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Math Help - Vector problem:

  1. #1
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    Vector problem:

    If Jack and Luke are on opposite sides of a canal pulling a barge with tow ropes, Jack exerting a force of 50 pounds at 20 degrees to the canal, and Luke at angle of 15 with enough force so that the resultant force vector is along the canal directly, what is the number of pounds with which Luke must pull and the magnitude of the resultant?

    CI said 50sin(20) = (Luke's Vector Magnitude)Sin(15)
    Divide both sides by Sin(15) and got 66. Can anyone confirm this?

    And also how do I find magntiude if the vectors are connected at their tail points? I assume you just add 50cos(20) and 50cos(15)? Is this right?

    And also, if you are pulling a wagon with a force of 20 pounds on a handle that maes 30 degrees with the horizontal, what is the work if its pulled 100 feet up an incline that makes an angle of 30 degrees with the horizontal?
    Last edited by Alan306090; February 3rd 2010 at 05:13 PM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Quote Originally Posted by Alan306090 View Post
    CI said 50sin(20) = (Luke's Vector Magnitude)Sin(15)
    Divide both sides by Sin(15) and got 66. Can anyone confirm this?
    Correct - as long as the angles are measured from the edge of the canal.

    Quote Originally Posted by Alan306090 View Post
    And also how do I find magntiude if the vectors are connected at their tail points? I assume you just add 50cos(20) and 50cos(15)? Is this right?
    Not quite - you want to add the horizontal components of the two vectors. You already know that the magnitude of the 15-degree vector is 66 pounds (not 50), so you get:

    F_Total = 50 * cos(20) + 66 * cos(15)

    Quote Originally Posted by Alan306090 View Post
    And also, if you are pulling a wagon with a force of 20 pounds on a handle that maes 30 degrees with the horizontal, what is the work if its pulled 100 feet up an incline that makes an angle of 30 degrees with the horizontal?
    This one is so poorly worded that the ambiguity makes it impossible to understand teh geometry of what's going on. Given the handle is at 30 degrees to the horizontal - are we to assume that the force being applied to the cart is at 30 degrees to the horizontal? And since the cart is on a 30 degree incline to the horizontal, does that mean that the handle - and force - are actually in-line with the cart (since they're all at 30 degrees to the horizontal)? If there's a figure for this problem it would help to post it. But to give you a hint here - the work done will be the product of the force applied times the distance traveled by the cart times the cosine of the angle between the force and the direction of travel.
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