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physics

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"Only two forces act on an object (mass = 3.90 kg), as in the drawing. (F = 76.0 N.) Find the magnitude and direction (relative to the x axis) of the acceleration of the object. " (The angle is 45 degrees and the horizontal force is 40N)

  • physics - ,

    Net force is F = 40 + 76•cosα =40 +76•0.707 = 93.74 N
    a =F/m =93.74/3.9 = 24 m/s^2

  • physics - ,

    I’ve found the figure to this problem, so the solution is following.
    x- and y- components of net forces are

    F1(x) = 40 N, F1(y) =0
    F2 (x) =76•cosα = 93.74 N,
    F2 (y)= 76•sin α = 93.74 N ,
    Net force
    F(x)=133.74 N,
    F(y) = 93.74 N.
    F = sqrt(133.74^2 + 93.74^2) = 163.33 N
    a =F/m =163.33 /3.9 = 41.88 m/s^2
    Net force makes an angle β with x-axis,
    tan β = 93.74/133.74 = 0.7
    The direction of the acceleration of the object is β = 35o above the x-axis

  • physics - ,

    +y| / F
    | /
    | /
    | /
    /a_ _ _ _ _ +x
    40N


    This is pretty much what the picture looks like. Your answer was incorrect!

  • physics - ,

    F1(x) = 40 N, F1(y) =0
    F2 (x) =76•cosα = 53.74 N, F2 (y)= 76•sin α = 53.74 N ,
    Net force F(x)=93.74 N, F(y) = 53.74 N.
    F = sqrt(53.74^2 + 93.74^2) = 108 N
    a =F/m =108 /3.9 = 27.7 m/s^2
    Net force makes an angle β with x-axis,
    tan β = 53.74/93.74 = 0.57
    The direction of the acceleration of the object is β = 29.8o above the x-axis

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