The direction of static friction?

In my book, the following are described to be the steps that one must follow if he wishes to find the direction of static friction force acting on an object:

"(1) Draw the F.B.D(free body diagram) of the object in question with respect to the other object on which it is kept.

(2) Include the pseudo force also, if the contact surface is accelerating.

(3) Decide the direction of the resultant force and resolve this force into two components; one along the surface of contact and the other along the normal to this surface.

(4) The direction of static friction is opposite to the component of the result force along the contact surface."

Firstly, are these steps correct? Will I always obtain the correct answer by following these steps? Secondly, consider the following problem:

In questions like these, when I try to find the resultant force and then assign the direction to static friction, I find that my answer is incorrect! Are there some other steps to follow when I know that the object is definitely accelerating in one direction or another?

Edit 1: I approached the above question by doing exactly what my book suggests; first, I identified all the forces acting on the smaller block, being the normal force applied by the block A and its weight mg downward. Adding these two forces, I get another force that actually acts on the body like so:

I wonder if I'm not supposed to take either of the contact forces into consideration when drawing the F.B.D to figure out the direction of static friction acting on the body. Also, are there other steps for finding the direction of kinetic friction?

Edit 2: From Judge's answer, I have been able to figure out my mistake, it being that I completely missed step (3) and didn't resolve the resultant force into components! Sheesh. Sorry!

Much thanks in advance! :) Regards.

• If you have a question about a concept in a specific problem, you need to show your work so that we can specifically answer your difficulty. Tell us your reasoning for the direction you chose, and we can help correct your reasoning. – Bill N Sep 2 '16 at 16:36
• What does F.B.D. Mean? – aventurin Sep 3 '16 at 8:49
• @aventurin I think it means "Free Body Diagram". – Judge Sep 3 '16 at 9:12
• Yes, I've edited my question to indicate that it does mean free body diagram. Sorry! – user106570 Sep 3 '16 at 9:40

Firstly, are these steps correct? Will I always obtain the correct answer by following these steps?

Yes and yes, those steps are correct :)

Are there some other steps to follow when I know that the object is definitely accelerating in one direction or another?

I don't think so. Let's work through your example using those steps:

Inertial frame of table

1. The 10kg block (B) has a weight force downwards, $\vec{W_B} = -mg \ \hat{y}$. A is accelerating into B, compressing it until B provides an equal and opposite reaction on A. We'll call A's force on B $\vec{N_{BA}}$ and it points right.
2. In an inertial frame (e.g. the table's) we skip this step.
3. Resultant force on B, $\vec{R_B}$ is the vector sum of all forces on B: $\vec{W_B} + \vec{N_{BA}}$, pointing down-right. Resolving $\vec{R}$ into components perpendicular and parallel to the slope yields \begin{align} \vec{R}_{parallel} &= \vec{W_B} \\ \vec{R}_{perpendicular} &= \vec{N_{BA}} \end{align}
4. The direction of static friction, $\hat{Fr}_{BA}$ is the opposite to $\vec{R_{parallel}}$ : $$\hat{Fr}_{BA} = -\hat{R}_{parallel} = -\hat{W_B} = -(-\hat{y}) = \hat{y}$$

To stop the block slipping downwards, the frictional force must be equal and opposite to the downwards force \begin{align} \vec{Fr}_{BA} &= - \vec{W_B} \\ \mu |\vec{N_{BA}}| \hat{y} &= -(-mg)\hat{y} = mg\hat{y} \end{align} If you consider the horizontal components you can see that $|\vec{N_{BA}}| = m|\vec{a}|$, so $$|\vec{a}| = \frac{g}{\mu} = 19.62 \ \text{ms}^{-2}$$

