I had a question which has always been of slight confusion to me. Say you are dealing with your typical block-on-an-angled-plane setup.

You have a block with mass m that is initially at rest. The block sits on a surface at angle θ to the ground, has coefficient of kinetic friction μk, and a coefficient of static friction μs. There is an external force F acting on the block parallel to the ground. I am wondering, if you are given this setup, how to determine the direction of friction without first determining the direction of motion?

Problem Diagram

In this case the external force opposes the force of gravity.

To avoid the static case, let's set μs = 0, so the block would move immediately.

Now say we take the relevant components, F|| and Fg,||, as the two forces as forces which will determine the direction of motion of the block. We now have two scenarios, if the residual force

Fnet = F|| - Fg,||

Say we defined up the plane as positive. We will now consider the two possibilities:

If Fnet > 0, the applied force is greater and the block moves up the plane with Fk < 0, opposing motion and directed down the plane.

If Fnet < 0, the gravitational force is greater and the block moves down the plane with Fk > 0, opposing motion and directed up the plane.

My question comes in at this point, if this analysis is not performed how is the direction of motion determined?

F = ma = Fk - Fg,|| + F|| = Fk + Fnet

ma = (+/-)Fk + Fnet


An attempt to clarify my question:

My question was directed only at the direction of kinetic friction, for simplicity's sake imagine that the coefficient of static friction is 0 (μs = 0) or that the force interaction is so great that it is relatively 0 (Fnet >> Fs). Or that we are concerned only with the block after it has overcome Fs, and then the question can be analogized to the static case, and finding which direction the block moves.

Under these conditions it can be imagined that the block will move immediately.

It can be simplified to "how to determine direction of motion" if it suits you better, as I am aware that Kinetic friction is a reactionary force, so my question does indeed boil down to this.

Given the setup provided: Block on an inclined plane, External force opposing gravity, no static friction, and block initially at rest. Which of the following three options is the most trivial:

  1. Solving for Fnet and inferring the direction of motion from the acceleration, which will then reveal the direction that kinetic friction will act

  2. Other solution that does not involve first solving for Fnet first, essentially my question asks if this option exists, determining direction of motion without going into any grainy details of force interaction. If it does, then I would love it if someone could explain it to me, if it doesn't so be it.

Is there a way to determine the resultant direction of motion using a physical argument and not a mathematical one?

  • $\begingroup$ the friction is a consequence of the movement of the body and it's always opposite to it. $\endgroup$ Commented Mar 23, 2014 at 2:03
  • $\begingroup$ @tcassanelli So, in regards to my question, you are saying that the sign applied to the friction would only be determined after you performed an analysis on the other forces, find the overall direction, and only then can the direction be determined? $\endgroup$
    – farmerjoe
    Commented Mar 23, 2014 at 2:06
  • 2
    $\begingroup$ If it is just a classical mechanic effect that would be it. $\endgroup$ Commented Mar 23, 2014 at 2:07
  • $\begingroup$ @tcassanelli Ok cool, thank you! Out of curiosity what effect could friction be thought of other than a classical mechanic one? $\endgroup$
    – farmerjoe
    Commented Mar 23, 2014 at 2:14
  • $\begingroup$ friction is an electrical effect between atoms and depending on that the effect could be different. But they are not important in this case, only in an atomic and sub atomic scale $\endgroup$ Commented Mar 23, 2014 at 2:17

2 Answers 2


Since we know that the block is restricted to move only along the inclined plane, all the force components perpendicular to the inclined plane come in action-reaction pairs( assuming that the inclined plane is stationary and perfectly rigid ) and therefore I shall not include those components in my free body diagrams in order to make things clear.

First, let us take a look at the free body diagram of the block when it is at rest in the absence of the external force F.

enter image description here

In this case, the components perpendicular to the inclined plane ( $mg\cos B$ and N ) cancel each other. Since we are only concerned about the motion along the inclined plane, I have not included them in the diagram. The components along the plane( $mg\sin B$ and $F_s$ ) also come in pairs. The net force on the block is therefore 0 N and the block remains at rest..

*Note: One cannot avoid the case of static friction. If one looks at the region of contact between two surfaces( requires a great extent of magnification ), one would find many ups and downs( mountains ) on each of the surfaces that are locked into each other. This mechanism prevents relative motion. Friction arises as a result of this mechanism. However, at the atomic level, it is the electromagnetic interaction that is responsible for various forces including friction. To assume $\mu_s = 0$ is to assume that the surfaces are smooth. In that case, μk will also be zero.*

Now, let us look at the free body diagram of the block at rest when an external force F is acting on it.

enter image description here

Figure 1 shows the components of the external force. $F\cos B$ acts along the inclined plane and $F\sin B$ acts perpendicular to the plane.

Figure 2 shows various forces acting along the inclined plane. Once again, we are bothered about the motion along the inclined plane and hence ignored the components of force perpendicular to the inclined plane as they always come in action-reaction pairs.

From Figure 2, one can observe that the magnitude of static friction has reduced by $|F\cos B|$.

Note: Static friction is a variable physical quantity.

Even in this case, a net 0 N force is acting on the block and it remains at rest.

Now, as the magnitude of the external force keeps increasing, the magnitude of static frictional force keeps decreasing( provided the block remains at rest ) and the component of the external force up the inclined plane keeps increasing so as to see that the block remains at rest. At one instant, the static frictional force becomes 0 N and $F\cos B = mg\sin B$.

enter image description here

So, as the magnitude of the external force increases, the component of the force up the inclined plane increases.

For $F\cos B > mg\sin B$, a net force $F\cos B - mg\sin B$ acts on the block up the inclined plane. Now, static friction reverses its direction (down the incline) in order to prevent the block from moving up the inclined plane. As $F\cos B$ increases up the inclined plane, static friction increases down the inclined plane in order to prevent relative motion thereby keeping the block at rest as shown below.

enter image description here

At one instant, when static friction reaches its maximum value, an infinitesimal increase in $F \cos B$ will make the block move up the inclined plane and this is when kinetic friction comes into play.

enter image description here

Kinetic friction

enter image description here

If $F\cos B - (mg\sin B + F_s) > F_k$ (kinetic friction), the block accelerates up the inclined plane.

Note: Kinetic friction is a constant physical quantity and it is always less in magnitude when compared to static friction.

Kinetic friction opposes relative motion. So, in this case, the direction of $F_k$ is down the inclined plane( opposite to the direction of motion as shown in figure 3 ).

Analysis can be made by first studying all the forces involved and depending on the direction and magnitude of all the forces involved, one can determine the direction of motion.

  • 1
    $\begingroup$ Thank you! You have answered my question, slightly indirectly, but nevertheless. I made an edit to slightly clarify what I am asking towards the end of my post in case you have any more advice you'd care to share. $\endgroup$
    – farmerjoe
    Commented Mar 24, 2014 at 14:25

First remember that it isn't the acceleration/force that matters. It's the velocity -- friction opposes motion, not acceleration. So if the block starts with a upward motion, the friction will point down, no matter what the forces are.

So you definitely need to know about the direction of the velocity. If you have an initial velocity, this isn't a problem because it's given anyway. But if you start from a standstill, you need to determine the direction in which the block will move. But in this case, you don't need to consider the friction at all! Just determine the direction of the other forces. If it is greater than the static friction, it will move and then the regular friction will apply to oppose the direction of motion (which is now equal to the direction of the non-friction forces). If it's not enough, it won't move at all.


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