Pre Calc

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Find the work done by a force F of 30 pounds acting in the direction (2,3) in moving an object 3 feet from (0,0) to the point in the first quadrant along the line y = (1/2)x.
The answer is 85.38 foot-pounds.
I just need to figure out how to get the answer.

  • Pre Calc -

    Find the angle of the Force F
    cos T = 2/sqrt13 = .555
    T = 56.3 deg above x axis
    or
    tan T = 2/3
    so T = 56.3

    Find the direction of motion from slope of line
    tan slope = 1/2
    slope angle = 26.6 deg

    angle between force and motion = 56.3-26.6 = 29.7 degrees
    so
    component of force in direction of motion = 30 cos 29.7
    = 26.06 pounds
    26.06*3 = 78.2 ft lbs
    beats me how to get 85.38

    It would be much closer if the force vector were direction(3,2)

  • Pre Calc -

    the vector is actually <2,2>
    and the line angle is 64.3

  • Pre Calc -

    the vector is indeed <2,2>, but the line angle (angle formed by y=1/2x and x-axis) is as Damon posted, 26.6 degree. As for why it isn't 64.3 degree is because slope is change in y over the change in x, thus this means horizontal change is 2 and vertical change is 1. When finding the line angle, simply use arc tan(opp/adj) or in this case arc tan(1/2) which will result in 26.6 degree.

    As for the answer, its actually 95.36 ft lb if the vector is <2,2>

  • Pre Calc -

    F1 -> 30 lbs, direction <2, 2>
    magnitude of <2, 2> = 2sqrt(2)
    Find unit vector of F1 (divide original components by magnitude)
    -> <30/sqrt(2), 30/sqrt(2)>
    F2 -> 3 feet, direction y = 1/2x
    3 = sqrt(x^2 + y^2)
    y = 1/2x
    3 = sqrt(x^2 + (1/2x)^2)
    9 = 5/4* x^2
    x^2 = 36/5
    x = 6/sqrt(5)
    y = 2/sqrt(5)

    F1 = <30/sqrt(2), 30/sqrt(2)>
    F2 = <6/sqrt(5), 3/sqrt(5)>

    Find dot product of F1 and F2

    (30/sqrt(2))*(6/sqrt(5))+(30/sqrt(2))*(3/sqrt(5)) = 85.38

    Answer in the back of the book is 85.38

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