Non-inertial frame of block B

1. Same as in inertial frame.
2. A's and B's frame is accelerating with respect to the table, so we add a pseudo force $\vec{P}$ pointing left. B is also accelerating downwards (due to gravity), so we add an additional pseudo force pointing up. We then add these 2 pseudo forces to every other object in the system (A and the table).
3. The resultant force on B is zero: $\vec{R}_B = 0$, which is not surprising in it's own frame. To get static friction, consider what happens to A from B's perspective. The resultant force on A is $\vec{P}_{vert}$, which accelerates it upwards.
4. The direction of static friction on A from B is in the opposite direction to $\vec{P}_{vert}$ $$\hat{Fr}_{AB} = -\hat{P}_{vert} = -\hat{y}$$ Every action has an equal and opposite reaction (cheers Newton!), so we also have a static friction on B from A $$\hat{Fr}_{BA} = -\hat{Fr}_{AB} = -(-\hat{y}) = \hat{y}$$ which is the same as in the inertial frame.

To stop A slipping upwards past B, the frictional force on A must be equal and opposite to the upwards force \begin{align} \vec{Fr_{AB}} &= - \vec{P_{vert}} \\ &= -(-W_B) \\ -\mu |\vec{N_{AB}}| \hat{y} &= -(--mg)\hat{y} = -mg\hat{y} \end{align} If you consider the horizontal components you can see that $|\vec{N_{AB}}| = m|\vec{a}|$, so $$|\vec{a}| = \frac{g}{\mu} = 19.62 \ \text{ms}^{-2}$$ which is also the same as in the inertial frame... nailed it! :) Now if you're like me, you're probably thinking "that was the worst piece of nasty, hateful, misery I've ever seen." And you'd be right! My advice would be just pick an inertial frame and be done with it.

Edit 1: Simple example

Let's walk through a more simple example; just block A and the table (no block B).

Inertial frame

1. See the diagram below. I start by adding all the original forces (left) and then I add all the action-reaction pairs (right).
2. Skip this step, because there's no pseudo forces in inertial frames
3. Take the vector-sum of all forces on A to calculate the resultant force on A, $\vec{R_A}$ \begin{align} \vec{R_A} &= \vec{D} + \vec{N_{AT}} + \vec{W_A} \\ &= D \hat{x} + (N_{AT} - W_A) \hat{y} \\ &= D \hat{x} + 0 \hat{y} \end{align} Now we resolve $\vec{R_A}$ into components that are parallel and perpendicular to the slope \begin{align} \vec{R}_{parallel} &= D \hat{x} \\ \vec{R}_{perpendicular} &= 0 \hat{y} \end{align}

4. We know the direction of static friction is opposite to $\vec{R}_{parallel}$, so $\hat {Fr_{A}} = -\hat{x}$. Done :)

Non-inertial frame

2. We add pseudo forces to all objects, such that the resultant force on A is zero, which is what one would expect in a frame, in which it's not moving. However, the resultant force on the table is now not zero.
3. The resultant force on the table is the vector sum of all forces on it \begin{align} \vec{R_{T}} &= \vec{P_{horz}} + \vec{N_{TG}} + \vec{W_T} + \vec{W_A} + \vec{N_{AG}} \\ &= -P_{horz} \hat{x} + (N_{TG} - W_T + W_A - N_{AG})\hat{y} \\ &= -D \hat{x} + 0 \hat{y} \end{align} Resolving this into components: \begin{align} \vec{R}_{T, \ parallel} &= -D \hat{x} \\ \vec{R}_{T, \ perpendicular} &= 0 \hat{y} \end{align}

4. The direction of static friction on the table is opposite $\vec{R}_{T, \ parallel}$:
$$\vec{Fr}_{TA} = -\vec{R}_{T, \ perpendicular} = -(-D) \hat{x} = D \hat{x}$$ Every action has an equal and opposite reaction, so as A is exerting a frictional force on the table rightwards, the table must be exerting a frictional force leftwards on A and voila $$\hat{Fr}_{AT} = -\hat{Fr}_{TA} = -\hat{x}$$

To be honest, pseudo forces make my brain explode. There's a helpful example over on wikipedia. In the example, the person is analogous to block B and the seat is analogous to block A. If you imagine being block B it's like being accelerated in a car; you get pushed back into your seat (which is the pseudo-force). Then at constant velocity you feel like your not moving, but the road is moving under you.

