# What is the actual reason for the effects of fictitious forces? [duplicate]

Suppose a person is standing in a bus moving with constant velocity. Assume that static friction between his feet and bus surface is very low (you could assume he is on roller skates) and we are observing this situation from the frame of reference of the bus. Then the driver applies the brakes and thus we feel accelerated or jerked in the forward direction.

We could explain this by saying that there is a fictitious force acting behind his back (which is opposite to the direction of acceleration of the bus) by using Newton's law in an accelerating frame.

But we know that there is no such force acting on a person standing inside the bus. What's the actual reason for him accelerating forward, or backward when the bus accelerates forward? (There would be no help from static friction for the motion of the person in this case.)

## marked as duplicate by sammy gerbil, Qmechanic♦Dec 30 '17 at 12:58

The simple answer is the the person on the roller skates is not accelerating and no force is acting on them.

Suppose I am standing on the pavement outside the bus when the bus starts to move. Clearly I am not accelerating because I am just standing there. If I watch the roller skater through the bus windows then when the bus starts to move I will see the roller skater remain stationary relative to me. Since I am not accelerating, and the skater remains stationary relative to me, that means the roller skater isn't accelerating either.

The reason it appears to you that the roller skater is accelerating is because you are accelerating.

Or we can look at this another way. Most mobile phones have a built in accelerator. If I look at my phone now it tells me that my acceleration is zero. If I looked at my phone while I was standing on the pavement watching the bus it would also tell me I wasn't accelerating. And if the roller skater looked at their phone it would tell them that they weren't accelerating. By contrast your phone, and the phone of everyone else on the bus, would show a non-zero acceleration.

• So what's the cause of jerk we feel when driver applies the break i mean why i moved forward when driver applies the break I want to understand this from the frame of reference of bus From earth inertia is the explanation but what from bus – parshyaa Dec 30 '17 at 9:43

When the bus accelerates forward the skater would begin to roll backwards as observed by the bus reference frame. The skater would appear motionless as observed from an outside viewer, as long as the coefficient of friction is zero, because there's essentially no forces acting on the skater.

Say it's a really long bus and the skater is still rolling backwards as viewed by the driver. Then the bus slows to a stop. As viewed by the driver the skater would merely slow to a stop. It would not begin moving forward towards the driver until the bus starts moving in reverse. Meanwhile to the outside observer it looks like the skater hasn't moved.

Ultimately there are two rules of thought involved in this example. First is Newton's laws of motion. You have to consider what forces are involved and where. Second is the concept of a frame of reference, what the object appears to be doing according to what the observer is seeing.

What's the actual reason for him accelerating

The reason is that when the bus slows down, he is not slowed down along with it.

The bus and him have the same speed to start with. When the bus slows down, it looks like it moves backwards from his perspective. He still continues forwards with the same speed, because nothing stops him.

The human brain might misinterpret this, though. It might be more intuitive to believe that the bus is stationary and that you are the one moving, rather than you being stationary and the bus moving backwards. But this is just a wrong interpretation of what the eyes see. It is the bus slowing down, and not himself speeding up.

[...] what's the cause of jerk we feel when driver applies the break

You don't feel any jerk. Everyone else do (backwards).

A person sitting in a seat with a seat belt on wants to continue moving at constant speed, just like the man on roller skates. But the seat belt catches him and pulls him backwards along with the bus. It is not him moving into the seat belt but rather the seat belt slowing down and moving backwards into him.

[...] why i moved forward when driver applies the break

You continue forwards since nothing slows you down. As you are explaining already, no force acts in you. Only the bus slows down. It looks like you are pushed forwards while it actually is the bus that is being pushed backwards.

I want to understand this from the frame of reference of bus

The explanation is the same regardless of the frame you see it from. The World works the same no matter where you are in it.

But Newton's laws do not hold true in non-inertial frames. So if you choose such a frame it all gets tricky. Then you must invent some "fictitious" forces or "pseudo" forces that do not exist in reality before Newton's laws make sense again.

These fictitious forces just represent this tendency to continue moving, that we have explained above.

No matter what frame you choose, the explanation about what in reality is taking place in this situation is the same.

From earth inertia is the explanation but what from bus

Actually, inertia is not the explanation. Inertia (mass) is a resistance to accelerate. But the reason that you fly forwards in a braking bus is that you are not accelerated (negatively) at all!

It is not that you are resisting deceleration, but rather that you are not being decelerated.

If you are decelerated, such as the passenger with a seat belt on, then it is inertia that determines how tough it is to slow you down (how much tension the belt must provide and how squeezed you feel).

• This explanation is still from earth's frame, i am moving from earth's frame not from bus's frame, but i need the explanation from the frame of reference of bus because in bus i am not moving, i am in rest but after applying breaks i moved forward(accelerated) and this is why i stated my problem as "question related to psuedo/fictitious force" – parshyaa Dec 30 '17 at 10:47
• @parshyaa Newton's laws don't apply in non-inertial frames. That's why we have to invent "fictitious" forces to make it fit again. See the two last parts of the answer. – Steeven Dec 30 '17 at 10:48
• Yes i agree what you are saying but that's not my question, my question is that since fictitious forces are non real forces then what is pushing me from my back to move forward when driver applied the break? Yes exactly Newton's law do not works in accelerating frame but to make it work we do some adjustment and that adjustment is fictitious force, – parshyaa Dec 30 '17 at 10:50
• @parshyaa The explanation is no different if we are looking at it from inside the bus. The world works the same, regardless of your point of view. Nothing is pushing you forwards when the brake is applied - inside the bus it just looks like it. – Steeven Dec 30 '17 at 10:54
• Explanation must be different, from earth's frame you were moving with some speed and when driver applied break u kept on moving while bus stops that is why we feel it like a jerk, but when we discuss this from inside the bus, you are not moving, you are being at rest and you must be at rest until and external force acts, but when driver applies break your position changes, you are no longer at rest but there was no external force acting on you then what made you accelerate in forward direction, if you say its psuedo force then i need a explanation becuause fictitious forces are non real force – parshyaa Dec 30 '17 at 11:05

If you want to look mathematically into it, it is just the consequence of moving terms in Newton's second law

$$\sum\boldsymbol{F}=m\boldsymbol{a}$$

If you are standing in the bus, you have

$$\boldsymbol{a}_{\rm you\: rel. \:ground}=\boldsymbol{a}_{\rm you \:rel. \:bus}+\boldsymbol{a}_{\rm bus\: rel. \:ground}$$

So Newton's second law in the ground frame is

$$\sum\boldsymbol{F}=m\boldsymbol{a}_{\rm you \:rel. \:ground}=m\boldsymbol{a}_{\rm you\: rel. \:bus}+m\boldsymbol{a}_{\rm bus\: rel. \:ground}$$

or equivalently in the bus frame

$$\sum\boldsymbol{F}-m\boldsymbol{a}_{\rm bus \:rel. \:ground}=m\boldsymbol{a}_{\rm you\: rel. \:bus}$$

Here

$$\boldsymbol{F}_{\rm fict}\equiv-m\boldsymbol{a}_{\rm bus \:rel. \:ground}$$

is a fictious force that is there only because you are working in non-inertial frame.