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When normal force isn't normal l

  1. Oct 30, 2013 #1
    When normal force isn't "normal"l

    1. The problem statement, all variables and given/known data
    A box with a certain mass is pulled across a horizontal surface by a rope at an angle. Figure out the components, the friction, the acceleration of the box, when the tension in the rope, its angle, the mass of the box, and mu are given.


    2. Relevant equations
    F=ma, sin (theta) hypotenuse, etc...


    3. The attempt at a solution
    I understand the solution but I'm having trouble explaining how the normal force is found. I am used to calculating the using m(g) to calculate the force of gravity. It seems as though the normal force should just be the same as gravity plus the upward component of the rope. I get that equilibrium plays a role but am just having difficulty explaining it.
     
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  3. Oct 30, 2013 #2

    ehild

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    What do you mean with the sentence "It seems as though the normal force should just be the same as gravity plus the upward component of the rope."? The rope has no "upward component". In what direction is the rope pulled?

    ehild
     
  4. Oct 30, 2013 #3

    SteamKing

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    Imagine a box of mass 'm' just sitting on the ground. What is the normal force exerted by the ground on the box?

    Now, imagine the box has a rope tied to it. Does the normal force on the box change?

    If you pull straight up on the rope so that there is a tension of 5 N and the box is still sitting on the ground, what is the normal force exerted by the ground on the box? If the tension is 10 N? When is the normal force = 0?
     
  5. Oct 30, 2013 #4

    HallsofIvy

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    If the rope makes angle [itex]\theta[/itex] with the horizontal and has tension T, then the "upward force" is [itex]Tsin(\theta)[/itex] and the horizontal force is [itex]Tcos(\theta)[/itex]. The "normal force" (normal to the surface) is the weight of the box, mg, minus that upward force: [itex]mg- Tsin(\theta)[/itex].
     
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