It's always worth bearing in mind these useful rules:

1. The magnitude of static friction is always less than or equal to the magnitude of the driving force
2. Frictional forces, like all forces, come in action-reaction pairs
3. You can take any frame of reference you like (in classical mechanics). After all, what's stopping you?
4. Pseudo-forces are just the result of coordinate transformations :)

I'll leave what happens in A's frame as a fun exercise for you.

• Yes, yes! That's correct! I just realized that my mistake is that I hadn't resolved the resultant force into components! Sheesh. Anyway, I have just one more doubt; how can we take the frame of reference of the block B itself when it is the body in question? Secondly, if we were taking the frame of reference of block A, we need to apply a pseudo force, yes? If we consider this pseudo force and the normal force too, things get more complicated. Can you please explain some more? – user106570 Sep 3 '16 at 9:38
• @KaumudiHarikumar "how can we take the frame of reference of the block B itself when it is the body in question?" The frame of reference can be taken anywhere. If you choose the body itself, the body just seems to be standing still (no acceleration) from this point of view. It is like sitting inside the trunk of a car; the car might be moving, accelerating, turning etc. but for your eyes it looks totally still. You just fell all these weird "pseudo-forces". – Steeven Sep 3 '16 at 10:41
• @KaumudiHarikumar "if we were taking the frame of reference of block A, we need to apply a pseudo force, yes?" Yes, of course, because this reference frame is accelerating, so pseudo-forces are felt. "If we consider this pseudo force and the normal force too, things get more complicated." Well, the pseudo-force simply appears as a consequense of the whole thing accelerating. The normal force is there as well, yes, because of the contact, but that's another thing. The problem should be solvable in this way nomatter if you look at object A or B (or the ground/table). – Steeven Sep 3 '16 at 10:46
• Okay, but it would be useless to take block B itself as the frame of reference if we wish to study the block B, no? Secondly, in his answer, Judge has not only taken B to be the frame of reference but he's also not considered the normal force N on the block. Why not? – user106570 Sep 3 '16 at 10:59
• Sorry for the slow response. I've made the mother-of-all edits to my answer, which will hopefully explain everything correctly :) Please let me know if you find anything incorrect or unclear. And I should also clarify, that my previous answer was technically incorrect, even though it wrangled the right answer somehow. Hopefully all good now, although I had to adopt a more complex notation to name all the bazillion forces. – Judge Sep 4 '16 at 13:48

There does not seem to be an error in those steps assuming that you do NOT include the static friction in steps 1-3. The static friction would be added after those steps.

Remember that the static friction will be parallel to the contact surface causing the friction on the object.

EDIT: There is a situation in which these steps can get tricky. That would be one in which there are no outside forces parallel to the contact surface if you ignore static friction. Consider a block resting on a piece of paper on a horizontal surface (a table). You gently pull horizontally on the paper and the block slides with the paper, relative to the table but at rest on the paper. Static friction between the paper and block makes the block move, but there are no other real horizontal forces to use in steps 3 and 4. But the reference frame of the paper is accelerating with respect to the table, so one would draw a pseudo-force opposite the acceleration of the paper.Then the direction of the static friction will be opposite the direction of the pseudo-force, which is in the "forward" direction. (And that is why I don't like to use pseudo-forces .... ;) ).

I much to prefer to play a thought experiment and see what direction the object will slide on a surface if there is zero friction. Then the friction is opposite that sliding direction.

• Hi Bill, I'm not an expert on this, but wouldn't you add a pseudo-force in step 2 from the motion of the block's frame relative to the table (the inertial frame). If you do this, you get your horizontal force with which you can find the direction of static friction. Sorry if I've misunderstood. – Judge Sep 2 '16 at 17:56
• Yes, you're correct. I simply never work with pseudo-forces that much, and I missed that. I'll come up with an edit to fix it. – Bill N Sep 2 '16 at 19:36
• I've not worked with them much either; I much prefer inertial frames and Newton's laws :) Nice edit, and good spot with the direction of the pseudo force pointing backward. I missed that subtlety, so I've edited my answer (which is hopefully correct now). – Judge Sep 2 '16 at 20:44

@Judge gave a quite good answer on how to follow your steps, but I will point out that there are other steps to follow than those you have listed, which will avoid pseudo-forces.

