1. Not finding help here? Sign up for a free 30min 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!

Torque and Cross Products

  1. Oct 23, 2013 #1
    First of all I want to thank you for taking your valuable time to help me out. I typically would not ask a homework question, but I have exhausted all other resources in search for an answer to my question. I will do my best to fully summarize my situation and its circumstances, seeing as this is my first post.

    I have already successfully answered questions (a)-(e). The key to (f) lies in question (b): It asks if there are multiple points that can produce a result half of the value found in question (a) and in the opposite direction. That would be a value of 9. It also dictates that you find this value as it lies on the y axis. This means (to me) that the i values will all be 0 in the cross product matrices.

    1. The problem statement, all variables and given/known data

    A particle is located at the vector position = (6.00i + 9.00j) m and a force exerted on it is given by = (4.00i + 3.00j) N.

    (a) What is the torque acting on the particle about the origin?
    -18k

    (b) Can there be another point about which the torque caused by this force on this particle will be in the opposite direction and half as large in magnitude?
    YES

    (c) Can there be more than one such point?
    YES

    (d) Can such a point lie on the y axis?
    YES

    (e) Can more than one such point lie on the y axis?
    NO

    (f) Determine the position vector of one such point. (Give a point on the y axis.)


    2. Relevant equations

    I used cross products to find the first solution which was -18k N m.


    3. The attempt at a solution

    I assumed that I would just rearrange the same cross product equation (RxFy - RyFx) (where R = radius, F = force) while substituting a variable for Ry, considering the Rx values to be equal to 0. I could not come up with a correct solution in this manner, so I tried another way.

    My logical reasoning (which was apparently flawed) was that given a particle with only a y-component, the only forces taken into consideration would be those perpendicular to that: the i forces. So I attempted the solution by stating that:
    RyFx = 9 --> 4Ry = 9 --> Ry = 2.25m

    This did not produce the correct solution either. Any information or guidance you could provide me with is much appreciated. Thank you all!
     
  2. jcsd
  3. Oct 23, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    R=(Rx,Ry) in T=RxF is a vector that points from the pivot point to the particle.
    Therefore Rx is not 0.

    If the vector from the origin to the particle is Q, and the vector from the origin to the pivot if P, then you should be able to get a better expression. Otherwise the strategy looks good. Px=0, Py=y, solve for y.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Torque and Cross Products
  1. Torque, Cross product (Replies: 4)

Loading...