# Friction direction when applied force and direction of motion not parallel

Clarification: My original description was not clear. The directions in the diagrams are NOT horizontal and vertical; they are all in a horizontal plane. Hopefully the new diagrams are more clear.

I am attempting to model a situation where a block is sliding on a horizontal frictional surface and has an externally-applied force. Traditionally, since the block is moving the friction force vector would be opposite the direction of motion, as shown in the first diagram below. The net force acting on the block would have a component perpendicular to the direction of motion, which would cause perpendicular acceleration. Experimentally, the mass does not accelerate perpendicular to the direction of motion if the applied force is small. This implies that the friction force also has a component that resists the applied force. My question is how to model the friction force component that is perpendicular to the direction of motion. What is its limiting magnitude?

To be clear, I am familiar with simple/traditional friction models that state the friction force vector must be parallel to and opposing the direction of motion. Like any model, that is a simplification of reality. Can anyone explain a more complete friction model?

The net force on the block will not always be perpendicular to the motion. It will only be if the resultant of the applied force and the frictional force is perpendicular to the motion,i.e, the component of F in the direction parallel to the velocity(and thus opposite to friction) cancels out the force of friction so that only the component perpendicular to motion remains. So if we see it from the top view :- Here, F is applied force, f is frictional force, Fy and Fx are components of force perpendicular and parallel to direction of velocity. For resultant to be perpendicular to velocity , $Fsin(\theta)= f$ Therefore for any particular magnitude of F there will be only two angles(the angle between the applied force F and the vertical) possible for which the resultant force is perpendicular to the motion. Friction will always act in the opposite direction of velocity and as the velocity changes direction , it will too. Hence, even if we apply force F such that the resultant is perpendicular to motion, the motion will be circular only for an instant as when the direction of velocity changes the resultant is no longer perpendicular to the motion. In the end, the velocity will be in the direction parallel to the applied force and continue that way.

• Thanks for the answer. However, I am not suggesting hat the net force is perpendicular to the motion. Rather, there is a component of the friction that is not parallel to the direction of motion, which is not consistent with the traditional/simple kinetic friction model. – JohnV2 Dec 2 '17 at 15:54

Since we're dealing with kinetic friction, the force of friction by definition points in the direction parallel to the relative slip velocity between the block and the plane at all times, so there will be no component of the friction force perpendicular to the direction of motion to model.

If you're still reading this response, then I'll assume you'd still like to know how to properly analyze this situation in full detail. I remember analyzing this exact situation when I took a course on Newtonian mechanics taught by Professor Oliver O'Reilly, who wrote and published Engineering Dynamics: A Primer as a supplemental textbook for physics and engineering students studying Newtonian mechanics. Much of what will appear in the following analysis can be found in the primer, but during the analysis I'd like to bring special attention to what O'Reilly has referred to as "The Four Steps" method, which can be implemented to analyze most situations encountered in Newtonian mechanics.

The four steps are:

1.) Define one or more coordinate systems and express the position, velocity, and acceleration vectors for each particle in the system, and state any kinematic constraints on the motion of each particle

Let's choose a Cartesian coordinate system defined by unit vectors $\{\mathbf E_x, \mathbf E_y, \mathbf E_z\}$ such that $\mathbf E_x$ points to the right, $\mathbf E_y$ points into the screen you're currently looking at, and $\mathbf E_z$, which defines the normal unit vector of the surface on which the block moves, points toward the top of this page. For the block,

$$\mathbf r = x\mathbf E_x + y\mathbf E_y + z\mathbf E_z$$ $$\mathbf v = \dot x\mathbf E_x + \dot y\mathbf E_y + \dot z\mathbf E_z$$ $$\mathbf a = \ddot x\mathbf E_x + \ddot y\mathbf E_y + \ddot z\mathbf E_z$$

are respectively the position, velocity, and acceleration vectors. Since the block is constrained on the horizontal surface, $\dot z = 0$ and $\ddot z = 0$. Also, since the surface itself is stationary, the slip velocity $\mathbf v_{rel}$ of the block relative to the surface is $\mathbf v$, simply the velocity of the block.

2.) Draw a free-body diagram (FBD) for each particle and label the forces acting upon them

Using the first diagram in your original post as our FBD, let's identify all of the forces acting upon the block. You have already labelled two forces: the applied force $\mathbf P$, which, in its most general form, is

$$\mathbf P = P_x\mathbf E_x + P_y\mathbf E_y + P_z\mathbf E_z$$

and the force of kinetic friction $\mathbf F_f$, which by definition is

$$\mathbf F_f = -\mu_k |\mathbf N| \frac{\mathbf v_{rel}}{|\mathbf v_{rel}|} = -\mu_k |\mathbf N| \frac{\dot x\mathbf E_x + \dot y\mathbf E_y}{\sqrt{\dot x^2 + \dot y^2}}$$

where $\mathbf N$ is the normal force of the surface acting on the block. Speaking of which, let's add that normal force into our FBD:

$$\mathbf N = N\mathbf E_z$$

The fourth and final force we need to consider is the weight of the block,

$$\mathbf W = -mg\mathbf E_z$$

3.) Use Newton's 2nd Law to construct the equations of motion (EOMs) for each particle