Are there some other steps to follow when I know that the object is definitely accelerating in one direction or another?

Yes, there are. And I personally find it much simpler.

Specifically, a simple method is to just randomly choose the direction of the static friction and add it to the free-body diagram. In the problem given, it must be either up or down, so let's say that we choose down. Then, when solved, if the result is positive then it was correct and the direction was indeed downwards; if the result turns out to be negative, you have chosen the wrong direction and it is upwards instead. In any case you get it right in the end, because a minus on a force just means that you should flip the force vector.

So, outlined, the steps would be:

a) Draw free-body diagram (for a stationary frame) and

b) draw static friction in a random direction along the contact surface (there are just two choices).

c) Solve the whole thing with Newton's laws as always.

d) When you have found the value of static friction, look at it:

• Is it positive then all is good;
• is it negative then you chose the wrong direction and you must flip it around.

In the problem given, the box on the side would have a free-body diagram with normal force $\vec n$, weight $\vec w$ and static friction $\vec{f_s}$. We set $\vec n$ rightwards, $\vec w$ downwards and let's choose $\vec{f_s}$ downwards as well as are arbitrary choice. Setting up Newton's 2nd law for the horizontal and vertical directions give: $$\sum F_x=m a_x\quad\Leftrightarrow\quad n=ma\\ \sum F_y=m a_y\quad\Leftrightarrow\quad -f_s-w=0 \quad\Leftrightarrow\quad f_s=-w=-mg$$

It here turns out that the value of $f_s$ is negative, so the chosen direction of the force vector $\vec{f_s}$ is wrong. It should rather be the other way: upwards.

If you had chosen upwards as the direction from the start, the equations would have become:

$$\sum F_x=m a_x\quad\Leftrightarrow\quad n=ma\\ \sum F_y=m a_y\quad\Leftrightarrow\quad f_s-w=0 \quad\Leftrightarrow\quad f_s=w=mg$$

The value of $f_s$ is positive, so the chosen direction is correct: upwards.

The given problem is quite simple since acceleration doesn't have any influence on the direction. If it was an incline moving instead of a wall, it would have. In that case you must know the direction of the acceleration - because you can't have two such unknowns where we don't know the directions. But usually you also do know or is able to easily guess the acceleration direction.

An extra comment to be aware of is that it is often easy to guess/see what the direction is right away before you begin. In the given problem that is possible.

Because, remember what static friction does: it tries to prevent sliding. So, it must always "hold back", when other forces try to start slide. It always "holds back". In your given problem, you can farely easily see that gravity pulls downwards with the weight $w$, so naturally static friction "holds back" by pulling upwards.

This easy and quick calculation-less method of "seeing" the direction is often possible because it often is clear in which direction the other forces (namely the resulting force) pull. But in some cases with many forces present in both directions this is not easy to see; for example if I was pulling up with a rope in the box in the problem. Then I would have an upwards pulling force, and the weight would still be downwards, and then the direction of static friction depends on which of these that "counts the most". If I pull a tiny bit, then the box would fall without static friction to help to hold up. But I pullvery much - more than gravity pulls - then the static friction must hold back downwards, to avoid the box from sliding up.

• Hi, thanks! I'm a little confused about one more aspect and I have described this in my comment on Judge's answer. Can you please help me to clear up that doubt? – user106570 Sep 3 '16 at 9:51
• @KaumudiHarikumar The aspects you ask about in the other answer are about pseudo-forces. My answer here avoids pseudo-forces all along, so that should clear it out. – Steeven Sep 3 '16 at 9:57
• Okay, but do you think you can answer that anyway? Just for my understanding..? – user106570 Sep 3 '16 at 9:58
• @KaumudiHarikumar I have done so to the answer above. I hope this makes it clear. – Steeven Sep 3 '16 at 10:47