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Net force, force diagrams, normal force

  1. Oct 8, 2008 #1
    In a problem whereby you want to calculate the net force of the rope lifting a box, and there is something in the box, whether you take into account the normal force.
    The forcees would be the tension of the rope, and the weight of the box and the object inside. To caculate the net force of the rope, do you take into account the normal force?
     
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  3. Oct 8, 2008 #2

    PhanthomJay

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    No. (and the force in the rope is called its tension, T, not the 'net force of the rope'). When you take a free body diagram of the box with its contents, the normal force between the box and its contents is internal to the system, and does NOT enter into your equation. So the the net force acting on the box/object system is T - (total weight of box and object). If you wanted to calculate the forces on the object within the box, then and only then do you look at the normal force on the object, and the net force acting on the object within the box is then N - the object's weight . ( and the tension force does not enter into that latter equation when you're looking at the forces on the object alone). I don't know if I have made this clear enough for you.
     
  4. Oct 8, 2008 #3
    its just that for a simple problem of that nature, in the student solutions manual it made use of the normal force for the solution.
     
  5. Oct 9, 2008 #4

    PhanthomJay

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    If you are just looking for the rope tension, and you know the acceleration, or whether or not there even is an aceleration, you don't need it. What is the exact wording of the problem?? Was the acceleration given??
     
  6. Oct 9, 2008 #5

    Redbelly98

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    Is the box resting on a surface, or suspended in mid-air? The normal force between the surface and box would be included in the calculations, since it is a force acting on the box (or box+contents).

    There could also be friction present.
     
  7. Oct 10, 2008 #6
    The problem is this:
    a 0.250 kg block of cheese lies on the floor of a 900 kg elevator cab that is being pulled upward by a cable through distance d1 = 2.40 m and then through distance d2 = 10.5 m. (a) Through d1, if the normal force on the block from the floor has constant magnitude FN = 3.00 N, how much work is done on the cab by the force from the cable? (b) Through d2, if the work done on the cab by the (constant) force from the cable is 92.61 kJ, what is the magnitude of FN?

    Do you count the normal force in your calculations?
     
  8. Oct 10, 2008 #7

    Redbelly98

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    Yes, when you're calculating the net force on the cab.
    No, when you're calculating the work done on the cab by the cable.
     
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