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Statics question

  1. Jan 29, 2004 #1
    a lever is resting on a valve. the valve is located 3 in from the end of the lever. the total length of the lever is 9.5 in. a 4 lb force is present at the opposite end of the lever. the force makes a 28 deg angle with the lever. the lever makes a 20 deg angle with the horizontal. im supposed to find the moment about B, which is the spot where the lever comes in contact with the valve.
    the force in the vertical direction is -1.9 and the force in the horizontal direction is -3.5.
    i got these by multiplying the magnitude of the force by the cos and sin of the angle. i ended up with an answer of 12.35 lb * in.
    is that correct?
     
  2. jcsd
  3. Jan 29, 2004 #2

    Doc Al

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    Staff: Mentor

    I assume you are trying to find the torque exerted by the 4 lb force about point B. A diagram would be helpful.

    I presume when you say "vertical" and "horizontal" you really mean perpendicular and parallel to the lever, right? You only need the perpendicular component of the force. If that's what you did, then fine! (Is the torque clockwise or counter-clockwise?)

    Another way to look at it:

    To find the torque exerted by the 4 lb force about point B you can use T=rFsinθ. r is the distance from B to the point of application of the force (r = 6.5"); F is the force (F = 4 lbs), and θ is the angle between them (θ = 28 or 152). So, I get T = 12.2 lb*in. (I can't specify the direction of the torque, since I don't know how you defined your angle.)
     
  4. Jan 30, 2004 #3
    heres a pic. the lever itself makes a 20 deg angle with the horizontal. alpha equals 28 deh. was my answer of 12.35 correct?
     

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  5. Jan 30, 2004 #4

    Doc Al

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    Reread my answer. Did you make use of the 20 deg angle in your calculations?
     
  6. Jan 30, 2004 #5
    heres the question from the text:
    a foot valve for a pneumatic system is hinged at B. knowing that alhpa = 28*. determine the momeny of the 4-lb force about point B by resolving the force into components along ABC and in a direction perpendicular to ABC.


    so just doin (4lb)(sin 28) X (6.5 in) is the correct answer?
     
  7. Jan 30, 2004 #6

    Doc Al

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    Yes. Note that (4lbs)x(sin 28) = 1.88 lbs, the component of the force perpendicular to ABC. This is the same thing you found, although you called it (incorrectly) the vertical component. Make sense?
     
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