Average Clamping Forces for Vehicle Brake Calipers

[revolution200]

If a brake caliper exerts 1000kg on a brake pad what is the force due to the this weight?

It can't be umg because there is no g in horizontal weight.

Does g = 1

Therefore F = m

 

[Hootenanny]

If a brake caliper exerts 1000kg on a brake pad what is the force due to the this weight?

It can't be umg because there is no g in horizontal weight.

Does g = 1

Therefore F = m

Your question does not make sense. A brake caliper cannot 'exert' 1000 kg on anything, since kilograms is a measure of mass and not force.

 

[revolution200]

I have a specification for a brake caliper that states its clamping force is equal to 4136kg. I know force doesn't equal 4136 but what does clamping force mean then

 

[Hootenanny]

I have a specification for a brake caliper that states its clamping force is equal to 4136kg. I know force doesn't equal 4136 but what does clamping force mean then

Could you attach or link to the specification document?

 

[alxm]

I'd interpret it at the equivalent of 4136 kg, that is 4136 kilogram-force or kiloponds. So the force would be 4136/g or about 422 Newtons.

 

[revolution200]

http://www.dcsint.nl/pdf/5020a.pdf

Thank you for your fast respones

 

[Hootenanny]

http://www.dcsint.nl/pdf/5020a.pdf

Thank you for your fast respones

In which case, I'd agree with alxm's interpretation,

I'd interpret it at the equivalent of 4136 kg, that is 4136 kilogram-force or kiloponds. So the force would be 4136/g or about 422 Newtons.

Bloody engineers

 

[revolution200]

does this mean the static frictional force is independent of vehicle mass?

 

[russ_watters]

The force between the brake pads and rotors is not a function of gravity. It is generated hydraulically. If that's the static friction you are talking about, then you are correct, it is not a function of vehicle mass. Judging by the question you asked yesterday, however, I'm guessing you now asking about tires and are still confused about this...

 

[russ_watters]

I'd interpret it at the equivalent of 4136 kg, that is 4136 kilogram-force or kiloponds. So the force would be 4136/g or about 422 Newtons.

That would be 4136*g or 40,500 Newtons.

http://www.google.com/search?sourceid=navclient&ie=UTF-8&rlz=1T4ADBF_enUS311US311&q=kilogram+force

 

[revolution200]

That can't be true. I have a car stopping with kinetic friction as u*m*g = 0.7 * 1000 * 9.8 = 6860N

I thought static (ABS) is more effective than kinetic

Static coefficient * force = 0.5 * 422 = 211N

This can't be right

 

[revolution200]

oh, cheers

 

[revolution200]

can anybody give me an example of average clamping forces for vehicles