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A few Motion Questions

  1. Mar 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A 2200kg airplane is pulling two gliders, the first of mass 310kg and the second of mass 260kg, down the runway with an acceleration of 1.9m/s^2. Neglecting the mass of the two ropes and any frictional forces, determine the tension force in the second rope.


    2. Relevant equations
    F=ma

    3. The attempt at a solution
    This is what i have done, however i do not understand the book's answers.

    Two forces acting on rope 2, so...

    T-494 = 4180

    T=4674N

    book's answers is 494N, wouldn't this be the net force on the second glider??

    1. The problem statement, all variables and given/known data
    A child pulls an 11kg wagon with a horizontal handle whose mass is 1.8kg, giving the wagon and handle an acceleration of 2.3m/s^2.
    a) Find the tension at each end of the handle?

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    Can someone check if my answer is correct.

    Net Force of Handle and Wagon are:

    F=4.14N and F=25.3N

    Tension from handle to wagon

    T1-4.14 = 25.3
    T1=29.22N

    Tension from wagon to handle

    25.3-T=4.14
    T=20.89N

    1. The problem statement, all variables and given/known data
    In a tractor pulling contest, a 2300kg trctor pulls a 4900kg sledge with an acceleration of 0.61m/s^2. If the tractor exerts a horizontal force of 770N on the ground, determine the magnitude of
    a) the force of the tractor on the sledge.
    b) the force of the sledge on the tractor and
    c) the frictional force exerted on the sledge by the ground.


    2. Relevant equations
    F=ma

    3. The attempt at a solution
    Like the last question can someone tell me if my answer is correct, if not what have i done wrong?

    a)Fnet = ma
    T - Fnet = ma
    T - 7700 = 2300 * 0.61
    T - 7700=1403
    T=9103N

    b) Fnet = ma
    T - Fnet = ma
    T - 48020 = 4900 * 0.61
    T=51009N

    c)
    T - Friction = ma
    9103 - Friction = 4900 * 0.61
    Friction = 6114N

    P.S
     
  2. jcsd
  3. Mar 27, 2010 #2

    Doc Al

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    I don't understand your reasoning. Where do these number come from?

    Hint: What forces act on the second glider? Apply Newton's 2nd law.

    Yes!
     
  4. Mar 27, 2010 #3
    There is two forces on the second glider and that is Tension and Fnet on glider so Tension is what i want to find.

    Fnet of second glider =ma
    Fnet of second glider = 260 * 1.9
    Fnet of second glider = 494N

    Now i know the Fnet i can sub into T-Fnet = ma (force of airplane)

    T - 494 = 4180
    T=4674N
     
  5. Mar 27, 2010 #4

    Doc Al

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    Fnet is not a force--it's the sum of all the forces. There's only one horizontal force acting on the second glider--the tension in the rope. (And if you are analyzing the glider, m would be the mass of the glider, not the plane.)
     
  6. Mar 27, 2010 #5
    but wouldn't there be a force acting from the second glider to first rope


    Tension Rope 1<---------glider 2 ------------->Tension Rope 2
     
  7. Mar 27, 2010 #6

    Doc Al

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    It depends on which glider is glider2. I took glider2 to be the last one in the chain, like this:

    glider2---(rope2)---glider1---(rope1)---plane

    So glider2 only has one rope pulling on it.
     
  8. Mar 27, 2010 #7
    yeh the book's doesn't have labels on the plane or glider it just has it like

    glider-----rope---glider----rope---plane

    How do you know which glider is which???
     
  9. Mar 27, 2010 #8

    Doc Al

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    They start counting from the plane.
     
  10. Mar 27, 2010 #9
    oh ic, i understand now thanks for the help.
     
  11. Mar 27, 2010 #10
    Can anyone answer my other question.
     
  12. Mar 27, 2010 #11

    Doc Al

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    Not quite right. And your method is a bit confusing. The handle has two ends: the child end and the wagon end. What force does the child exert on the handle? That's the tension at the child end. What force does the wagon exert on the handle? That's the tension at that end. (Hint: Newton's 3rd law.)
     
  13. Mar 28, 2010 #12
    Well, i calculated that if there is only one force on each of the object, then logically speaking the tension on both ends are equal and opposite according to Newton's Third Law.

    However if the masses are different on each object would the force exerted be different or the same?

    If they are different would the Tension at the child's end be 4.14N and the wagon's end be 25.3N?
     
  14. Mar 28, 2010 #13

    Doc Al

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    Which exerts the greater force: the child on the handle, or the handle on the wagon?

    It may help to draw a force diagram showing all the forces acting on the wagon and on the handle.
     
  15. Mar 28, 2010 #14
    The child exerts one force

    child--------->handle

    Handle exerts two forces, one on the child and the another one on the wagon

    child<--------handle--------->wagon

    To answer your question i would say the handle on the wagon exert more force.
     
  16. Mar 28, 2010 #15

    Doc Al

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    Good.

    No. The child must exert enough force to accelerate both wagon + handle, while the handle only needs to exert enough force on the wagon to accelerate the wagon.

    (wagon + handle)---> child

    (wagon)---> handle
     
  17. Mar 28, 2010 #16
    so the child will exert the total force of the wagon and handle, which is 4.14 + 25.3 = 29.44N

    so the wagon end will exert 25.3N
     
  18. Mar 28, 2010 #17

    Doc Al

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    Right!
     
  19. Mar 28, 2010 #18

    Doc Al

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    You messed up the direction (and thus the sign) of the forces. (Also: Don't mix up an individual force--the 7700N--with the net force.)
     
  20. Mar 28, 2010 #19
    so the tractor on the sedge would exert two forces

    sledge<--------tractor--------->7700N

    7700-T = 2300 * 0.61
    T=6297N

    The sledge exerts one force on the tractor

    sledge------>

    F=ma
    F=4900*0.61
    F=2989N
     
  21. Mar 28, 2010 #20

    Doc Al

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    I'd phrase that this way: There are two forces acting on the tractor:
    (1) The force from the sledge, called T.
    (2) The force from the ground, which we know is 7700 N.

    Good.

    That's the net force on the sledge. To answer (b), consider Newton's 3rd law.
     
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