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Homework Help: Moments about two points of an object

  1. Apr 26, 2014 #1
    1. The problem statement, all variables and given/known data
    hi guys this is an image of the rear wheel of a push bike going into a corner as well as what i think are relevant forces,
    i need to calculate the moments or torque at A (the contact patch) and B (the hub centre)
    this is to calculate the force on the left hand side of the frame and right hand side of the frame while in a corner
    Fsr = the force going through the suspension
    Fc = lateral cornering forcer on the rear wheel
    phi = lean angle
    MG = mass & gravity
    t = the cross sectional radius of the tyre
    CoG(y) = centre of gravity height

    2. Relevant equations
    this is my question what is the relevant equation to find the moments for A and B??

    i can and have calculated the above parameters through lot of prior equations that take into account the wheelbase, velocity, friction, radius of the turn ect. but i think that the above are the require ones to calculate this... if there is an easier way please let me know

    i said thought im not 100% these are the relevant forces but i am fairly confident also to point out this is assuming that it is a 2d diagram should be a simple moments equation really (i think) also it can be assumed the bike is stationary and that it is simple leant over with a machine applying the forces if it makes it easier to calculate...
    thats the torque or moments about point A ( contact patch ) and B the wheel spindle centre
    but any help would be great whether in motion stationary or anything, if you ignore the forces i have written and know another way very happy to listen to that too thanks alot

    3. The attempt at a solution

    ive tried looking at the centripetal forces as uniformaly distributed loads, along the height of the bike,
    ive tried looking at it as basic moments but as the main force im considering (Fsr) is going from CoG to contact patch through A&B the torque would equal 0
    and ive also tried looking at it as though Fc (cornering force) is acting at the CoG this gave moderatly good results but exceed the torque of A in regards to MG meaning the bike would be rolling and self righting as such the bike would not be in equilibrium so also wrong

    there are no definitive numbers to input as it would take the best part of an hour if i gave all prior equations, forces, geometric aspects velocities ect. but if you could tell me what i need and where i need to be looking (if you could explain your reasoning too would be useful) i will figure the relevant forces ect. myself... thanks very much guys

    Attached Files:

  2. jcsd
  3. Apr 26, 2014 #2
    It appears your whole purpose is to find the small bending moment in the fork. Your free body diagram, however, is a little confusing:

    1) Fc should be acting at the CG.
    2) There should be a friction force acting on the tire equal to Fc but in the opposite direction.
    3) There should be a normal force acting on the tire.
    4) The diagram implies that there is some deformation of the tire. Are you taking this into account?
    5) Are we neglecting the mass of the wheel itself and the gyroscopic effect of the spinning wheel?

    Why would you do this?

    Not necessarily true. The torque would depend on the relative magnitudes of Fsr and Fc.

    If your assuming the bike is in equilibrium then the moment at any point must be zero. But this has to do with the relative magnitudes of Fc and Fsr. And Fc depends on how fast the bike is moving and the cornering radius.

    In order to get the magnitude of the bending moment in the fork you'll have to draw another free body diagram with the specific dimensions.
  4. Apr 27, 2014 #3
    thanks very much
    1) i thought that Fc should be acting on Cg but the results did not equate as i later stated,
    2) although not drawn on yes i am aware that there should be an equal friction force at the contact patch opposing Fc
    3) im going to sound dense here but, normal force, what do you mean the they are Mg, Fc and Fsr are they not??
    4)i was planning on taking into account deformation of the tyre but since im struggling already and my deadline approches i going to dismiss that.
    5) and although it sounds silly yes i am ignoring gyroscopic affect although if the mass of the wheel or (MoI) is (in your opinon) a significant part to the solution despite gyroscopic effect then i dont think it would be to hard to consider it

    because i assumed that as allowing the force to act on CG did not give the answers i was looking for in my desperation i was looking for any other feesable method of find the results (needless to say it also gave an incorrect answer)

    could you elaborate on why the relative magnetudes would allow the torque to be anything other than 0 if it is running through the cog B and A (this is why i considered the deformation of the tyre however it proved extremly hard to calculate accuratly and using approximations did not give the expected results either)

    and yes for the whole thing i am assuming the bike being in equilibrium i have all the prior equations to calculate Fc ect. i am not looking for a worked answer but the generic equation or formula the rest i can calculate for myself if it helps for your responce purposes the one a under the same cornering conditions
    Fc = 701.98
    Fsr = 1819.45
    and MG = 2609.45
    and just if it is relavant the traction co efficient (mu) is 1.1

    i really hope this helps you help me, and i would like to thank you so much for your time it really is appreciated
  5. Apr 27, 2014 #4
    1) Fc can only act at the CG since this is the resultant of all the forces acting on the system. If the system is in equilibrium then the torque caused by Fc is offset by the torque by MG.

    2) OK, but it might help you if you include all the forces on the system and locate them at the correct locations at which they are acting.

    3) By normal force I am talking about the reaction force of the ground that acts perpendicular to the ground surface.

    4) OK, this makes it easier. The thickness then of the tire causes a small bending moment in the fork of the bike and this is what you are trying to solve for.

    5) Given the small bending moment in the fork that you are trying to solve then you should probably include everything possible. So I would include the mass of the wheel and the gyroscopic effect.

    Not sure why you think it gave you an incorrect answer. You are probably doing something else wrong.

    Fc acting at the CG at a distance y from the ground causes a torque of Fc*y about A.
    MG acting at the CG at a distance x from the contact point causes a torque -MG*x about A.

    If the system is in equilibrium then the two torques are equal:
    Fc*y - MG*x = 0
    →Fc*y = MG*x

    You can show mathematically that if a system is in equilibrium then the moment at any point anywhere on the planet is also zero.

    I would think this number has to be less than 1.

    Also what is the value of [itex]\varphi[/itex]?
    Last edited: Apr 27, 2014
  6. Apr 27, 2014 #5
    WoW thanks ever so much
    this has helped alot i have doing some revision of my previous calculations now realising that the tyre radius is alot more important and i am re-jigging them accordingly, thanks for all your help :D
    have been at work while writing this and; have attached a picture the moments are close but still not the same and im still unsure why but they are alot closer than previous attempts
    although im going to have attempt to ignore the wheel mass and gyroscopic effect as i am running so short on time
    also in the same scenario;
    [itex]\varphi[/itex] is now Δ[itex]\varphi[/itex] = 46.442 degrees
    i now have [itex]\varphi[/itex] this is including the tyre radius (they are draw on the wrong way on the diagram sorry)
    so using CG (y) as circle radius
    0.4263*(sin(46.422) to give the new CG (y)= 0.3361

    also the traction co efficient can be higher than the friction co efficient due to the way that hot rubber interacts with the ground (so im told lol )

    Attached Files:

    Last edited: Apr 27, 2014
  7. Apr 27, 2014 #6
    Interesting, I didn't know about that.

    Your FBD looks better.
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