## Main Question or Discussion Point

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

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Hootenanny
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
Gold Member
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.

http://www.dcsint.nl/pdf/5020a.pdf [Broken]

Thank you for your fast respones

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Hootenanny
Staff Emeritus
Gold Member
http://www.dcsint.nl/pdf/5020a.pdf [Broken]

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 :tongue2:

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

russ_watters
Mentor
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...

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

oh, cheers

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

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