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I would like to know what the dynamic force load is...

  1. Jul 18, 2016 #1
    1. The problem statement, all variables and given/known data
    I am currently doing an internship and as part of my assignemnt I have to calculate the dynamic load on the bearing of a certain part that is rotating about a fixed point.
    https://www.physicsforums.com/attachments/chute-png.103400/ [Broken]
    This is the part that moves ( the brown part rotates about the back shaft)https://www.physicsforums.com/attachments/paint-chute-png.103401/ [Broken] The Ww is 8 kg,
    The Ws = 6 kg
    the weight of the blue boxes (Wp) on it is total 50 kg and it is centerized
    Tetha is 20-60 degrees

    I would like to what the dynamic force load is

    w = 0.5 rad/s

    2. Relevant equations
    Sum of moments = Moment of inertia* angular acceleration
    F= M*a


    3. The attempt at a solution
    I assumed all the masses are combined to become 64 Kg acting at the center of the wood
    but I got the force to be tiny.
    I am lost on how to start
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Jul 18, 2016 #2
    Please explain :
    The system is rotating about which point, namely A or B?
     
  4. Jul 18, 2016 #3
    Assuming the system is rotating about point A(as far as i understood) you can approach the problem in following way:
    Calculate the spatial position of the centre of mass.
    There will be several forces acting on this point.
    The forces acting here will be: 1. Gravitational force due to the combined weight of the system acting downwards
    2. Centrifugal force due to rotation acting along the line joining the point A and centre of mass of the system which will be mrw2
    Resolve the forces along the direction of Ax and Ay.
    There you have it.
    Hope it helped
     
  5. Jul 18, 2016 #4
    Hello Divya,
    Thank you for your reply,

    Yes your assumption is correct,the rotation is about point A but the force in order to rotate is provided by shaft B

    In order to simplify the question , I have made a new drawing.

    About your point 1 : Don't you account for the reaction forces?
    About your point 2 it's not clear which r I should take and do I then add the gravitational to the centrifugal force ?

    How about the reaction forces ?
     

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  6. Jul 18, 2016 #5
    I am writing the eqbm equations:
    Ay= Fb(Force at B) - Wcosθ
    Ax= Mr(w2)2-Wsin
    θ

    The radius would be the distance between the point A and centre of mass of the system.

    You will have to calculate w by the equation
    w2=w1+αt..........................(1)
    w1 = intial angular velocity

    θ= (w1)t + 0.5αt2..................(2)

    Replace these two equations in the above equations of Ax and Ay
    The net resultant force at the point A will be
    A= √(Ay2 + Ax2)
    Note you will have to find the maxima of the above equation by differentiating it with respect to t.
    You will have to put that value of the t in eq (2) for solve for θ.
    If the value of θ lies between 20 to 60 degrees then thats your answer. If it does not lie in that range then just put the value of θ=20 and θ=60 and solve for A(net force). Compare the two value that you obtain and the greater one will be your answer.

    Please note that the angular velocity of the system will change at every point because of the acceleration provided by the force at B.
     
  7. Jul 18, 2016 #6
    Hello ,
    Thank you very much for your effort.
    I assume that everything is a contact expect tetha right ?
    Here is where I get stuck in integration
     

    Attached Files:

  8. Jul 18, 2016 #7
     
  9. Jul 18, 2016 #8

    haruspex

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    Compared with the diagram, I think you have the signs wrong on both Ax and Ay.
    For the mrω2 term, what is your r? Remember it is a lamina, not a point mass.
    That is only valid for constant angular acceleration. If the force at B is constant then the acceleration will not be.
    In your first equation you have made the same mistake as DSS with regard to moment of inertia. It is not a point mass at distance 220mm (r2 ?) from A.
    I don't understand how you got the second equation from the first. If I differentiate the second equation wrt t I do not get back to the first equation.
     
  10. Jul 18, 2016 #9
    Yes my last solution is wrong. The correct way would be solving the following eq:
    d2θ/dt2= (Fbr1 - Wcosθ.r2)/I
    where r1 and r2 are the distances of the force Fb and weight from the point A respectively.

