Friction in a brakepad/brakedisc scenario,

[KristianKL]

Homework Statement

Hello.
I'm working on a project here in Denmark, where we want to know which force, with which the brake pads on a mountain bike push against the brake disc, is needed to prevent the wheel from rotating. (the scenario is depicted here: http://imageshack.us/photo/my-images/522/mountainbikediscbrake.jpg/")

We´re setting the coefficient of friction to μ=0,4

Homework Equations

I know that we should use this equation:
F=μ*N

where N equals the force with which the brake pads pushes against the brake disc.
We'll name it Fb. (N=Fb)

The Attempt at a Solution

I'm not sure how to take in consideration that there is two brake pads pushing against the brake disc from both sides.
My guess is that in this case the pressure delivered to the brake pads are equally distributed between the two brake pads.
In this case the required force from each brake pad would be:
F=μ*0,5Fb

And the required force for both brake pads together (determined by which pressure is delivered to the caliber) would just be:
F=μ*Fb

Are these considerations correct or are there other factors to take in consideration when there are two forces pushing from each side. Does μ change ? does the surface area of the brake pads have anything to say?

Thanks for reading, hope you can help me

Greetings

Kristian K. L.

 

[PeterO]

Homework Statement

Hello.
I'm working on a project here in Denmark, where we want to know which force, with which the brake pads on a mountain bike push against the brake disc, is needed to prevent the wheel from rotating. (the scenario is depicted here: http://imageshack.us/photo/my-images/522/mountainbikediscbrake.jpg/")

We´re setting the coefficient of friction to μ=0,4

Homework Equations

I know that we should use this equation:
F=μ*N

where N equals the force with which the brake pads pushes against the brake disc.
We'll name it Fb. (N=Fb)

The Attempt at a Solution

I'm not sure how to take in consideration that there is two brake pads pushing against the brake disc from both sides.
My guess is that in this case the pressure delivered to the brake pads are equally distributed between the two brake pads.
In this case the required force from each brake pad would be:
F=μ*0,5Fb

And the required force for both brake pads together (determined by which pressure is delivered to the caliber) would just be:
F=μ*Fb

Are these considerations correct or are there other factors to take in consideration when there are two forces pushing from each side. Does μ change ? does the surface area of the brake pads have anything to say?

Thanks for reading, hope you can help me

Greetings

Kristian K. L.

Is that coefficient of friction the friction between the pad and the disc, or the friction between the tyre and the ground?

Looks like it is pad to disc.

The key to brakes is that when braking there is one kinetic friction scenario and one static friction scenario.

While riding along, the tyre is not slipping along the surface, so static friction applies to the tyre/road interface.

When the brakes are applied lightly, we start to get kinetic friction between the pad and the disc - the disc is sliding between the pads.

As we increase the pressure, we can reach a point that sufficient force is applied, through the brake system, [hydraulic or cable?] so that the wheel stops. We then have static friction between the pad and the disc, and kinetic friction between the tyre and the road.

Co-effiicient of kinetic friction is generally less than the coefficient of kinetic friction. Tha is why we usually try not to "lock-up" the wheels when braking.

Of course it is never that simple. if you get onto a soft surface, the trenches dug by a sliding tyre can sometimes be a better use of the energy of the bike/car than the brakes, which is why a rally car can often stop quite efficiently by locking up the wheels on a gravel road.

 

[KristianKL]

Thank you for answering :)

It's the coefficient between the pad and the brake disc. We´re looking at the static scenario. The disc is standing still and we want to know, if the breaks are activated while standing still, which force is required to make the wheel rotating even though the brakes are activated..

(sorry for my limited vocabulary.. )

 

[KristianKL]

btw. The braking system is hydraulic

 

[PeterO]

btw. The braking system is hydraulic

How do you know the coefficient is 0.4 ?

ps: Langauge fine. [I have Danish ancestry]

 

[KristianKL]

Good, As long as you understand what I'm saying ;)

My friend found it by lookup, but its not that important if it's correct, what is important is whether my assumptions about the forces involved are correct:

1. That the force F delivered by hydraulic pressure to the caliber is the sum of the two forces with which the brake pads push against the brake disc.

2. That the fact that the brake pads are pushing from each side at once doesn't change the value F, as it is still just the sum of the two powers with which the brake pads pushes against the brake disc.

3. That the force F is the minimum force that is needed to make the wheel start rotating.

If these statements are true, my calculations should be correct.

 

[PeterO]

Good, As long as you understand what I'm saying ;)

My friend found it by lookup, but its not that important if it's correct, what is important is whether my assumptions about the forces involved are correct:

1. That the force F delivered by hydraulic pressure to the caliber is the sum of the two forces with which the brake pads push against the brake disc.

2. That the fact that the brake pads are pushing from each side at once doesn't change the value F, as it is still just the sum of the two powers with which the brake pads pushes against the brake disc.

3. That the force F is the minimum force that is needed to make the wheel start rotating.

If these statements are true, my calculations should be correct.

If there was only one brake pad, the disc itself would have to provide the reaction force, and you way well distort the disc.
By pushing from both sides, the other pad is effectively supplying the reaction force.
This has the advantage of "lightness" for the disc.

example.

have someone hold a sheet of paper, hanging vertically.

