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Homework Help: Vectors with appl of dot product

  1. Oct 10, 2006 #1
    I'm trying to solve this problem with different approaches. Link is here

    http://img138.imageshack.us/img138/2268/problem1eb1.th.png [Broken]
    the answer is 0i + 0.0927j + 0.0232 k lb


    * Means DOT

    Method one: Find the force in scalar components. Then Take the cross product of two position vectors to get a normal vector (V) to the surface. THen take the unit vector of V. Then F(normal) = (V*F)V
    THis method worked F(normal)= 0i + 0.0927j + 0.0232 k lb


    Method 2: Find the force in scalar components. Draw a perpendicular line to the surface and by similar triangles the normal vector V is 8j + 2k. Take the units vector of V. Then F(normal) = (V*F)V
    THis method worked F(normal)= 0i + 0.0927j + 0.0232 k lb

    Method 3: This one i tried taking a position vector along the surface like from point A to point J. Position vector AJ = 0i-1j+4k. THen getting of units vector, unit vector eAJ = 0i - 0.2425j + 0.9701k.

    THen F(parallel)= (eAJ*F)eAJ

    F(normal) = F- F(parallel)
    F(normal) = -0.1231 i + 0.0927j + 0.0232 k lb

    THe only difference is the x-component, which differs a little. Is this answer still right? I didn;t round any numbers when calculating.
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 13, 2006 #2

    berkeman

    User Avatar

    Staff: Mentor

    The normal to that surface would by definition have no x-component, so the Method 3 answer is wrong. I'm not understanding what you are trying to do in that method, so I'm not able to figure out what is wrong.
     
  4. Nov 5, 2006 #3
    Almost forgot about this.

    Here's the work that I did for method 3.
     

    Attached Files:

  5. Nov 7, 2006 #4
    The manner in which you obtain the F (parallel) is incorrect. The direction of F (parallel) is not simply AJ just because AJ lies on the surface of the slope.

    To visualize F (parallel), try imagining the shadow of the string on the sloped surface if a light source is at some point above (though not exactly) the string.
     
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