Need a quick answer please!

  • #1

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
 

Answers and Replies

  • #2
Hootenanny
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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.
 
  • #3
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
 
  • #4
Hootenanny
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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?
 
  • #5
alxm
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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.
 
  • #6
http://www.dcsint.nl/pdf/5020a.pdf [Broken]

Thank you for your fast respones
 
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  • #7
Hootenanny
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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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  • #8
does this mean the static frictional force is independent of vehicle mass?
 
  • #9
russ_watters
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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...
 
  • #11
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
 
  • #12
oh, cheers
 
  • #13
can anybody give me an example of average clamping forces for vehicles
 
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