1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook