic friction is less, so it may well be that now the object will slide downhill.

• But what if the object was initally already moving with a small downwards velocity? with a kinetic friction of the same magnitude of the projected gravity on the slope. If we give it a large sideways impulse, according to the model, it should suddenly start accelerating downwards because the direction of kinetic friction will no longer oppose gravity Commented Apr 27, 2021 at 19:55
• i.imgur.com/hapvc9b.png might make it more clear Commented Apr 27, 2021 at 20:03
• @xcrypt To answer to your comments: If you give it "a large sideways impulse", so the object slides sideways on the slope instead of downwards or upwards, then a kinetic friction will appear in the other sideways direction (opposite). Now that the object is in motion, static friction no longer holds it stationary against gravity, so it will also start sliding downwards in the next moment - then the motion is not entirely sideways but now at an angle. Kinetic friction will push opposite to this new direction, whichever it is. Commented May 1, 2021 at 11:29
• @xcrypt The image you link to in the comment does not show a "large sideways impulse" but rather a perpendicular impulse (perpendicular to the slope). If this happens, then the object is lifted off of the surface - then there is no friction at all. Friction requires contact. Then gravity will cause it to accelerate back down again, but now it has been given som initial speed upwards at an angle. Should this happen, then the object will be falling through the air in a curve, a projectile motion, which does not involve friction. Commented May 1, 2021 at 11:31

$$\underline{\textit{Qualitative analysis}}$$

It is easily observed that the contact force referred to as kinetic friction is applied by one surface on another in the direction opposite to that of the relative velocity of the latter surface.

Let us conduct the following static friction related thought experiments. These empirical thought-proofs show that this type of contact force is indeed, applied in the opposite direction of 'impending' relative velocity. Let $$f$$, $$\mu_s$$, $$W$$, $$f\leq F$$ and $$\tau$$ denote the relevant component of force of friction, relevant component of static friction coefficient, weight of the object and the externally applied force and torque on the body.

$$\textit{Friction block thought experiment:}$$ In the figure below, we know that in the static case, $$f=F$$ and that if $$F=f\leq\mu_sW$$, then the block has vanishing acceleration. Clearly, the 'impending' relative velocity of the block is in the direction of $$F$$ towards the right of the figure and the force of static friction is acted by the ground surface on the block surface in the opposite direction of this relative velocity.

$$\textit{Rolling without slipping thought experiments:}$$ In the case of the circular body rolling without slipping in the figures below, we know that the translational velocity (in the direction of $$F$$ or towards the right of the figure), $$v$$, of the center of the circle is given as $$v=-\omega R$$ due to the assumption of rolling without slipping, where $$\omega$$ is the non-vanishing $$Y$$ component of the rotational velocity (with the other components necessarily being vanishing due to the assumption of planar motion) and $$R$$ is the radius of the circle. Therefore, assuming that the center of the circle is also the center of mass (COM) of the body, we obtain the translational acceleration of the COM as $$a = -\alpha R$$, where $$\alpha$$ is the non-vanishing $$Z$$ component of the rotational acceleration (with the other components necessarily being vanishing due to the assumption of planar motion) of the body. Further, the analysis of the angular momentum implies that $$\tau_\text{ext}=I\alpha$$ where $$\tau_\text{ext}$$ is the externally applied torque on the body and $$I$$ is the moment of inertia about the axis passing through the center of the circle. Further, the rolling without slipping phenomenon implies that the relative velocity of the contact point of the circular body with respect to (w.r.t.) that of the ground surface is vanishing. In both situations depicted in the figure below, this assumption implies that $$f=\mu_s W$$.

In both thought experiments shown below, the rotational velocity and acceleration of the body are measured positive in the direction of a right hand screw being screwed out of the screen. The coordinate system used is $$XYZ$$ with the $$X$$ axis pointing to the right of the screen parallel to the ground surface and the $$Z$$ axis pointing vertically downwards. In both cases below, the COM of the body will (using our physical intuition in the thought experiment) accelerate towards the right of the page, that is, in the direction $$+X$$.

• Force driven wheel (figure on the left): The Newton's laws of motion imply that $$F-f=\frac{W}{g}a=-\frac{W}{g}\alpha R$$ and $$-fR=I\alpha$$ which implies that $$0\leq a$$, $$\alpha\leq 0$$. We observe that if the direction of the static friction force is reversed, we would obtain a contradiction since the rolling without slipping condition would be violated (because the direction of acceleration obtained would be opposite to that required in the known relationship $$\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in the direction of the applied force $$F$$ which points in the $$+X$$ direction, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $$m:=\frac{W}{g}$$, then $$I=m\frac{R^2}{2}$$, so that the equations of motion yield $$F=f$$. The static friction condition $$f\leq \mu_s W$$ therefore implies that $$F=3f\leq \mu_s W=3\mu_s mg$$, which provides the upper bound on the driving force which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $$a=\frac{2f}{m}$$ allowable under the rolling without slipping regime. In fact, this is the underlying reason why circular wheels are more efficient than non-circular ones.
• Torque driven wheel (figure on the right): The Newton's laws of motion imply that $$-\tau+fR=I\alpha$$ and $$f=\frac{W}{g}a=-\frac{W}{g}\alpha R$$, which implies that $$0\leq a$$, $$\alpha\leq 0$$. Clearly, assuming that the direction of the friction of force is opposite to that shown in the figure will lead to a contradiction violating the rolling without slipping condition (because the direction of acceleration obtained would be opposite to that required in the known relationship $$\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in opposite to the direction of the applied force $$F$$, that is in the direction $$-X$$, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $$m:=\frac{W}{g}$$, then $$I=m\frac{R^2}{2}$$, so that the equations of motion yield $$\tau=-\frac{3}{2}{f}{R}$$. The static friction condition $$f\leq \mu_s W$$ therefore implies that $$\tau\leq \frac{3}{2}\mu_s WR=\frac{3}{2}\mu_s mgR$$, which provides the upper bound on the driving torque which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $$a=\frac{f}{m}$$ allowable under the rolling without slipping regime.

$$\underline{\textit{Conclusions}}$$

1. The contact force referred to as kinetic friction is applied by one surface on another in the direction opposite to that of the relative velocity of the latter surface w.r.t. the former surface.
2. The contact force referred to as static friction is applied by one surface on another in the direction opposite to that of the 'impending' relative velocity of the latter surface. The direction of the impending velocity, which is a fictitious quantity, is in the direction of the relative acceleration (w.r.t. the surface applying the force) of the point of contact resulting from the dynamics in which the friction force of interest is fictitiously assumed to be vanished w.r.t. the former surface.

Kinetic friction always acts opposite to relative motion, while static friction acts opposite to the tendency of motion. Kinetic friction is a dissipative force that turns kinetic energy into non-useful waste heat. If kinetic friction could act in the same direction as relative motion, that would imply that friction could increase the magnitude of an object's velocity relative to the surface, imparting useful kinetic energy from somewhere. But kinetic friction only turns movement into heat, it cannot turn heat back into motion. Friction can never make an object move faster, only slower - it always acts opposite to relative velocity.

• Static friction (from the road) acting on a powered wheel can cause or maintain motion in the direction that the friction is acting. Commented Apr 27, 2021 at 19:31
• @R.W.Bird I don't quite follow. A car on a frictionless road would have the tires spinning in place, moving backward at the contact patch. Friction opposes that motion and pushes forward so that the contact patch stays stationary relative to the ground. Perhaps I should have been more specific that friction opposes relative motion between the touching surface - the fact that the body of the car moves forward relative to the ground doesn't really matter, since the body of the car and the road are not in contact, so there's no frictional force between them. Commented Apr 27, 2021 at 19:52
• You can have kinetic friction acting in the direction of the velocity of the object. e.g. imagine dropping a box onto a moving conveyor belt. The box will slide at first, picking up speed in the direction of the friction force. The key is not to focus on the velocity of the object as seen from an arbitrary frame, but rather the relative motion between the surfaces, which is not frame-dependent Commented Apr 27, 2021 at 20:12
• @BioPhysicist I've made the notion of relative velocity more explicit. I didn't think that was totally necessary, as all motion is relative - you can't change the way friction acts simply by walking toward or away the object under study. There isn't really any such thing as the universal "direction of velocity", as it all depends on the observer. In friction problems, it's usually simplest to treat the surface providing friction as the observer so that relative motion is built-in to the problem formulation. Commented Apr 27, 2021 at 20:37
• Nuclear Hoagie: I did not mention friction-less. The drive wheels on a car are pushed forward by static friction from the road. Commented Apr 28, 2021 at 16:00

A better phrasing is:

Friction always pulls in the direction that prevents sliding (often called relative motion).

Remember, there are two types of friction:

• Kinetic friction when the object slides. To prevent sliding (to stop sliding), kinetic friction pulls exactly opposite to the velocity. Regardless of any forces acting.

• Static friction when the object is stationary but there are forces trying to make it slide. To prevent sliding (to keep the object stationary), static friction pulls opposite to whichever direction the net force pulls.

Gravity is an external force. When standing still on flat ground, gravity doesn't try to initiate sliding, so there is no need for a static friction to act opposite to gravity. When standing on a slope, a component of gravity pulls along with this slope and tries to start sliding down - static friction thus must pull opposite to this, so up the slope. But not if I at the same time push on the object up along the slope - then static friction might have to pull downwards to prevent the object from sliding upwards.

So static friction is not related to gravity in general. Gravity is just one possible force to hold back against. There is no requirement for frictions to act opposite to gravity.