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Static equilibrium. Friction example

  • Thread starter J-dizzal
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  • #1
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Homework Statement


A woman pedals her bicycle up a 8-percent grade on a slippery road at a steady speed. The woman and bicycle have a combined mass of 76 kg with mass center at G. If the rear wheel is on the verge of slipping, determine the coefficient of friction μs between the rear tire and the road. If the coefficient of friction were doubled, what would be the friction force F acting on the rear wheel? (Why may we neglect friction under the front wheel?)
20150721_204521_zpsgpr2j4pv.jpg

Homework Equations


FssN

The Attempt at a Solution


20150721_204444_zpsasubifac.jpg
 

Answers and Replies

  • #2
Dr. Courtney
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Static equilibrium?
 
  • #3
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Static equilibrium?
Im not sure if i should approach these problems as static equilibrium or not.
 
  • #4
SammyS
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Homework Statement


A woman pedals her bicycle up a 8-percent grade on a slippery road at a steady speed. The woman and bicycle have a combined mass of 76 kg with mass center at G. If the rear wheel is on the verge of slipping, determine the coefficient of friction μs between the rear tire and the road. If the coefficient of friction were doubled, what would be the friction force F acting on the rear wheel? (Why may we neglect friction under the front wheel?)
[ IMG]http://i1164.photobucket.com/albums/q562/falsovero/20150721_204521_zpsgpr2j4pv.jpg[/PLAIN]

Homework Equations


FssN

The Attempt at a Solution


[ IMG]http://i1164.photobucket.com/albums/q562/falsovero/20150721_204444_zpsasubifac.jpg[/PLAIN]
Come on. Talk to us.

Why do you think they gave you the location of the center of gravity?
 
  • #5
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Come on. Talk to us.

Why do you think they gave you the location of the center of gravity?
I guess so that static equilibrium moment equations can be made.
but its a wheel so how can there be moments at the contact point?
 
  • #6
SammyS
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I guess so that static equilibrium moment equations can be made.
Yes.

Not all the weight is supported by the rear wheel.

As regards your tread title: The friction is static. There is equilibrium, but I would not call it static, although others may disagree. The bicycle is moving. It's at a constant speed (and presumably at a constant velocity), so acceleration is zero, therefore, forces cancel & torques cancel.
 
  • #7
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Yes.

Not all the weight is supported by the rear wheel.

As regards your tread title: The friction is static. There is equilibrium, but I would not call it static, although others may disagree. The bicycle is moving. It's at a constant speed (and presumably at a constant velocity), so acceleration is zero, therefore, forces cancel & torques cancel.
Im really confused on how to find weight distribution, I must of forgot how to do this already. Is it found by doing solving the static equilibrium equations?
 
  • #8
394
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Yes.

Not all the weight is supported by the rear wheel.

As regards your tread title: The friction is static. There is equilibrium, but I would not call it static, although others may disagree. The bicycle is moving. It's at a constant speed (and presumably at a constant velocity), so acceleration is zero, therefore, forces cancel & torques cancel.
I'm setting up moment equations to find the weight distribution but its not going well, can you give me any tips on how to find the weights at each tire?
 
  • #9
501
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Consider variables N1 and N2 for normal at rear and front wheel . Use two equations - from moments and and balancing forces ⊥ to incline .
 
  • #10
394
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Consider variables N1 and N2 for normal at rear and front wheel . Use two equations - from moments and and balancing forces ⊥ to incline .
I think ive found N1 and N2 correctly, If N1 is the normal force at the rear tire, the next task is to find Fs the static frictional force, would i be correct to assume that it is equal to the force of gravity pulling the bike down the ramp (parallel)?
 
  • #11
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Yes - the biker pedals at a steady speed .
 
  • #12
haruspex
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I think ive found N1 and N2 correctly,
would you please post what you got, preferably symbolically in terms of W=weight, h=height of CoM, horizontal distances d1, d2 from CoM to points of contact of wheels (all measured in terms of when bike is on the flat), and the angle theta.
 
  • #13
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would you please post what you got, preferably symbolically in terms of W=weight, h=height of CoM, horizontal distances d1, d2 from CoM to points of contact of wheels (all measured in terms of when bike is on the flat), and the angle theta.
So far,
ΣM2=Wcosθ(d2) - Wsinθ(h) - N1(d1+d2). This solves for N1, the normal force at the front tire.
Fsfriction=Fgravity on bike parallel to ramp.
μs=Fsfriction / N1. This gives the coefficient of static friction for part (a) of the problem statement.

part (b) says to double μs, then I substitute that value into the equation; μs=Fsfriction / N1. Solves part (b).

Note, my coordinate axis did not get into the picture i took sorry, x axis being parallel to the ramp.
 
  • #14
501
65
So far,
ΣM2=Wcosθ(d2) - Wsinθ(h) - N1(d1+d2). This solves for N1, the normal force at the front tire.
Fsfriction=Fgravity on bike parallel to ramp.
μs=Fsfriction / N1. This gives the coefficient of static friction for part (a) of the problem statement.

part (b) says to double μs, then I substitute that value into the equation; μs=Fsfriction / N1. Solves part (b).
I'd rather not comment on your entire solution , but it seems to me as though you have just equated Fs as = μs*N1 , and not taken into account friction offered by back wheel .
 
  • #15
haruspex
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ΣM2=Wcosθ(d2) - Wsinθ(h) - N1(d1+d2).
Check your signs. Imagine a very large h.
This solves for N1, the normal force at the front tire.
You mean rear tire, right?
 

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