How can deceleration due to friction be positive?

We really need to see your FBD and equations for this because the issue is you are either not being consistent in your FBDs when you assign polarities to the directions of the vectors or you are being consistent but don't understand it.

Nothing says the force you exert has to be positive. Nothing says friction has to be negative. Positive and negative are just math conventions. Positive and negative are arbitrary. All that is required is that they are opposite. The math is dumb and has no context to the problem. It just knows positive and negative are opposite and it is up to you to assign meaning to them. And you have to be consistent since the math has to be consistent.

For a falling object problem, I could define everything pointing down as negative and everything pointing up as positive. Or I could do the opposite and define everything pointing up as negative and everything down as positive. It doesn't matter as long as I am consistent.

If your pushing force was to the right, you could define everything pointing right as positive and everything right as negative. Since friction would be opposite of your pushing force then friction would be negative. But you could also do the opposite. You could define left as positive and right as negative. In that case the pushing force would be negative and the friction force would be positive. Does it matter? No, it doesn't. What matters is the friction and pushing are opposite of each other.

The math only cares about consistency.

In fact, in some problems you don't actually know what direction a force points ahead of time. So you have to "guess" by just assigning it a direction and therefore a polarity based on your convention and working the problem out. And if the number you calculate is positive it means that you guessed correctly. But if the number is negative then that means you guessed incorrectly and it is actually in the opposite direction of what you guessed.

---

For example:

If my convention is

• Down = positive
• Up = negative

The usual convention is up is positive and down is negative but I chose a convention that is unusual on purpose to prove the point that it doesn't matter and to try and break you out of rigid thinking if you happen to be stuck in it.

Let's say I have a block supported on a table with weight of m. I know there's the acceleration of gravity pulling the weight down. We intuitively know that the table has to exert a force up to stop the block from falling.

$Sigma F = ma$

$(mg) - (N) = m(0frac{m}{s^2})$

When the forces are on the same side of the equal sign, everything in one direction has one sign and the everything in the opposite direction has the opposite sign based on our sign convention. The same holds true for the acceleration terms (the 'ma' terms but there is only one in this problem).

If we calculate the above out, we get:

$N = mg$

Great. We guessed correctly because our N is positive.

---

But let's pretend we DO NOT know that the normal force will ultimately point up to opposite gravity. In that case, we have to guess the direction and calculate things out. The math will give us the magnitude of the force and the sign will tell us whether we guessed the correct direction or not.

$Sigma F = ma$

$(mg) + (N) = m(0frac{m}{s^2})$

Like before, we assigned positive to the mg term since it is down and in our sign convention down is postive. And we guessed (wrong on purpose) that the normal force will also be down so it is positive too.

If we solve the equation, we get:

$N = -mg$

Notice that the number we get for N is negative. That means the direction is the OPPOSITE of what we guessed. So the magnitude is correct but the direction is the opposite because we guessed wrong.

If we take our result and plug it back into the original equation we get:

$(mg) + (N) = m(0frac{m}{s^2})$

$(mg) + (-mg) = m(0frac{m}{s^2})$

$(mg) - (|N|) = m(0frac{m}{s^2})$

This is identical to the equation we had when we guessed the direction of the normal force correct:

$(mg) - (N) = m(0frac{m}{s^2})$

$(mg) - (|N|) = m(0frac{m}{s^2})$

If we had guessed correct, our calculated N would have been positive and not changed anything in the equation (hence the absolute sign just to try and make things clearer). Our result always gives us the correct magnitude but the sign tells us whether we guessed our direction correctly or not because when you fit the entire number into the original equation, the sign on the number will change the sign on the term in the equation if the number is negative.

So you can see whether something is positive or negative doesn't actually matter. What matters is whether they are opposite or not because the math only knows if something is opposite or not. It's stupid. It has no context to the problem. You're the one who has to assign meaning to positive or negative and it doesn't matter which you pick as long as you are consistent because the math must be consistent.

And back from B to A on that same surface, why is the acceleration then changed to positive? In fact, friction slows things down so it should be negative as well.

Kinetic friction always slows things down. But static friction can both slow things down and speed things up (cause acceleration) depending on the circumstances. I'm not sure how that would apply to your scenario because you have not provided sufficient details.

For a car it is static friction between the drive wheels and the road that causes a car to accelerate. See the Fig below. Per Newton's third law the static friction force acting forward on the drive wheel is equal and opposite to the force the drive wheel exerts on the road. Then, since the static friction force is the only external force acting on the car (ignoring air resistance) per Newtons second law the static friction force equals the mass of the car times its acceleration.

On the other hand, either static or kinetic friction can cause the car to decelerate depending on whether or not skidding occurs during braking. But only static friction can cause acceleration.

Hope this helps.

You are talking about components of motion in a 1D system (as along an x axis). There must be a choice for the positive direction. An object moving in the positive direction has a positive velocity. An object loosing speed while moving in the negative direction has a positive acceleration. (If continued, it will eventually stop and then gain speed in the positive direction.)

This might be due to your axes.

• If the axis points from A to B, then with positive velocity, friction causes opposite, thus negative, acceleration.

• With negative velocity (we are moving backwards from B to A) friction pulls backwards, meaning in the positive direction, so the acceleration is positive.

The signs might relate to an axis direction like this and not to other properties.

The whole idea of acceleration being negative is false. Acceleration is a vector quantity, it has direction and magnitude. A vector's magnitude cannot be negative, $|vec{a}| ge 0$ and direction is not something you can specify as negative or positive. If the aforementioned acceleration vector is 1-dimensional it can be written as:

$$ vec{a} = |vec{a}|hat{i} $$

The most you can say about this is that if it points through the negative direction or the positive direction.

If the net force (friction in your case) independent of time let's say:

$$ vec{F}_{fr} = langle -3 rangle $$

Since $ vec{F}_{net} = mvec{a} $:

$$ frac{1}{m}langle -3 rangle = vec{a} $$

We know that mass is scalar and positive so the acceleration vector can point through to the negative direction if the net force vector points through to the negative direction.

Very simple question to understand. Let us say you have your car and putting it in the reverse gear and going back. Friction will act in forward direction. Now , you slowly try to increase the acceleration of your car. Then , deceleration will tend to decrease and then become positive after some time. After acceleration is positive now , friction direction will also change.

Check on direction of friction online if you have difficulty.

Therefore , it is all a matter of Cartesian coordinates you choose.