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I can't study on untill I get this-please do help me

  1. Sep 16, 2005 #1
    I can't study on untill I get this--please do help me

    hello

    Perhaps a little introduction. I've quit highschool 7 years ago and recently decided to finish it and as such I'm learning physics at home by myself.
    I hope you guys can answer all three questions in (gory details) since they are kinda related and knowing just part of the complete answer won't do me much good

    A) Suposse we have a box with some mass and a rope attached to the left side of a box. If some external force F pulls this box to the right then box will aplly same force F to the rope.
    And like wise, rope will exert force -F to the box. Now box has two external forces acting on it : one is F that is trying to pull box to the right, and -F that is trying to pull it to the left. Which would suggest that the two external forces acting on box cancel eachother out

    I'm confusing something here



    B)I'm not even shure how to ask it. If I want to pull an object with mass m and I apply force F to it, then ( due to third newton law ) object will apply force -F on me. But due to inertia of an object, how much smaller will the net force be ?

    C)Two objects with

    m1 = 2 kg
    m2 = 4 kg

    are connected with rope. We pull the heavier of the two objects with force F = 24 N
    Find a. What is the force F1 in rope ?

    F = ( m1 + m2 )*a - > a = F/( m1 + m2 )

    the above formula gives the correct result but as usual my study book has to complicated the matters


    Code (Text):

      __  F1         -F1  __     F
     |m1|----->----<-----|m2|--------->
     |__|                |__|  

     


    According to my study book an object with mass m1 is being pulled by the force of rope F1 to the right with acceleration a

    An object with mass m2 is being pulled by force

    F - F1 = m2 * a

    Due to newton's third law object with m1 should also be pulled with force F, but here we have force F1 instead. HUH?!!!


    I hope someone can finally clear these thigs up for me since I'm truly stuck and can't move on until I get this

    thank you very much
     
  2. jcsd
  3. Sep 16, 2005 #2

    HallsofIvy

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    Inertia does not affect "net force". The net force on the system (both the object and you) is F-F= 0 so it does not move. Of course, the net force on the object alone is F so it moves with acceleration a= F/m. If you had nothing holding you down (your feet on the floor for example), you would move toward the object with acceleration F/your mass. The center of mass of the system would not move.

    First, think of the two objects as a single system. Since they are connected, they must move with the same acceleration, just as if they were one mass.

    Together they have mass 2+ 4= 6 kg. Since you are pulling with force 24N, they move with acceleration F/m= 24/6= 4 m/s2. Since they are connected, the must move with the same acceleration, just as if they were one mass.

    Now look at the single 2kg mass. Since it is moving with acceleration 4 m/s2, it must have force F= ma= (2)(4)= 8 N. That is "F1".

    Now look what is happening to the 4 kg mass. It is also moving with acceleration 4 m/s2 so the net force on it is F= ma= (4)(4)= 16 N. What is happening is that you are pulling on it with a force of 24 N but it is pulling the other weight with force 8N. Of course (equal and opposite reaction), the other weight is pulling back with force 8N. The net force on it is 24- 8= 16 N.
     
    Last edited: Sep 16, 2005
  4. Sep 16, 2005 #3
    Excellent questions for someone new to the subject. You are however mis-applying some of newtons rules.

    Equal and opposite does _not_ imply that every object experiences balanced forces

    Balanced forces do not imply that something is not moving

    if you are draging a box of rocks at constant speed the tension on the rope is balanced by the friction of the ground (you need to be certain you are clear about the object you are concidering)

    The tension at one end of a rope is often balanced by the same tension at the other end of the rope, so if you are pulling a box of rocks the box pulling back on the rope is balanced by you pulling on the rope so the rope move with constant speed. the rope pulling on you is balanced by the grounds friction on your feet so you move at a constant speed.

