Results 1 to 4 of 4

Math Help - angles and the distance between weights

  1. #1
    Senior Member
    Joined
    Jan 2007
    Posts
    477

    angles and the distance between weights

    One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of 4.00 N and is applied perpendicular to the stick at the free end. The other force has a magnitude of 6.00 N and acts at an angle θ = 68.3, measured with respect to the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the axis of rotation.

    not sure why this wouldn't work, but i know that i could take the sin68.3 * 6.00 N to get the other sides, and Cos if i needed to.

    i'm not sure how to find what i am looking for.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    You seem to be pushing the wind, rather than the stick.

    4 N is at the free end of the 1 m stick. This is "1 m".
    6 N is somewhere along the stick. Call it "x m".

    Try:

    (4 N)*(1 m) = (6 N)*(sin(68.3))*(x m)


    Thought Question:
    Does it matter which way we measure the 68.3? Is that 68.3 from the ray containing the fixed end or 68.3 from the ray containing the free end?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,100
    Thanks
    379
    Awards
    1
    Quote Originally Posted by rcmango View Post
    One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of 4.00 N and is applied perpendicular to the stick at the free end. The other force has a magnitude of 6.00 N and acts at an angle θ = 68.3, measured with respect to the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the axis of rotation.

    not sure why this wouldn't work, but i know that i could take the sin68.3 * 6.00 N to get the other sides, and Cos if i needed to.

    i'm not sure how to find what i am looking for.
    The stick is not rotating so the sum of the torques on the stick is 0 Nm rad. (Typically this unit is given as merely "Nm," but the rad should technically be in there.)

    I am going to take the axis of rotation as the physical axis (the left hand point of the stick in my diagram) and I'm going to call a counter-clockwise angle as positive.

    There are two forces we need to consider: the force F applied at the right-most point of the stick, and the force A applied somewhere on the stick at a 68.3 degree angle. (See the diagram below.)

    So
    \sum \tau _{net} = \tau _F + \tau _A = 0

    So what are the torques from each force? Well for the 4.00 N force:
    | \tau _F | = rF \cdot sin(\theta)
    where \theta is the angle between the moment arm r and force vector F when placed tail to tail. So for the force F this angle is 90 degrees. Thus
    | \tau _F | = rF \cdot sin(\theta) = (1)(4.00) \cdot sin(90^o) = 4.00~Nm

    As for the direction, what direction would this force rotate the stick about the given axis? Counter-clockwise, of course, so this is a positive torque.

    We don't know the location of the other force, so I'm simply going to call it "d" and say that it is measured from the axis. So
    | \tau _A | = dA \cdot sin(\theta)

    What is \theta in this case? It will be 68.3 degrees. (And remember you are putting the force vector A tail to tail with the moment arm vector r.) So
    | \tau _A | = dA \cdot sin(\theta) = (6.00)d \cdot sin(68.3^o) = 5.5748d~Nm

    This torque is going to be negative because the force A will make the stick rotate clockwise.

    So:
    \sum \tau _{net} = \tau _F + \tau _A = 0

    4.00 - 5.5748d = 0

    Now solve for d.

    -Dan
    Attached Thumbnails Attached Thumbnails angles and the distance between weights-torque.jpg  
    Last edited by topsquark; November 27th 2007 at 09:17 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2007
    Posts
    477
    later today, i'm going to sit down and really look this problem over closer, thanks for the help and work.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding distance between two points using two angles
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: January 18th 2011, 11:32 AM
  2. Replies: 3
    Last Post: September 5th 2010, 09:22 PM
  3. Combinations of weights
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 11th 2010, 03:52 AM
  4. Weights
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 4th 2009, 04:08 AM
  5. R-K weights?
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 26th 2009, 05:15 AM

Search Tags


/mathhelpforum @mathhelpforum