    After double integration with respect to time we will have two unknown variables whose value we can calculate by:
    at t=0 , θ = 20
    at θ=20, w=0.5rad/s
     
  11. Jul 18, 2016 #10

    haruspex

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    How are you doing that? There is a theta on the right hand side.
     
  12. Jul 18, 2016 #11
    There's a way to integrate it to some extent which is just enough for us to get us the answer:
    0.5(dθ/dt)2 = aθ - bsinθ + c
    is the integrated equation
    where
    a= By.r1/I r1 is the distance between point A and B
    b= Wr2/I r2 is the distance between point A and Centre of mass

    Now in the above equation dθ/dt is w.
    putting the boundary conditions as...........
    w=0.5 when θ=20 degree will help us calculate the value of c
    Once we have the value of c , we can easily calculate the angular velocity at any angle between 20 to 60 degrees from the same equation
    Now we write the expression of A= √(Ax2 + Ay2)
    In this equation put the following values of θ= 20, 30, 40, 50, 60 and calculate the value of A
    I am very sure that the maximum value of A will be at θ=60° since A is most likely to be an increasing function in the domain θ ∈ [20,60]
    Please do calculate and verify.
     
  13. Jul 18, 2016 #12

    haruspex

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    Ok. You previously said to integrate twice with respect to time.
     
  14. Jul 19, 2016 #13

    Delta²

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    yes instead he integrates with respect to ##\theta## and uses ##\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\frac{d\theta}{dt}=\frac{d\omega}{d\theta}\omega##.

    But let me ask another thing, in what frame of reference we do the calculations? Isnt the centrifugal force appearing in a rotating frame of reference?
     
  15. Jul 19, 2016 #14
    There is a rotating mass whose point of rotation is about point A. Now in order to write the eqbm equation i applied the corresponding inertial centrifugal force mrw2. So the frame of reference i am using is ground(stationary).
    Let me ask one more thing: while application of a corresponding inertial force whose direction is faced radially outwards wouldn't there be a force which is perpendicular to this inertial force?
     
  16. Jul 19, 2016 #15

    Delta²

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    What do you mean, if we have tangential acceleration along with centripetal acceleration? In this problem yes we have, but it isn't necessary to happen always.
     
  17. Jul 19, 2016 #16

    Nidum

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    Some of the help that you have received so far is infinitely unhelpful .

    You do not have enough information to work out the forces acting on the platen .

    To get any further you need information about how the angular position of the platen varies with time .
    or
    How the magnitude and direction of the driving force varies with angular position of the platen .

    Get the information from the project specification or experiment or use experience and common sense to derive a plausible profile .

    If you cannot obtain this information then you will have to use a simple static analysis and apply an appropriate g or usage factor multiplier .
     
  18. Jul 19, 2016 #17

    First I would to stay that all the help received from all the members is greatly appreciated .

    How would you suggest I find that information by experiment ? ( experiment set up )
    The only information I found from testing is that it takes 1.6 seconds to travel 40 degrees,
    I am not sure on how to get an equation relating the angle and time due to the short time (1.6s)

    Finally I am just an intern and dont have much experience yet so please be patient
     
  19. Jul 19, 2016 #18

    Nidum

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    No difficulty with that at all .

    Let's see if we can solve your problem using sound engineering principles .

    We are already much better off than we were before since you have now given us the information that the complete movement takes 1.6 seconds .

    What do you think is happening during that 1.6 seconds ?
     
  20. Jul 19, 2016 #19

    Nidum

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    Think about acceleration , intermediate motion and deceleration .

    We need to get at the maximum input force available as well .

    How is this device operated - by hand or by a powered mechanism ? Can you estimate the maximum force that can be applied ?
     
  21. Jul 19, 2016 #20
    Capture.PNG
    As you can see here , the arm is operated by a pneumatic cylinder.
    The output of the cylinder is 754 N

    The acceleration seems to be constant as the angle of cylinder change as the arm rotates about point A
     
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