    If you want to accelerate the objects involved then the net force on each object has to be out of balance. The force of friction on your legs has to be greater than the tension in the rope to accelerate you. (technically the force on one end of the rope has to be a tiny bit more on one end than the other to accelerate the rope but this difference is so small it is assumed to be zero) The tension pulling the box has to be greater than the friction for the box to accelerate. Since all the accellerations have to be the same it is just a matter of working out the net force needed by each object based on F=ma
     
  5. Sep 16, 2005 #4
    This I understand . I understand C example if I think ob both objects as one. But if I look at them separately, confusion arises.

    So net force redistributes itself between the two objects? How is that possible ?

    *In order to ask the following questions I must first understand question A !

    1-Let us think of the two objects as separate entities, and take object with mass m2 into consideration - I pull on it with force F. It pulls on rope with
    force F and rope pulls back on it with force F. So two external forces cancel eachother out. Then object with mass m1 pulls via rope on m2 with Force F1 and m2 pulls me with force F1 ????
     
  6. Sep 16, 2005 #5

    StatusX

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    The center of mass of a system cannot move, but the individual objects within the system can if they exert forces on each other. For example, if there are two electrons near each other, they will each experience a force from the other, and will change positions. But because these forces are always equal and opposite, the center of mass does not move. This can be proven pretty easily, but I think it's intuitive enough.

    One problem might arise if you consider external forces. For example, if you have an object in a uniform gravitational field, clearly the center of mass is changing as it falls. So you might say the center of mass does not move as long as there are no net external forces on the system. But this is really only a way to cope with simplifications we make to problems, and in fact the center of mass law is always valid if we are treating a real system. In this case, we are approximating the gravitational field of a large object, say the earth, by a uniform field. But in reality, the smaller object is accelerated towards the earth AND the earth is accelerated towards the smaller object (equal and opposite forces), so the center of mass does not move. But the earth moves so little it is safe to ignore it, so long as we remember that the principle is always valid.

    In your first example, you pull the block and the block pulls you. But you don't move. Why? Because of friction. But what is the opposing force to the friction? Well, you push the earth in the opposite direction. Again, this is a negligibly small effect, but if you consider as the system the combination of you, the block, and the earth, the center of mass does not move, but the objects in the system exert equal and opposite forces on each other and get rearranged.
     
    Last edited: Sep 16, 2005
  7. Sep 16, 2005 #6

    Doc Al

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    A good statement of Newton's 3rd law is this: Forces always come in pairs. If body A exerts a force on body B, then body B must exert an equal and opposite force on body A.

    Let's use that to analyze the forces on the connected masses:

    Start with force F. We are not told what body is exerting that force, so let's just assume another rope is pulling on m2 with a force F to the right. Newton's third law says that m2 must then be pulling back on the rope with a force F to the left. Note that F acts on m2: it does not act on m1! The only force acting on m1 is the tension in the connecting rope. (Of course, the force F will determine what that tension will be, but the tension will not merely equal F! To figure out what the tension force F1 is, use Newton's 2nd law.)

    Now we can deal with the tension in the rope connecting the massing. If the rope pulls on m2 with force F1 to the left, then m2 pulls back on the rope with a force F1 to the right. Since the tension in the rope is the same throughout, the rope is also pulling m1 to the right with a force of F1; thus m1 pulls back on the rope with a force of F1 to the left.

    If you like, you can think of the connecting (massless) rope as merely transmitting the force F1 between m1 and m2: m1 pulls m2 to the left with a force F1; m2 pulls m1 to the right with a force F1.

    Note that m2 has two forces acting on it; m1 only has one.
     
    Last edited: Sep 16, 2005
  8. Sep 16, 2005 #7
    So basicly force will "figure out" how much of its force must distribute over a rope to object m1 in order to accelerate m1 with a ?

    So if F is pulling on object on the right side,this object or force F will somehow figure out how much of force F must it distribute over to the left side of object and pull with that force ( call it F1 ) whatever is on object's left side?
    Main point being, that object never pulls with main force F whatever is on its left side ?

