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Reaction force to tension

  1. Oct 11, 2006 #1

    fro

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    The problem involves drawing a free-body diagram:

    A book lies on a flat table. It is given that the weight of the book is 20N. I assume that since the book is not moving, the Fn (normal force) should also be 20N. The next part of the problem states that a string tied to the middle of the book pulls the book upward with a force of 15N. The question asks for the reaction forces in this case. I am confused to what the reaction force to the tension force of 15N. Any ideas?
     
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  3. Oct 11, 2006 #2

    OlderDan

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    As stated, the problem involves only an understanding of action and reaction forces. Newton III tells us that whenever one object exerts a force on another object, the second object exerts an equal and opposit force on the first object. This is called an action-reaction pair. It makes no difference which force is considered the action, and which is the reaction because you can never have one force without the other.

    String exerts 15N force on book, so _____ exerts ___N force on _______
     
  4. Oct 11, 2006 #3

    fro

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    So would the net force be 15N, upwards?
     
    Last edited: Oct 11, 2006
  5. Oct 11, 2006 #4

    OlderDan

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    The idea of a net force applies to the sum of all forces acting on one object. An action force and its partner the reaction force NEVER act on the same object. They can NEVER cancel.

    The book does not move because the sum of the forces actin on it are zero. There are three forces acting on the book. What are they?
    There are two forces acting on the string. What are they?
    The only way you can put a string under tension (assuming it is massless) is to apply at least two foces to the string. One force acting on a string can never put it under tension.
     
  6. Oct 11, 2006 #5

    fro

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    Really confused here.

    The three forces working on the book are the Fn (normal force) = 20N, Fg (gravitational force) = -20N, and Ft (tensional force) = 15N. So should not the net force be 15N, upwards?
     
  7. Oct 11, 2006 #6

    OlderDan

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    The normal force was 20N before you started pulling on the string. The book is not being pulled hard enough to lift it, so it is still in the same place and the sum of the forces has to be zero. Does the weight change? Does the normal force change?
     
  8. Oct 16, 2006 #7
    I'm confused as well.

    First of all...
    Are the 2 forces acting on the string be tention force upwards and tention force downwards?

    And the 3 forces acting on the book be: W(force downwards), normal force and tention force upwards!?
     
  9. Oct 16, 2006 #8
    Consider Newton's First Law. Since the body is not moving, the net force should be zero.
    The total force acting downward is only contributed by the weight of the book. The total force acting upward is the reaction force and the tension.
     
  10. Oct 16, 2006 #9

    OlderDan

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    A string will be under tension only if being pulled at both ends. The two forces acting on the string are the agent (you perhaps) pulling up on the string, and the book pulling down on the string. By Newton's third law, those two forces have associated equal and opposite reaction forces. The two forces acting on the string are equal and opposite, unless the string has mass. The string pulls down on the agent with a force equal to the force applied by the agent, and the string pulls up on the book with a force equal to the force the book exerts downward on the string. We call the forces exerted by the string the tension forces. Tension pulls up on the book, and tension pulls down on the agent. The book and the agent exert forces equal in magnitude to the tension in the string.

    If you look inside the string at one tiny piece of the string, it is being pulled in both directions by the pieces of string next to it. Those two forces are tension forces because they are being supplied by the string. Tension forces are exerted on every little piece of string. If one piece of string is weaker than the others, and the tension is increased by pulling harder at the ends, the weak point is where the string breaks.

    If the string has mass, and is accelerating, then every little piece of string has mass and must be experiencing a net force, which means the tension on one side must be greater than the tension on the other side. In this case, there is a gradual change in the tension throughout the string. The difference in tension from one end to the other would be the mass of the string times the acceleration. This is why in most problems you are told to assume the string has no mass.

    Your last statement is correct.
     
    Last edited: Oct 16, 2006
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