Newton's 2nd Law states that,

$$\mathbf F_{net} = m\mathbf a$$

where $\mathbf F_{net}$ is the sum of all forces acting on a particle with mass m, and $\mathbf a$ is the acceleration of the particle. Let's add up the forces from our FBD and break Newton's 2nd Law into 3 equations -- one for each basis vector of $\{\mathbf E_x, \mathbf E_y, \mathbf E_z\}$:

$$P_x - \mu_k N \frac{\dot x}{\sqrt{\dot x^2 + \dot y^2}} = m \ddot x \tag{\mathbf E_x}$$ $$P_y - \mu_k N \frac{\dot y}{\sqrt{\dot x^2 + \dot y^2}} = m \ddot y \tag{\mathbf E_y}$$ $$P_z + N - mg = 0 \tag{\mathbf E_z}$$

4.) Perform mathematical analysis on the EOMs, and introduce any other fundamental laws of physics that may be relevant to the analysis

We see from the equation for $\mathbf E_z$ that

$$N = mg - P_z$$

Plugging this into the equations for $\mathbf E_x$ and $\mathbf E_y$ and then rearranging, we end up with

$$\ddot x + \mu_k \left(g - \frac{P_z}{m}\right) \frac{\dot x}{\sqrt{\dot x^2 + \dot y^2}} = \frac{P_x}{m}$$ $$\ddot y + \mu_k \left(g - \frac{P_z}{m}\right) \frac{\dot y}{\sqrt{\dot x^2 + \dot y^2}} = \frac{P_y}{m}$$

a set of coupled, second-order nonlinear ODEs of functions $x$ and $y$ , with boundary conditions $\mathbf r_0$, the initial position of the block, and $\mathbf v_0$, the initial velocity of the block.

This is the point where I would go find my friendly neighborhood computer programmer, who would then kindly do all the work of writing out a code that would numerically find approximate solutions to these ODEs for me. However, if you're more experienced with coding than I am, feel free to write up a code for the equations and try specifying various values for the vectors $\mathbf P$, $\mathbf r_0$, and $\mathbf v_0$, and see what you come up with. To verify your code is working properly, make sure to give the code some basic test examples such as

$$\left(\mathbf P, \mathbf r_0, \mathbf v_0\right) = \left(\mathbf 0, \mathbf 0, v_0 \mathbf E_x\right)$$ $$\left(\mathbf P, \mathbf r_0, \mathbf v_0\right) = \left(mg \mathbf E_z, \mathbf 0, v_0\mathbf E_x\right)$$ $$\left(\mathbf P, \mathbf r_0, \mathbf v_0\right) = \left(\mu_k mg \mathbf E_x, \mathbf 0, v_0\mathbf E_x\right)$$

which all result in fairly simple closed-form analytical solutions to the ODEs.

• Thanks for the detailed answer. However, my question is not about solving the motion equations. Rather, it is about explaining the observed friction component perpendicular to the direction of motion. Simple friction models state that friction is entirely parallel to the direction of motion but that is not what we are measuring. – JohnV2 Dec 2 '17 at 15:51
• And that simple friction model, which I have used in the analysis, is completely valid to use in this situation. The force of friction will be entirely parallel to the direction of motion, and you will not observe any component perpendicular to the direction of motion. This is validated by the fact that the EOMs I derived using the simple friction model give accurate predictions of the behavior of the object. – JM1 Dec 3 '17 at 1:14
• Yes, your equations of motion appear to be correct given the simple traditional friction model. My question, which I am not stating very well, is about the friction model not the equations of motion. We have empirical data that strongly suggests the simple friction model is not adequate in our situation. The block is not accelerating perpendicular to the direction of motion despite an applied perpendicular force. Therefore, there appears to be some friction perpendicular to the direction of motion. What could physically cause that? – JohnV2 Dec 3 '17 at 2:23
• Do you have a link or reference to the empirical data in question? I would like to take a look at it in order to clarify exactly what you're talking about. – JM1 Dec 3 '17 at 2:36
• @JohnV2 As per my comment above, if you are making a claim which contradicts the traditional friction model, you must provide the evidence for that claim. You must describe your experiment and the results you got. – sammy gerbil Dec 7 '17 at 12:31

Friction force is by definition always against velocity. That is: $$\vec{F}=- \mu \vec{v}$$ On another side, acceleration is always in direction of force, so your pictures are simply nonsense, an has nothing in common with equations of motion. In order to build model, just try to put all vector forces in your problem, add them as vectors and put it in Newton equations $$\vec{F}= m \vec{a}$$ where a us acceleration. Take into account every force important for your model. If all other forces except friction are constant, you will obtain differential equations with friction force depending on unknown velocity and acceleration ( first deferential of velocity). If you solve it ( maybe working in certain frame of reference) you would get general solutions, which incorporated with initial conditions will get you solution of the problem.

If force H is applied, it causes acceleration. If there is no acceleration, you have another force, probably something you called friction F, which cancel applied force. In that case you there may be several cases.

