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Free body diagram of two different objects

  1. Oct 6, 2016 #1
    • Homework posted in wrong forum, so no template
    I was wondering if anyone can provide feedback on my free body diagram of two objects acting on each other. The question is asking for two separate free body diagrams. A force is applied to object 1, and object 1 pushes against object 2. Object 1 is on wheels and object 2 is in contact with the ground. The two objects accelerate. Find the constant for the kinetic friction.

    I can confidently find the constant, as long as my free body diagram is correct. But, I would like some feedback on whether it's correct it or not.

    On object 2, the vertical force is in equilibrium, F2N - w2. The horizontal force is F1 ON 2 - fk.

    On object 1, the vertical force is also in equilibrium F1N - w1. The horizontal force for object 1 is a bit confusing. I'm assuming it's nearly frictionless, since it's on wheels. So is it, F - F2 ON 1? I guess it would be incorrect to subtract fk from the horizontal force on object 1, since it's only acting on object 2, however how does that friction force affect object 1?
     
  2. jcsd
  3. Oct 6, 2016 #2
    Hello Dear,

    I do not know if they have asked for Both Rolling Friction for Object on Wheel and Sliding Friction for Object on Ground.

    But if Rolling Friction is to be Neglected then f1 = 0 will be used.

    See the Picture Attached.
     

    Attached Files:

  4. Oct 6, 2016 #3
    Thank You for the response. At the moment, we're ignoring rolling friction. The question had a hint, "only one of the objects experiences the friction force". So I assumed, the kinetic friction is only applied to object 2. However, I feel like I'm missing something on my free body diagram for object 1.

    Also, there is a force, F, being applied to object 1, and object 1 is pushing against object 2. So, if m1a = F - F2 ON 1 is true, is the force F on object 1, related to the action-reaction forces?
    Two+objects+are+in+contact+a+force+is+applied+to+_a654e09acbdb41e3b4c70fdaad166284.jpg
     
    Last edited: Oct 6, 2016
  5. Oct 6, 2016 #4
    If m1a = F - F21 then F will not be in m2 but you will have to Consider acceleration of M2.

    Similar to This picture.
     
  6. Oct 6, 2016 #5
    The acceleration should be the same for both objects.
    a = m1/(F - F2 ON 1)
    a = m2/(F1 ON 2 - fk)

    So F = (m1(F1 ON 2 - fk)/m2) + F2 ON 1

    These equations are derived from my diagram. So is my diagram correct? Ignoring rolling friction.
     
    Last edited: Oct 6, 2016
  7. Oct 6, 2016 #6
    Sorry Brother,

    I could not send the picture in previous answer.
    Here it is
     

    Attached Files:

  8. Oct 6, 2016 #7
    Oh so, friction is applied to each individual system? For instance, if rolling friction is ignored, there would be no friction force on object 1, and the kinetic friction would be only present for object 2? If both objects were treated as one system, the friction force on the system would essentially be the kinetic friction from object 2 alone?
     
  9. Oct 6, 2016 #8

    Dale

    Staff: Mentor

    The free body diagrams are fine.
     
  10. Oct 6, 2016 #9
    I don't Understand, Sir.

    If they are all fine... for object 2: F1 on 2 = fk = F2 on 1 (Newton's 3rd Law) = F - m1a
    then, how will I take account of the fact that body of mass m2 accelerating with a.

    Please help me Understand.
     
  11. Oct 6, 2016 #10
    No, actually in the picture I considered Both the Frictional Resisting forces to be same as f .

    But if you take individual Free Bodies and Ignore Rolling Friction, then there will be No Friction Force on the Object with Wheels in its FBD.

    [and sorry for the Emoticon... I was in Hurry and my Browser somehow added that... I know you can understand even with it]
     
    Last edited: Oct 6, 2016
  12. Oct 6, 2016 #11

    Dale

    Staff: Mentor

    I am not sure why you believe this. It is not correct. ##f_k## is the friction force and is proportional to the normal force. It does not have any particular relationship to ##F_{1on2}##
     
  13. Oct 6, 2016 #12
    But that is What Free Body Diagram of Object 2 Shows on using Equilibrium along Horizontal axis.
     
  14. Oct 6, 2016 #13

    Dale

    Staff: Mentor

    It is not in equilibrium:
     
  15. Oct 6, 2016 #14

    Dale

    Staff: Mentor

    @Ehden, your diagrams are correct. I do hope that you do not get confused from this thread.
     
  16. Oct 6, 2016 #15
    I had drawn all the Diagrams.... I don't think i considered any body is in Static Equilibrium. They are all in Dynamic Equilibrium and All Inertia Forces should be Shown.

    And even if Dynamic Equilibrium is not drawn, at least accelerations with Direction is shown, which was not In diagram making it look like a Static Condition.

    Ehden is Correct with his Equations and Diagrams, both.
    But not with his Diagram alone.
     
    Last edited: Oct 6, 2016
  17. Oct 6, 2016 #16
    These Equations are Correct.

    And my Apologies, if I made anything Complicated.
     
  18. Oct 6, 2016 #17

    Dale

    Staff: Mentor

    I dislike this approach and prefer to use an inertial frame, but regardless of whether you use an inertial or a non inertial frame you never get ##f_k=F_{1on2}##
     
  19. Oct 6, 2016 #18
    Thank you for your response!

    Don't worry, a learning experience for both of us!

    Also, the question was asking for free body diagram after the force is applied, so I made a mistake with object 1. There should've been three forces on object 1, as the force isn't being applied after the object is already accelerating. So from that point of view, I derived the constant.

    uk = (m1a - m2a)/m2g
    = (a/g) ((m1/m2) - 1)

    Is my work correct?
     
  20. Oct 6, 2016 #19
    Yes, that is True,it never holds unless it is a Static Equilibrium.
    That is why I always Prefer to Draw the accelerations when Inertial Frame is not Used, it avoids confusion.

    By the way Thank you Sir, for Guidance.
     
  21. Oct 6, 2016 #20
    They seem to be Very correct if there is no F.
     
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