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Forces (Equilibrium).

  1. Nov 21, 2005 #1
    The question is this.

    "Is it possible for an object to be in a state of equilibrium when forces of 10N, 20N and 40N act upon it?"

    I'm thinking yes, because you can always have the 10 N and 20 N force acting in the same direction and the 40N acting on both at a slight angle, equalling 30. But I don't know if I'm right. This seems like a really simple question, yet I don't know how to answer it. Any help would be greatly appreciated.
  2. jcsd
  3. Nov 21, 2005 #2

    Doc Al

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    Staff: Mentor

    Think of it this way: In order to completely balance the 40 N force, what must the other two forces add up to? Is that possible?
  4. Nov 21, 2005 #3


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    Homework Helper

    Your speculation here is very thoughtful. But remember, if you apply the 40N force at an angle, then there will be some of it not in the direction of the first two forces, meaning you will need to offset one of those as well, which will mean....etc etc.

    All forces can always be resolved into their orthagonal components, so to keep it simple, work on it there. Apply the forces only in either the x or y directions.
  5. Nov 21, 2005 #4

    Ah, so I guess the answer would be no. As if I keep it simple, like making all of them act on the x direction, there is no way that the 10N and 20N force can balance out the 40N force. Is that correct?

    Oh, and thanks to the both of you.
  6. Nov 22, 2005 #5
    The maximum force that can be applied with the help of 10N and 20N in a specified direction is 30N. that is their actual sum when the angle between them is 0deg. so you can't get a force of 40N with the help of other two forces in the direction opposite to the direction at which 40N force is applied.

    when is the sum of three forces 0? AB + BC + CA = 0 in vector notations if they form a triangle. sum of two sides of a triangle can never be greater than the third side. it is the tht you get three vctors which can be properly placed such that their sum is 0.
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