If you push on one side with your hand, the paper easily is displaced.

If the paper was instead a hanging block of steel, you would be able to push quite hard, without the steel moving appreciably.

Now, if you push on the paper from both sides - with each hand - you can apply as much force as you are capable of without deflecting the paper.

That is why the brakes have a pad on each side.

Left pad pushed with force F_B to the right
Right pad pushes with force F_B to the left.

The "reaction" force to each pad is [of course] the same size, but is being supplied by the other pad, rather than the disc structure.

This is similar to tying a string to each end of a spring balance, then pulling on each string with a force of 100N. What does the spring balance read? The answer is 100N.
Many people think it should be 200N, but when the spring balance is used normally - with one end attached to a support, it reads 100N when a 100N load pulls one way with a force of 100N, and the support pulls the other way with a force of 100N.

 

[KristianKL]

So, is this what you´re saying?: that there are two brake pads pushing against the brake disc at the same time from both sides, doesn't mean that there is a greater force applied to the brake disc.

Would that mean that the overall force which pushes against the brake disc is only F_B? that is, half the force which is delivered to the caliber.?

If that's correct, that the force applied is the same as if there were only one brake pad, I don't really know how to describe why it will brake more efficient when you're pushing from each side with two brake pads..

How ever, I do understand the example you put up with the spring.

 

[PeterO]

So, is this what you´re saying?: that there are two brake pads pushing against the brake disc at the same time from both sides, doesn't mean that there is a greater force applied to the brake disc.

Would that mean that the overall force which pushes against the brake disc is only F_B? that is, half the force which is delivered to the caliber.?

If that's correct, that the force applied is the same as if there were only one brake pad, I don't really know how to describe why it will brake more efficient when you're pushing from each side with two brake pads..

How ever, I do understand the example you put up with the spring.

A brake caliper is designed so that each brake pad pushes with a force F_B.
It could be a piston on each side, or a sliding caliper with a single piston. [or 4 pistons/6pistons/8pistons in some racing cars - and Ferraris]
Just because they both push with force F_B does not mean the force on the disc is 2F_B

The force is the same as if there was only one pad, however if there was only one, the disc would have to have a mechanism [strength and fitting] to be able to "push back" - the reaction force.
By having a pad pushing from the other side, the disc does not have to be constructed so robustly. [it can be thinner and lighter]. The reaction force is being provided by the other pad.

Force is not delivered to the caliper. The hydraulic system is pressurised, meaning each pad pushes with force F_B.

Consider this:

You take a G-clamp - the ones with the screw mechanism for tightening - and clamp it gently onto your hand.
Now you tighten the screw.
Does the clamp press harder on your hand on the side with the screw mechanism, or the frame side?
In case you don't get the right answer to that follow the following sequence:
If the screw side pushes with a force of 50N and the frame only pushes with a force of 40N, what is the net force on your hand?
What would that net force do to your hand?
Does your hand behave that way?
What does that mean about the way a G-clamp applies force.

A disc brake on a mountain bike is just a hydraulic G-Clamp.

12:40 am here in Australia; off to bed.

 

[KristianKL]

Thank you again, for answering, it's very helpful :)

I've tried to answer the questions:

Does the clamp press harder on your hand on the side with the screw mechanism, or the frame side?

- They're equal.

If the screw side pushes with a force of 50N and the frame only pushes with a force of 40N, what is the net force on your hand?

- Would that be 10N? I'm not sure about that.. wouldn't they equalize?

What would that net force do to your hand?

- I guess it would move it in the direction force of the screw side is pointed?

Does your hand behave that way?

- No, it's locked in between by the clamp

What does that mean about the way a G-clamp applies force.

- It means that the force of one of the sides i the force applied to the hand, the force of the other side is the reaction force to this. I think I partly understand it now..

So my calculations for the brake system would then be:

F=μ*N

where F is the force needed to make the wheel start rotating.
μ is the coefficient of friction.
N is the force applied from one of the brake pads against the brake disc.
The force N is determined by the hydraulic system (which pistons, which pressure etc.).

 

[PeterO]

Thank you again, for answering, it's very helpful :)

I've tried to answer the questions:

Does the clamp press harder on your hand on the side with the screw mechanism, or the frame side?

- They're equal. Correct! You don't need the other questions - they were to lead you back to this answer in case you didn't get it right first time!

If the screw side pushes with a force of 50N and the frame only pushes with a force of 40N, what is the net force on your hand?

- Would that be 10N? I'm not sure about that.. wouldn't they equalize?

What would that net force do to your hand?

- I guess it would move it in the direction force of the screw side is pointed?

Does your hand behave that way?

- No, it's locked in between by the clamp

What does that mean about the way a G-clamp applies force.

- It means that the force of one of the sides i the force applied to the hand, the force of the other side is the reaction force to this.


I think I partly understand it now..

So my calculations for the brake system would then be:

F=μ*N

where F is the force needed to make the wheel start rotating.
μ is the coefficient of friction.
N is the force applied from one of the brake pads against the brake disc.
The force N is determined by the hydraulic system (which pistons, which pressure etc.).

Good luck with the analysis.

 

[KristianKL]

Thank you :)