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A few statics problems

  1. Feb 8, 2004 #1
    ive been wrestling with this for almost an hour and dont even know where to begin.

    A)The .732M X 1.2M lid of ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. if the tension in the cord is 54, determine the moment about each axis of the force exerted by the cord at D. the lid is opened .132 m along DC.
    here are the coordinates of each point
    B(1.2 i)
    C(1.2 i + .132j + .732 k)
    D(.732 k)
    E(.36 i + .132 j +.852 j)

    B)to lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. knowing that the man applies a 200 lb force to end A of the rope and that the moment about of the force about the y-axis is 175 ft-lb, find the distance a.

    the force is going along AB. in components it is -6i + 16j + Xk. the distance, a, is the z coordinate of point B.
  2. jcsd
  3. Feb 8, 2004 #2


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    First you have a copying error (or the lid is horribly warped!): D must be at 0i+ .132j+ 0.7k The hook, E, is exactly between and above C and D. To find the moment of the force about each axis:
    About the x-axis. Find the distance from the x-axis to D (that is is simply √((.132)2+(0.73)2)= .732 m, the length of that side of the lid, of course. Now find the component of the tension perpendicular to the lid. You might do that (I don't know what mathematics you have to work with) by finding the vector components of the tension force (It has magnitude 54 and is parallel to (.36-0)i+(.132-.132)j+(.852- .7)k= .36i+ .152k), determine the line perpendicular to the lid at D (0i-.7j+.132k), and find the projection of the force onto that line.
    About the y-axis. Since D is directly above the y-axis, the moment of the force about that axis is obviously 0.
    About the z-axis. The distance from the z-axis to D is just the y-component: 0.132. Project the tension force onto a vector perpendicular to the lid: that should be given by 0i+ j+ k.

    In problem 2, you say "the force is going along AB. in components it is -6i + 16j + Xk. the distance, a, is the z coordinate of point B." I take it that "Xk" should be "ak".
    The distance from the y-axis to point B is √(62+a2) and the force perpendicular to that is 200 cos(θ) where θ is the angle made by the perpendicular with the vertical: but cos(θ)= a/√(a2+ 62) so the moment is just 200(a/√(a2+ 62))(√(62+a2))= 200a= 175. a= 175/200= 7/8 m.
  4. Feb 8, 2004 #3
    yeah, youre right about point D. but point E is .36 m from the point A and .84 m from point B.
  5. Feb 8, 2004 #4
    figured this one out.
    Last edited: Feb 8, 2004
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