    If that is the case ... why or how is that so ?

    EDIT-this is vital . Why did object m2 redistribute force F so that anything on left side of object only was pulled by force F1, while rope didn't redistribute force F1 and if object m1 was pulling it with F1, then rope pulled with same force object m2 that was on the other side of the rope ?



    And if I may one more question ...

    What is net force here ? If we look at the tow objects separately then Net force for m2 would be

    Net force = F - F1 ?

    But if we look at them as one object ( m1 + m2 ) then net force = F ?

    uh, why?
     
    Last edited: Sep 16, 2005
  9. Sep 16, 2005 #8

    Doc Al

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    The force doesn't need to "figure out" anything (I trust you are speaking figuratively); the objects just do what they do, all of which is described by Newton's laws.

    Force F pulls on m2. If m2 were not connected to anything, it would just accelerate at the rate a = F/m2. But it can't since it's connected to m1 via that rope. So, in "trying" to accelerate, m2 must pull on m1. And that means (from Newton's 3rd) that m1 pulls back on m2, slowing down it's acceleration. The resulting acceleration is only a = F/(m1 +m2).




    Object m2 has two things pulling on it: The force F trying to accelerate it and the tension in the rope (F1) trying to de-accelerate it. But object m1 only has the tension pulling on it. You can think of it like this: F must accelerate both m1 and m2. So when force F is "transmitted" through m2 to m1 it gets reduced by the amount needed to accelerate m2 (F - F1) leaving only F1 to accelerate m1.

    If you think about it carefully, this kind of thing even happens when you attach a rope to a single object and pull with a force F. After all, the force really only acts directly on the part of the object it's attached to: a piece of one side of the object. But, since the molecules of the object are stuck together, that little piece can't accelerate without dragging the rest of the object with it. The bottom line, the force ends up accelerating the entire object; just like the force F ends up accelerating both m1 and m2 in your example.


    The beauty of Newton's Laws is that they describe how forces affect any object or collection of objects. You are free to define your "system" or object any way you want. If you take both m1 and m2 together, then the net force on the composite "object" is F. But you are free to choose m2 alone as your object: in which case the net force is F - F1. And if you chose m1 to analyze, the net force on it is just F1. It works the same way, no matter how you slice it.

    So if someone asks "What's the net force?", you must ask them "Net force on what?".

    And in some problems (like this one) you'll need to analyze it several ways at once! The way I would solve this problem is this: I immediate see that F pulls on both m1 and m2, so I know that a = F(m1 + m2). (Since they are connected, I know that both masses must move together with the same acceleration.) Then I look at m1 and immediately realize that the force on it must be F1 = (m1)a. And so on...

    I hope this discussion helped a little. (These ideas are quite subtle.)
     
  10. Sep 16, 2005 #9
    If I may ask something - if I picture m1 and m2 as one object then according to newtons third law whatever force I will apply to an object ( m1 + m2 ) I will get back an equal force pulling on me . But if I picture them separately, then won't the only force I get pull on be one from m2 ( F - F1 ) , and m2 will get pull back by m1 ( F1 ) ? So if machine was somehow attached on me that was able to detect reaction forces from objects I pulled , would the machine detect only F - F1 or F ?



    cheers
     
    Last edited: Sep 16, 2005
  11. Sep 16, 2005 #10

    Doc Al

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    Exactly right. You are pulling on the combined "m1 + m2" object with force F; so the "object" pulls back on you with force F. (Of course, you are really only directly pulling on m2, which is a part of "m1 + m2".)
    Even if you picture them separately, the force is the same. You pull on m2 with a force of F, so it pulls back on you with a force F. (Note that you do not directly pull on m1. m2 pulls on m1.)
    The machine would detect the actual force that you exert: F, not F - F1.
     
  12. Sep 17, 2005 #11
    thanx a million. I really appreciate your help
     
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