1. You may have some bounds in the system. For example you applied force but on the other side of the body there's a wall, so effectively you are pushing moving body to the wall. Wall has some strength and as long as you do not apply force strong enough to break it, there would be no effect of your force ( in ideal case if contact of a body with wall do not produce friction) or there would be addition force, friction from contact of the body and the wall. Equations of motions are the same as above, friction force have to be included, in this case. Suppose it is proportional to applied force H, and of course it is opposite to the speed. In this case I will add the following component ( where |°| means the norm, length of the vector). $$\vec{F} = \frac{\vec{v}}{|\vec{v}|} |\vec{H}| \mu$$

1. You may have some step effect: applied force H have effect if it is strong enough to break some initial resistance. Below such value it has no effect. It reminds moving body in the snow and in fact has the same equations as above but with term with Heviside step function theta, with step at $H_0$ value, which defines value of the force where acceleration will appear: $$\vec{F} = \frac{\vec{v}}{|\vec{v}|} | \vec{H}|\mu ( \theta( | \vec{H}| - H_0 ) )$$

Od course it is the simplest model where F and H are linearly dependent. Case 2 is the same as case 1 when $H_0$ is wall strength ( and in most cases it means it is infinite). You may have a case where F depends on F in much complicated way, where combination of several regimes occur. Relationship between F and H may be from some mechanical properties of surface structure, and reminds Young graph of mechanical deformation against force applied ( there is linear, nonlinear, plastic and break deformation on the graph. )

• That's an aggressive response. I understand the equations of motion and that friction should oppose the direction of motion in simple friction models. However, we have a physical system that demonstrates a friction component perpendicular to the direction of motion, which means there is underlying physics not incorporated into the simple friction model. – JohnV2 Dec 1 '17 at 21:47
• No it is not. Your idea is in contradiction to Newton F = ma law, which is very basic background of mechanics. Do not be upset, just read something about it. – kakaz Dec 2 '17 at 8:27
• My question must not have been written very clearly. It has nothing to do with acceleration. There is a block moving in one direction. A force with a component perpendicular to that direction is applied. Despite what simple friction models predict, the block does not accelerate perpendicular to its direction of motion. Therefore, there is a component of the friction force opposing the perpendicular component of the applied force. The aggressive part of your response is "your pictures are simply nonsense". – JohnV2 Dec 2 '17 at 17:41
• I have to upgrade my answer. – kakaz Dec 3 '17 at 10:24
• Thanks for the updated answer. The analogy of pullIng a sled through snow is useful. Now I need to figure out how large the perpendicular force could be. For that I’ll need to expand my model beyond friction. – JohnV2 Dec 7 '17 at 16:17

In the second case, decompose the force $\vec{F}$ into horizontal ($x$) component and a vertical ($y$) one:

$$F_x=F\cos\theta$$ And:

$$F_y=F\sin\theta$$

The $x$ component is what might overcome friction and thus cause motion acc. Newton's second.

The vertical one reduces the Normal force, caused by the weight $mg$:

$$F_N=mg-F_x=mg-F\cos\theta$$

This then gives rise to the friction force $F_f$, opposing $F_x$:

$$F_f=\mu F_N$$

If:

$$F\cos\theta > \mu(mg-F\cos\theta)$$

there will be acceleration to the right.

But there is no vertical friction at all.

Should it be the case that:

$$F\sin\theta > mg$$

there will be acceleration in the $y$ direction (upward).

• My description was not clear enough. See clarification at the top of the question. – JohnV2 Dec 1 '17 at 21:42
• Sorry, I should have also thanked you for the answer, which was absolutely appropriate and correct for the question that it seemed I was asking. – JohnV2 Dec 1 '17 at 21:55
• Better than thanking is upvoting, that's how this site works! ;-) – Gert Dec 1 '17 at 22:13
• I didn't up-vote because it doesn't answer the question I was actually asking (vs. the question it seemed like I was asking). – JohnV2 Dec 1 '17 at 22:31

Maybe... (only maybe) you are wrong with the concept of friction force. Your pictures are not enough clear (I have read your clarification but I don't see yet how are the directions of the forces)

Only to clarify : fricction is, by his own definition, opposite to the moviment direction, that's it. If you apply a force $\vec F1$ then the friction force will have the direction of $\vec F1$ and the opposite sense. Now, if you apply another external force $\vec F2$ which has not the moviment direction then you can easily add up the forces (vectorially, of course) and to obtain the net force $\vec F = \vec F_1 +\vec F_2$ then, the new friction force will have the new direction of movement with the opposite sense, no more.

With respect to a perpendicular component of the friction force that you mentioned, only to say that :

Friction is generated due to the imperfections, mostly microscopic, between the contact surfaces. These imperfections make the perpendicular force $\vec R$ between both surfaces is not perfectly, but that it forms an angle with the normal $\vec N$ (the angle of friction). Therefore, the resultant force is composed of the normal force $\vec N$ (perpendicular to the contact surfaces) and of the friction force $\vec F$, parallel to the contact surfaces.

What I'm telling you what you can see graphically in this image :

http://www.fotoseimagenes.net/imagenes/full/0/5/2/friccion-4.jpg