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Why don't I get these free body diagrams?

  1. May 22, 2005 #1
    A man sits in a bosun's chair that dangles from a massless rope, which runs over a massless, frictionless pulley attached to the ceiling and back down to the man's hand. The combined mass of the man and chair is 90.0 kg.

    Ok, I found a solution that had T=Mchair*a + Fnormal + Mchair*g. How did they know to separate the man and the chair when the problem only gives you a mass for the combined mass of the man and the chair?

    Then the problem asks for the weight exerted by the ceiling onto the whole system. Why is this answer not just all of the weights added together?
     
  2. jcsd
  3. May 22, 2005 #2

    quasar987

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    I'll give it a shot but I'm not excellent at FBD problems either.

    They don't claim to have a numerical value for M_chair. they just wrote the equation of motion for the chair.

    Because the normal force "relieves" that weight. Consider the whole rope+chair+man as a single body [itex]\beta[/itex]. Then the equation of motion reads

    [tex]\sum F_{\rightarrow \beta} = T+T-T-T-M_{man}g+F_{normal}-M_{chair}g = F_{normal} - (M_{man}+M_{chair})g[/tex]

    That's the total force on the system. So in order to keep it in equilibrium at all times, the "ceiling" (more like the ceiling on the nail, the nail on the pulley, the pulley on the rope) must exerts a force equal and opposite.

    Note that in the case where the man is not sitting in the chair, that force reduces to just the total weight.
     
    Last edited: May 22, 2005
  4. May 22, 2005 #3
    so how did you get the floor exerts a force when the pulley is attached to the ceiling? :cry:
     
  5. May 22, 2005 #4

    quasar987

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    I meant ceiling, not floor. lol
     
  6. May 22, 2005 #5
    So where did all of the T's come from? One from the top of the chair to the pulley and one from the pulley to the man's hand. And where did the other two come from?
     
  7. May 22, 2005 #6

    quasar987

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    The two -T are actual the normal forces exerted by the chair on the rope and by the man on the rope according to newton's third law.
     
  8. May 22, 2005 #7
    Ok, so we have a positive T coming from the chair to the pulley and a Positive T coming from the man to the pulley and then a negative T as the normal force from the chair to the pulley and a negative T as the normal force from the man's hand to the pulley? I thought you use normal force as in when two surfaces come in contact with each other eg the chair pushing up on the man's butt. And if the -T's are the normal forces, then how come there's a "Fnormal" in the equation too?

    I'm sorry to rip apart your response, but I'm really confused here. Thanks for helping though! :!!)
     
  9. May 22, 2005 #8

    OlderDan

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    What is the question? Your solution seems to still contain unknowns. If the question is only about the tension in the rope, there is no need to separate the man and the chair, and no need to worry about the normal force. Without knowing the breakdown of the mass into mass of chair and mass of man, you cannot find the normal force.

    If you just look at the man/chair group of mass M, the forces acting on M are gravity, Mg, and twice the tension in the rope, 2T. If the mass is acceleratiing downward, it is because the tension is less than half the weight, Mg. If the man is pulling himself upward at constant speed, there is no acceleration and the tension is half the weight. If the man is accelerating upward, the tension would have to be more than half the weight.

    The force provided by the ceiling is twice the tension in the rope (massless rope and pulley). If the man is moving at constant speed, that force is just the sum of the weight of the man and the chair, but if they are accelerating, it has to be less for downward acceleration and greater for upward acceleration. In the limiting case, if the man let go of the rope the acceleration would be -g (downward), the tension would be zero, and the ceiling would provide no force.

    The man and the chair share a common acceleration. If you knew ther separate masses you could write two equations

    [tex]F_{man} = T + N - M_{man}g = M_{man}a[/tex]

    [tex]F_{chair} = T - N - M_{chair}g = M_{chair}a[/tex]

    without knowing the separate masses, the best you can do is add the two equations to get

    [tex]F_{total} = 2T - M_{total}g = M_{total}a[/tex]
     
  10. May 22, 2005 #9
    ok the question is (a) With what force must the man pull on the rope for him to rise at constant speed? (b) What force is needed for an upward acceleration of 1.30 m/s2? (c) In the cases above, what is the force exerted on the ceiling by the pulley system?

    For a, I got 441. So then by that last equation, 2T-Mg=Ma. A=0 because it's constant velocity. So 2(441)-90(9.8) and that comes out to be 0. So there's no force being exerted on the ceiling?
     
