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Measuring Vectors

  1. Dec 17, 2014 #1
    vector.jpg
    1. The problem statement, all variables and given/known data
    This is an example problem in my calculus textbook. I don't get how they relate the position vector to the force vector. I have taken a calculus based physic course and I don't remember establishing such relationship between the position and the force vectors.

    Note: I have attached a picture of the situation.

    A television camera weighing 120 pounds is supported by a tripod, as shown in Figure 11.23. Represent the force exerted on each leg of the tripod as a vector.

    2. Relevant equations
    u=cv
    v + u= v1 + u1, v2 + u2, v3 + u3
    cv=cv1, cv2, cv3

    3. The attempt at a solution
    Let the vectors and represent the forces exerted on the three legs. From Figure 11.23, you can determine the directions of and to be as follows:


    PQ1= <0,-1,-4>
    PQ2= <√3/2, 1/2,-4>
    PQ3= <-√3/2, 1/2,-4>


    Because each leg has the same length, and the total force is distributed equally among the three legs, you know that the magnitude of F1=F2=F3. So, there exists a constant such that

    F1= c<0,-1,-4>=F2= c<√3/2, 1/2,-4>=F3= c<-√3/2, 1/2,-4>
    Let the total force exerted by the object be given by F= <0,0,-120> Then, using the fact that

    F= F1 + F2 + F3


    you can conclude that F1, F2, and F3 all have a vertical component of -40. This implies that c(-4)= -40 and c is equal to 10.
    Therefore, the forces exerted on the legs can be represented by

    F1= <0,- 10, -40>
    F2= <5√3, 5, -40>
    F3= <-5√3, 5, -40>
     
    Last edited by a moderator: Dec 18, 2014
  2. jcsd
  3. Dec 17, 2014 #2

    Orodruin

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    The assumption is that the legs only take up forces in the longitudinal direction and therefore must take a force in the direction they are pointing.
     
  4. Dec 17, 2014 #3
    I'm sorry, but I am still not very clear on this.
     
  5. Dec 17, 2014 #4

    Orodruin

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    It is an assumption (which will tend to be well founded). If the legs only take forces parallel to themselves, then the force in a leg must be proportional to any vector which is parallel to the leg. In particular, the difference vector between the camera position and where a leg stands is parallel to the leg.
     
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