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Homework Help: Basic Vector Problem

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data
    In a plane, find the resultant of a 300-N force at [itex]30^o[/itex] and a -250-N force at [itex]90^o[/itex]. Also, find the angle [itex]\alpha[/itex] between the resultant and the y axis.

    2. Relevant equations
    [itex]x = F cos \theta[/itex]
    [itex]y = F sin \theta[/itex]
    [itex]r^2 = x^2 + y^2[/itex]
    [itex]\theta = tan^{-1} (\frac {y} {x})[/itex]

    3. The attempt at a solution
    [itex]x_1 = 300 cos 30^o = 259.80, x_2 = -250 cos 90^o = 0, x_{TOT} = 259.80[/itex]
    [itex]y_1 = 300 sin 30^o = 150, y_2 = -250 sin 90^o = -250, y_{TOT} = -100[/itex]
    [itex]r = \sqrt {259.80^2 + (-100)^2} = 278.38 \sim 280[/itex]

    [itex]\theta = tan^{-1} (\frac {259.80} {-100}) = -68.947^o \sim -69^o[/itex]
    Since this is the angle between the x-axis and resultant, then the angle [itex]\alpha[/itex] between the resultant and the y-axis must be [itex]-90 - (-68.947^o) = -21.053^o \sim -21^o[/itex]. However, the book solution for the angle [itex]\alpha[/itex] is given as [itex]69^o[/itex]. Why?

  2. jcsd
  3. Jun 16, 2015 #2


    User Avatar

    Staff: Mentor

    Nice work. I've bolded the part where you transposed the y & x components when calculating theta... :smile:
  4. Jun 21, 2015 #3
    Is the problem that you mixed up your terms in the formula for θ.
    I think the formula for θ, the angle with respect to the x axis is: tan -1( y/x).
    It looks to me like you have x and y transposed, as y=-100 and x = 259.8, so should the expression be:
    θ= tan -1(-100/259.8)?
    If i use that expression I get θ ≈ -21° so with respect to the y axis ∠ = -69.
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