  11. May 22, 2005 #10

    OlderDan

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    The force the man applies to the rope is the tension in the rope. 441N is the correct answer for a) and that is the tension in the rope. The ceiling in all cases must provide a force that is twice the tension. I think you will find the correct answer for the ceiling in case a) to satisfy your intuition. For case b), it might not be so satisfying, but it will still be twice the tension, and it makes sense if you think about it. Ultimately, the force acclerating the man and chair has to be provided by the ceiling; the rope is just a mechanism for transferring the force from the ceiling down to the man and the chair.
     
  12. May 22, 2005 #11

    quasar987

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    Newton's third law is more universal than juste when two surfaces are in contact with one another. Take two electric charges. Their mututal force of attraction is the same. Take two chunk of matter, for instance you and the earth. The grav. force of the earth on you in the same as that of you on the earth. In the case of surfaces coming in contact, the normal force is issued from the atomic interaction of the two constituants. But it holds also if you pull on a rope. Imagine you're in the chair and you've got the rope on your hand. I'm sure you can feel a force; or a sensation that the rope seems to pull the flesh of your hand up. ouch. That's the static friction. It's the force you use to pull on the rope, but it's also the force the rope uses to pull your skin up.

    The force labeled F_normal stands for the chair pushing on the man's butt.


    (Hopefully someone will interupt if I'm talking nonsense)
     
    Last edited: May 22, 2005
  13. May 22, 2005 #12

    arildno

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    Yes, you're talking nonsense..:wink:
     
  14. May 22, 2005 #13

    quasar987

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    :rofl: rofl :rofl:
     
  15. May 22, 2005 #14
    I THINK I got it.

    Ok, so the reason why the forces pulling on the ceiling are 2T is because if we isolate the ceiling, the only think pulling on it are the "two" ropes, one part going down to the chair and the other going down to the hand. Then whatever the tensions are in each rope are the only forces pulling on the ceiling.

    And just to correlate it to another problem I did, someone check my reasoning here.

    (a) In Fig. 5-43b a 7.5 kg salami is hung off of a table, supported by a cord that runs around a pulley attached to the corner of the table and then to a scale. The opposite end of the scale is attached by a cord to a wall. What is the reading on the scale?
    (b) In Fig. 5-43c the wall has been replaced with a second 7.5 kg salami on the left, and the assembly is stationary. What is the reading on the scale now?

    Ok the answer to both of these is 73.5, and the reason is that if we isolate the forces acting on the scale, there is tension T1 acting on one side and tension T2 acting on the other side... Ok, nevermind, I dont know why I got the 73.5 for parts a and b...
     
  16. May 22, 2005 #15
    haha I love you guys
     
  17. May 22, 2005 #16

    quasar987

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    Easy now. :wink:
     
  18. May 22, 2005 #17

    OlderDan

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    You've got it. You could break it down a bit further. If you look at the ceiling, the only force acting on it is the attachment to the pulley. If you look at the pulley, there are three forces: The upward force from the ceiling and the two equal tensions downward (the tensions would be unequal if the pulley had mass and was accelerating of if there were friction). The pulley never moves up or down, so the net force acting on the pulley is zero, which means the upward force acting on the pulley is twice the tension. The force up on the pulley is the force downward on the ceiling by Newton's third law; they are an action-reaction pair)

    I don't know why either :rolleyes: Maybe because the scale reads in Newtons? You could do a better job keeping track of your units. It will help you in the long run. I assume in part b) the second salami is hanging over a pulley also, and that is why the system is stationary. Each salami has two forces that must be equal and opposite: their weight, mg, and the tension in the cord.

    The scale has two opposing tensions. If they were not the same, the scale would be acceleratiing. A scale can only register a force if there are two opposing forces acting on it. A spring scale being pulled at both ends registers the magnitude of the equal opposing forces, or, if the forces are unequal, it registers the lesser force and accelerates in the direction of the greater force. A bathroom scale has two equal and opposite forces acting. The downward weight of the person standing on the scale, and the equal force of the floor pushing upward. In an acclerating elevator, the scale would read the smaller of the the two forces acting on it, and it would be accelerating.

    In most problems we say to ignore the mass of the scale, or, as in this problem, the mass of the rope. In that case the opposing forces must be the same, even if the scale or rope is accelerating. If a force difference were applied to any massless object, the acceleration would be infinite. It is because we assume the rope has no mass that we can say the tension at both ends of the rope is the same. So even when the man/chair is accelerating, the tension in the massless rope is the same at both ends.
     
  19. May 22, 2005 #18
    perfect!

    thank you all so much!
     
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