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2 Forces problems (Halliday/Resnick/Walker, 7th Ed. book)

  1. Feb 25, 2007 #1
    1. The problem statement, all variables and given/known data
    In chapter 5, problems 26:
    A provided figure shows an overhead view of a 0.0250 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F_1 has a magnitude of 6.00 N and is at θ_1 = 30.0° (and, btw, is in quadrant II). Force F_2 has a magnitude of 7.00 N and is at θ_2 = 30.0° (in quadrant IV). In unit-vector notation, what is the third force if the lemon half (a) is stationary, (b) has constant velocity v= (13.0 i- 14.0 j) m/s, and (c) has varying velocity v= (13.0t i- 14.0t j) m/s^2, where t is time?

    2. Relevant equations
    F_net = m*a
    F_net = F_1 + F_2 + F_3
    F_3 = ?
    F_3 = F_net- F_1 - F_2
    needed: F_3x = m*a_x - F_1*cosθ_1 - F_2sinθ_2
    F_3y = m*a_y - F_1*sinθ_1 - F_2cosθ_2

    3. The attempt at a solution
    (a) acceleration = 0, therefore:
    F_3 = 9.00 N i- 9.06 N j

    (b) acceleration is still zero. so i assume part (a) = part (b)

    F_net= m*a
    m = 0.0250 kg
    a = (sqrt) [13.0^2 + (-14.0)^2] = 19.1 m/s^2
    F_net=0.4776 N
    F_3= -8.37 N i- 9.41 N j

    1. The problem statement, all variables and given/known data
    (Second problem) Chapter 5, problem 36:
    An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of 2000 kg. When that occupant drops a coin, its acceleration relative to the cab is 8.00 m/s^2 downward. What is the tension in the cable?

    2. Relevant equations
    I used:
    T= mg-ma

    3. The attempt at a solution
    T= 2000 kg (9.80 m/s^2)- 2000 kg (8.00 m/s^2)
    = 3600 N

    ---please HELP!:surprised :eek: :bugeye: :cry:
  2. jcsd
  3. Feb 25, 2007 #2
    I can't really visualize your first problem from your description, but from what I picked up I think you did it right.

    Did you derive your tension equation from Newton's second law? What does galilean relativity say about acceleration?
  4. Feb 26, 2007 #3
    Thank you Mindscrape.

    Yes the tension equation is from Newton's second law. I had to research the Galilean relativity....ermmm. I'll just ask my professor later. Thank you though!
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