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Statics problem

  1. Nov 18, 2013 #1
    Hi guys. I have one problem from competition that I can't solve. I'll try to translate it.

    "rod of mass m and length l is lying on a ball of radius R in state of equilibrium. The rod makes angle θ with the horizontal. There is friction in every point of contact, and it's assumed that it is large enough to hold the system in equilibrium. Find the friction between surface and a ball.

    I tried to sketch it in sketch toy, it looks like this: http://sketchtoy.com/55802563
     
  2. jcsd
  3. Nov 18, 2013 #2

    Simon Bridge

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    Your diagram appears to have the rod touching the ground - is this what the problem describes?

    Please show us your best attempt with your reasoning at each stage.
    The approach to these is almost always the same - draw each component in isolation and draw the linear forces at their points of action on each in turn.
     
  4. Nov 18, 2013 #3
    Yes, the rod is touching the ground. So, my attempt is :
    Just to say that the length of a rod is not given but it's easy to find it over R and θ. l=Rcot(θ/2)
    And the mass is not given but linear density ρ( I think) so m=ρl

    Okay I first started with the fact that if the system is in equilibrium that the sum of forces and moments is 0.
    First for rod : ƩF=0
    Forces acting on rod are gravitational force from middle of a rod mg,friction between rod and surface, and force that the ball is acting on the rod, let call it Fr(newtons 3rd law, its normal on the rod)
    Now in x and y directions I get

    ƩFx=Frsinθ-fr (fr- friction)
    ƩFy=mg-Frcosθ (1)

    For the ball

    ƩFx=Frsinθ-fr-frcosθ
    ƩFy=Mg-N-Frcosθ ( N is the force that surface is acting on the ball)

    ƩM (sum of moments should be 0)
    for rod
    mg cosθ *l/2=Nl (2)

    for ball fr*R=fr*R so frictions between rod and ball and ball and surface are the same.

    But from (1) I get mg=Frcosθ, and from (2) mg=2Fr/cosθ, which leads to contradiction, so obviously I am doing something wrong. Hope you can me show me what.
     
    Last edited: Nov 18, 2013
  5. Nov 18, 2013 #4

    Simon Bridge

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    Did you remember: reaction on rod from the floor, friction from ball and friction from the floor.

    This may help talk about things:
    Rod goes from A to B.
    The center of the ball is at point C.
    The Rod touches the ground at point A and the ball at point D.
    The ball touches the ground at point E.

    So you can define some lengths, eg.|AB|=l, |CD|=|CE|=R and describe relationships, eg.|AC|=|AE|, and so on.
    The ball does not slide or roll ... the friction at point D ##f_D## opposes the rolling, and the friction at point E ##f_E## stops the sliding.

    The triangle ADE may be useful to you here.
     
  6. Nov 19, 2013 #5
    Ok. Rod goes from A to B. It touches the ball in point B ( so it ends where it touches the ball).
    Center of the ball is in C and it touches the floor in point D.

    Forces acting on the rod : gravitational force(mg), reaction from floor(N2, reaction from the ball(N1, friction in point B(fb), and friction in point A(fa.)

    ƩF=0

    ƩFx=0

    fbcosθ+N1sinθ-fa+N2sinθ=0 ( we can't get anything from here)

    ƩFy=mg-N1cosθ+fbsinθ-N2cosθ=0 (same here)

    ƩM=0

    mg cosθ *l/2=N1 ----> N1=(mgcosθ)/2


    ƩM=0 on the ball

    fb*R=fd*R ---> fb=fd

    ƩF=0

    ƩFy= Mg-N3+fbsinθ-N1cosθ (can't get anything from here)

    ƩFx=N1sinθ-fb-fbcosθ --->

    fb=Nsinθ/(1+cosθ)=mgsinθcosθ/(2(1+cosθ)

    (using m=ρl=Rρcot(θ/2) and fact that tan(θ/2)=sinθ/(1+cosθ) )

    fb=1/2*(ρgRcosθ)
     
  7. Nov 19, 2013 #6
    And by the way, did you had something particular in mind when you said "The triangle ADE may be useful to you here." ?
     
  8. Nov 19, 2013 #7
    What is the direction of N2 ? Does it have a component in x-direction?
     
    Last edited: Nov 19, 2013
  9. Nov 19, 2013 #8
    Yes, it does N2 sinθ, it says there. I could be wrong but I made the direction of N1 the same as N2, that is normal on the rod.
     
  10. Nov 19, 2013 #9
    That is incorrect .The direction of N2 i.e normal force on the rod from the floor is vertically upwards .
     
  11. Nov 19, 2013 #10

    Thanks for your correction, I'll keep that in mind next time but I don't think that affects the final result.
     
  12. Nov 19, 2013 #11
    The signs of the components of forces are also wrong.

    1) What is the direction of friction on the bottom tip of the rod from the floor?
    2) What is the direction of friction on the upper tip of the rod from the ball?

    Assume rightwards direction to be +x, and vertically downwards to be +y.What is the equation of forces acting in the x-direction on the rod ?What is the equation of forces acting in the y-direction on the rod ?
     
  13. Nov 20, 2013 #12
    1.)to the right,+x direction

    2.)In the line of a rod, so it has x and y components.


     
  14. Nov 20, 2013 #13
    You are messing up with the signs.

    Please upload a picture clearly marking all the forces.
     
  15. Nov 20, 2013 #14
    It would be something like this. http://sketchtoy.com/55942979

    If it's still unclear how I marked the forces acting on the rod, I'll write it on a paper and then post it.
    Thanks for you time, Tanya.
     
  16. Nov 20, 2013 #15
    Oops it seems that I marked friction f_b in wrong direction..
     
  17. Nov 20, 2013 #16
    Yes...the direction of fb is opposite to what you have marked in the figure.

    What about the sign of component of normal force from the ball ,on the rod in x-direction ? From the figure it looks in the -x direction .But in your equation it is positive.

    Please reconsider your two equations and write them again (force equations on the rod in x and y direction)
     
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