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Expert please help me solve this question,

  1. Aug 5, 2006 #1
    Expert please help me solve this question, urgent..

    [​IMG]

    My steps are stucked as below

    3 T cos @ = 400, T = tension of each spring, @ = angle of spring with vertical axis
    L sin @ = 0.5
    L cos @ = h
    tan @ = 0.5/h

    with L = length of spring

    T = 1000 times x where x is the stretch length

    3000x cos @ = 400
    x = L - 0.5
    3000 (L-0.5) cos @ = 400
    3000 (h) - 1500 cos @ = 400
    3000 (h) = 400 + 1500 cos @
    h = (400 + 1500 cos @) / 3000

    and the I don't know how...
     
    Last edited: Aug 5, 2006
  2. jcsd
  3. Aug 5, 2006 #2

    Astronuc

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    Staff: Mentor

    You are using the right approach. The Tension, T, in the spring is given by the product of the spring constant, k (= 1000 N/m) and the displacement from equilibrium, x, where x = stretched length - unstreteched length = L - 0.5 m.

    But if one has two variables, then one needs two equations to solve for the variables.

    Edit: deleted ref to lateral forces.
     
    Last edited: Aug 5, 2006
  4. Aug 5, 2006 #3
    I'm sorry but can I have the solution to get the final value for the question?
     
  5. Aug 5, 2006 #4

    Astronuc

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    Staff: Mentor

    One has both a geometric relationship, e.g. L2 = (0.5 m)2 + h2, or tan @ = 0.5/h, or h = L cos @.

    And one has the force equilibrium equation in the vertical direction, as you indicated. T in the spring is simply T = k (L-0.5).

    See where that get's you.
     
  6. Aug 5, 2006 #5
    I tried.. but I'm not able to solve the question because I couldn't have the value of @...
     
  7. Aug 5, 2006 #6

    Astronuc

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    Staff: Mentor

    OK, let's try again.

    One has a force equilibrium equation (relationship) in which there are two unknowns, h and @. So one needs another independent relationship between h and @.

    So looking at the force equilibrium in the vertical direction, one has

    3 Ty = mg, or the vertical force components of the springs = weight of the ball.

    Now, mg = 400 N, the ball's weight (it's mass would be about 40.8 kg), and T = k (L-0.5), where k = 1000 N/m and L is a function of h and @, and Ty = T cos @.

    So the force equation is:

    3 [1000 N/m *(L-0.5)] cos @ = 400 N, but

    cos @ = h/L, so

    3 [ 1000 N/m *(L-0.5)] h/L = 400 N

    I'll leave the algebra for one to finish.

    But from the Pythagorean theorem - L2 = h2 + 0.52 m2, or

    [tex]L\,=\,\sqrt{h^2\,+\,0.5^2}[/tex].

    So one then has an equation for h in terms of the known dimension, weight of ball, and spring constant.

    Using lateral forces would not help since one simply shows that Tx = Tx, or T sin@ = T sin@.
     
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