Results 1 to 3 of 3

Math Help - direction of resultant of forces (vectors & bearings)

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    2

    Question direction of resultant of forces (vectors & bearings)

    I've been given the problem: "Find the direction and magnitude of the resultant of forces of 23.75 lb acting S 56.14 degrees W and 48.61 lb acting S 33.86 degrees E.

    I have found the magnitude (54.10 lb) but I don't know how to find the direction. I know the answer is S 7.82 degrees (that's what the back of the book says), but I don't know how to get it and I don't understand the textbook's explanation.

    Can someone help me, please?
    Last edited by mhunn5260; September 29th 2008 at 09:54 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by mhunn5260 View Post
    I've been given the problem: "Find the direction and magnitude of the resultant of forces of 23.75 lb acting S 56.14 degrees W and 48.61 lb acting S 33.86 degrees E.

    I have found the magnitude (54.10 lb) but I don't know how to find the direction. I know the answer is S 7.82 degrees (that's what the back of the book says), but I don't know how to get it and I don't understand the textbook's explanation.

    Can someone help me, please?
    I don't know your textbook's explanation but it could be like this.

    You must draw the figure so that you can see approximately the position of the R.

    For the horizontal components, going West is negative and going East is positive.
    For the vertical components, going North is positive and going South is negative.

    Summation of horizontal components.
    H = -23.75sin(56.14deg) +48.61sin(33.86deg) = 7.36178 lbs ----positive, so going East

    Summation of vertical components,
    V = -23.75cos(56.14deg) -48.61cos(33.86deg) = -53.59850 lbs ----negative, so going South.

    Meaning, the resultant R is going S some degrees E.

    This "some degrees", or theta, is arctan(H/V)
    Since you can see the figure, then disregard the signs of H and V. Remember that the signs are only for directions. Since you know already that R is due SE, just use the absolute values of the H and V-----or their positive values.
    So,
    theta = arctan(7.36178 /53.59850) = 7.82067 degrees

    Therefore, the bearing of R is S 7.82deg E --------answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    2
    Quote Originally Posted by ticbol View Post
    I don't know your textbook's explanation but it could be like this.

    You must draw the figure so that you can see approximately the position of the R.

    For the horizontal components, going West is negative and going East is positive.
    For the vertical components, going North is positive and going South is negative.

    Summation of horizontal components.
    H = -23.75sin(56.14deg) +48.61sin(33.86deg) = 7.36178 lbs ----positive, so going East

    Summation of vertical components,
    V = -23.75cos(56.14deg) -48.61cos(33.86deg) = -53.59850 lbs ----negative, so going South.

    Meaning, the resultant R is going S some degrees E.

    This "some degrees", or theta, is arctan(H/V)
    Since you can see the figure, then disregard the signs of H and V. Remember that the signs are only for directions. Since you know already that R is due SE, just use the absolute values of the H and V-----or their positive values.
    So,
    theta = arctan(7.36178 /53.59850) = 7.82067 degrees

    Therefore, the bearing of R is S 7.82deg E --------answer.
    Thank you for your reply, but you are using terminology that I'm not familiar with and I don't understand your summations.

    I have drawn my visual aid to show dotted north/south lines and dotted east/west lines; drew a solid line from origin S 56.14deg W and labeled it 23.75 lbs; drew a solid line from origin S 33.86deg E and labeled it 48.61 lbs (which are at right angles); and drew my hypotenuse - which I found to be 54.10 lbs (according to the answer in the back of the book, this is correct). Then, I used tangent to find both angles inside my newly formed triangle. (I'm not sure why I did that.) I don't understand how to find the direction. Your answer, S 7.82deg E, is close to the answer in the back of the book, S 7.82deg, but you've added E where the book does not. And I don't understand what it means by direction or how to get it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Magnitude and direction of a resultant vector.
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: June 1st 2011, 04:50 AM
  2. Question about resultant forces
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 9th 2008, 03:44 PM
  3. magnitude forces/resultant/etc.
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 30th 2008, 09:41 PM
  4. mechanics question, resultant forces.
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: February 13th 2008, 12:37 PM
  5. Resultant Forces
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: December 16th 2007, 11:16 AM

Search Tags


/mathhelpforum @mathhelpforum