Friction on an accelerating train

In summary: I don't know what the free body diagram will look like. Two... I don't understand the nature of the question.In summary, the conversation is discussing a physics homework question about a person standing on a train accelerating at 2.0m/s^2 and trying to determine the minimum coefficient of static friction needed to prevent them from sliding. The conversation includes a discussion about forces, free body diagrams, and different ways of approaching the problem, ultimately arriving at the correct solution of a coefficient of static friction of 0.2.
  • #1
beetlegauss
6
0
I have a question for homework, and this is a high school physics class. The question reads:

Suppose that you are standing on a train accelerating at 2.0m/s^2. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

I've drawn a free body diagram, and motion in the Y dimension doesn't really matter.
In the X dimension, I THINK there are 2 forces: Force of the train, and Force of friction. What I've come up with is this:

Fnetx=0 (Since I'm not accelerating in relation to the train itself. I'm hoping this assumption is right, or everything else is wrong.)
Fnetx=Ft-Ff
Ft=Ff
Ff=9.8mu, and Ft=2m (This is where I'm getting sketchy. Is the force of the train on the person the same as acceleration x the persons mass?)
9.8m/2m=u
u=.2

I'm basically posting to verify my answer, since I'm not entirely sure if the route I took to get to it was right. Thanks for reading!
 
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  • #2
Looks good to me except this bit;
9.8m/2m=u
It should read; 2m/9.8m , however, your final answer is correct so I assume its a typo.
 
  • #3
Oh yeah, thanks. So the coefficient is .2, and the force on the person is his mass times the trains acceleration, then?
 
  • #4
I don't really understand your reasoning. What's Ft?

Rather than try to describe things from the accelerated frame of the train, simply describe it from the frame of the tracks. Answer these questions:
(1) For you not to slide, what must your acceleration be?
(2) What force provides that acceleration? How does it relate to your weight and the coefficient of static friction?
(3) Use that to deduce the minimum coefficient of static friction needed.
 
  • #5
Doc Al said:
I don't really understand your reasoning. What's Ft?

Rather than try to describe things from the accelerated frame of the train, simply describe it from the frame of the tracks. Answer these questions:
(1) For you not to slide, what must your acceleration be?
(2) What force provides that acceleration? How does it relate to your weight and the coefficient of static friction?
(3) Use that to deduce the minimum coefficient of static friction needed.

Ft is, um, Force of the train. I didn't know what else to label it as. The only thing I was confused about is whether or not the Force of the train on me is it's acceleration times my mass.
 
  • #6
beetlegauss said:
Ft is, um, Force of the train. I didn't know what else to label it as. The only thing I was confused about is whether or not the Force of the train on me is it's acceleration times my mass.
There's no such thing as "force of the train". The only force acting on you (horizontally, at least) is the friction from the floor of the train.

(You can view things from an accelerating frame of reference, such as the train. But that introduces so-called "fictitious" forces that are artifacts of using such a non-inertial viewpoint. Things end up being close to what you did, but for very different reasons. I suggest forgetting about such advanced methods and sticking to real forces for now.)
 
  • #7
beetlegauss said:
Ft is, um, Force of the train. I didn't know what else to label it as. The only thing I was confused about is whether or not the Force of the train on me is it's acceleration times my mass.

What Doc Al is saying, is that you can also consider this problem from an inertial reference frame. This could be someone stood at the side of the tracks watching you accelerate past at 2m/s2. Now, the force required to accelerate you at this rate is twice your mass, yes? Now, this force is provided by friction between the floor of the train and your shoes; so we can say that [itex]2m = mg\mu[/itex], which is equivalent to your expression and you will obtain the same answer.

There is nothing wrong with solving problems using non-inertial (accelerating) reference frame provided you understand the implications. In general, it is always advisable to use inertial (non-accelerating) frame of reference to avoid confusion.

Edit: Looks like Doc Al's got it :wink:
 
  • #8
Doc Al said:
There's no such thing as "force of the train". The only force acting on you (horizontally, at least) is the friction from the floor of the train.

(You can view things from an accelerating frame of reference, such as the train. But that introduces so-called "fictitious" forces that are artifacts of using such a non-inertial viewpoint. Things end up being close to what you did, but for very different reasons. I suggest forgetting about such advanced methods and sticking to real forces for now.)

Yes, actually, I did the problem two different ways, both of which ended up with the same answer. The first way, which i thought was wrong, was this.

Ff=2m (so now I'm accelerating, and there is no force-of-train)
9.8mu=2m
u=2m/9.8m
u still equals .2
 
  • #9
Hootenanny said:
What Doc Al is saying, is that you can also consider this problem from an inertial reference frame. This could be someone stood at the side of the tracks watching you accelerate past at 2m/s2. Now, the force required to accelerate you at this rate is twice your mass, yes? Now, this force is provided by friction between the floor of the train and your shoes; so we can say that [itex]2m = mg\mu[/itex], which is equivalent to your expression and you will obtain the same answer.

There is nothing wrong with solving problems using non-inertial (accelerating) reference frame provided you understand the implications. In general, it is always advisable to use inertial (non-accelerating) frame of reference to avoid confusion.

Edit: Looks like Doc Al's got it :wink:

Alright, coolio, I think I've got both of them now. lol, thank you both.
 
  • #10
beetlegauss said:
Yes, actually, I did the problem two different ways, both of which ended up with the same answer. The first way, which i thought was wrong, was this.

Ff=2m (so now I'm accelerating, and there is no force-of-train)
9.8mu=2m
u=2m/9.8m
u still equals .2
Nothing wrong with that. And no need for any mysterious "force of train"!:wink:

One thing to realize is that [itex]\mu mg[/itex] is the maximum value of static friction. Using that, like you correctly did, gives you the minimum value of the coefficient of friction. (It can always be greater.)
 
  • #11
Are you Alex, by any chance? Because my physics teacher's name is Alex...
 

FAQ: Friction on an accelerating train

1. How does friction affect the acceleration of a train?

Friction is a force that opposes motion. In the case of an accelerating train, friction can slow down the train's acceleration by creating resistance against the train's movement. This resistance is due to the contact between the train's wheels and the tracks, which creates friction that must be overcome by the train's engine.

2. Can friction cause a train to slow down?

Yes, friction can cause a train to slow down. As mentioned before, the contact between the train's wheels and the tracks creates friction, which creates resistance against the train's movement. This resistance can cause the train to decelerate or even come to a stop if the force of friction is greater than the force of the train's acceleration.

3. How does the type of surface affect friction on an accelerating train?

The type of surface that the train is traveling on can greatly affect the amount of friction present. For example, a train traveling on a smooth, flat surface will experience less friction compared to a train traveling on a rough, bumpy surface. This is because there is less contact between the train's wheels and the surface on a smooth surface, resulting in less friction.

4. How does the speed of the train affect friction?

The speed of the train can also affect the amount of friction present. As the train's speed increases, so does the force of friction between the train's wheels and the tracks. This is because the contact between the two surfaces increases, creating more resistance against the train's movement. This can ultimately slow down the train's acceleration if the force of friction becomes too great.

5. How can engineers reduce friction on an accelerating train?

Engineers can reduce friction on an accelerating train by using materials that have lower coefficients of friction, such as lubricants or smoother surfaces. They can also design the train's wheels and tracks to have less contact or use mechanisms that can minimize the effects of friction, such as anti-lock brakes. Additionally, regular maintenance and cleaning of the train's wheels and tracks can also help reduce friction and maintain optimal acceleration.

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