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Homework Help: Static friction between 2 blocks and tension of string

  1. Jun 26, 2015 #1
    1. The problem statement, all variables and given/known data

    the maximum force is F=F1 +F2 + T as shown in the textbook....
    I cant understand why the tension of spring is also = F1...Is it because of the writer assume the spring is directly attached to the part P (upper part of the block) ??
    2. Relevant equations


    3. The attempt at a solution
     

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  2. jcsd
  3. Jun 26, 2015 #2

    haruspex

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    It says string, not spring.
    Consider the horizontal forces on block P. Given equilibrium, what equation can you write?
     
  4. Jun 26, 2015 #3
    F= umg
    Is it because of the string is attached to the block p ? So we should only consider the tension acted on block p ?
     
  5. Jun 26, 2015 #4

    haruspex

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    That's only one of the two horizontal forces acting on P. What's the other one?
     
  6. Jun 26, 2015 #5
    The other one is static friction force
     
  7. Jun 26, 2015 #6
    How about q ? There's only static friction force? No tension of string acted on q ?
     
  8. Jun 26, 2015 #7

    haruspex

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    I'm trying to lead you to the answer to your question. Stick with P. There are two horizontal forces on P. Look at the diagram. What are they?
     
  9. Jun 26, 2015 #8

    haruspex

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    No, ##F=\mu mg## is the static frictional force. What is the other horizontal force on P?
     
  10. Jun 26, 2015 #9
    Tension of string
     
  11. Jun 26, 2015 #10

    haruspex

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    Right. If there's static equilibrium, what does that tell you about those two forces?
     
  12. Jun 26, 2015 #11
    The object is stationary until the applied force larger than the static frictional force, the object starts to move
     
  13. Jun 26, 2015 #12

    SammyS

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    That almost answers haruspex's question.

    How do the two forces on P (static friction and tension of string) compare when the objects are stationary?
    By the way, here is a portion of the image you provided:

    upload_2015-6-26_10-6-54.png
     
  14. Jun 26, 2015 #13
    static friction = tension of string when the object is stationary
     
  15. Jun 27, 2015 #14

    haruspex

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    Right. Does that answer your question in the OP?
     
  16. Jun 27, 2015 #15
    No, I still dont know why the tension of string isnt exist for block q ...
     
  17. Jun 27, 2015 #16

    haruspex

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    You asked why the tension is equal to the frictional force. That is now explained.
    What do you mean by the tension of the string "not existing" for Q? Do you mean it does not act on Q? Clearly it does. What makes you think it does not?
     
  18. Jun 27, 2015 #17
    ok , i am confused now . for block Q to start to move( block p will remain stationary on top of the block q when block q starts to move ) , why the force F isnt equal to F2[ (u )(mass of p + mass of q)(g) ] only ??
     
  19. Jun 27, 2015 #18
    I thought the tension of spring is not existing of Q simply because i saw the max force for Q is T+F1 +F2 , which the tension of string is equal to F1 (only mass of block P involved , but not mass of block Q)
     
  20. Jun 27, 2015 #19

    haruspex

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    As os clear from the diagram, the horizontal forces acting on Q are af1 (but this is opposite to the F1 acting on P), F2, and T.
    From the earlier analysis of block P, T is equal and opposite to the F1 acting on P, and therefore completely equal to the F1 acting on Q.
    If block Q moves to the right, block P will move to the left. It's an inextensible string, not a spring.
     
  21. Jun 27, 2015 #20

    SammyS

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    The force that block Q exerts on block P is equal in magnitude, but opposite in direction to the force that block P exerts on block Q . (Newton's 3rd Law) The magnitude of that force is labeled F1 in the diagram. When acting on P, the force is to the right. When acting on Q the force is to the left.
     
  22. Jun 28, 2015 #21
    Can someone draw me a diagram so that i can understand better . Since the question say both blocks are connected by a string . So , i assume the blocks are connected in this way...Correct me if i'm wrong...
     

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  23. Jun 28, 2015 #22

    haruspex

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    That's wrong. Read the question again. It's just one piece of string that goes off to the left from P, around a pulley with a fixed axis, and doubles back to connect to Q. That's why the tension is the same.
     
  24. Jun 28, 2015 #23

    haruspex

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    That's wrong. Read the question again. It's just one piece of string that goes off to the left from P, around a pulley with a fixed axis, and doubles back to connect to Q. That's why the tension is the same.
     
  25. Jun 28, 2015 #24
    Something like this? uploadfromtaptalk1435502046398.jpg
    If this is true , then which part of rope should I pull ? the upper part connected to the block P or the lower part connected to the part Q ? How can be the tension be the same ?
    If I pull the upper part , then I agree the tension of the rope should be = F1
    If I pull the lower part of the rope , then I think that the tension of the rope should be = F2
     
    Last edited: Jun 28, 2015
  26. Jun 28, 2015 #25

    haruspex

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    Neither part of that string is being pulled directly. Q is being pulled to the right by force F.
    If the tensions in the upper and lower parts of the string are not the same then there will be a net torque on the pulley, making it turn.
    Consider the forces on each block separately.
    Write down the static equilibrium equation for the horizontal forces on P, then do the same for Q.
     
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