Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 decade ago

Three forces?

Three forces in the x-y plane act on a 8.10 kg mass: 11.70 N directed at 72o, 8.80 N directed at 131o, and 6.20 N directed at 211o. All angles are measured from the positive x-axis, with positive angles in the Counter-Clockwise direction.Calculate the direction of the resultant force using the same sign convention as above (in degrees).

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  • JSAM
    Lv 5
    1 decade ago
    Favorite Answer

    You need to break up all the forces into x and y-components and sum them up to get the resultant force (F_r).

    Force 1 = 11.7 N, angle 72

    x-component ---> 11.7*cos(72) = 3.62 N

    y-component ---> 11.7*sin(72) = 11.13 N

    Force 2 = 8.8 N, angle 131

    x-component ---> 8.8*cos(131) = -5.78 N

    y-component ---> 8.8*sin(131) = 6.64 N

    Force 3 = 6.2 N, angle 211

    x-component ---> 6.2*cos(211) = -5.31 N

    y-component ---> 6.2*sin(211) = -3.19 N

    Thus:

    F_rx = F_1x + F_2x + F_3x = 3.62-5.78-5.31 = -7.47 N

    F_ry = F_1y + F_2y + F_3y = 11.13+6.64-3.19 = 14.58 N

    Now find the magnitude and angle to find the resultant force:

    |F_r| = sqrt[F_rx^2 + F_ry^2] = sqrt[(-7.47)^2+(14.58)^2] = 16.38 N

    angle(F_r) = arctan(14.58/7.47) = 62.87 deg

    Since the resultant force has a negative x-component and positive y-component, the force lies in Quadrant II. Thus, to get the exact angle using your notation:

    angle = 180-62.87 = 117.13

    Thus: resultant force = 16.38 N, angle 117.13

    ------------

    Hope this helps

  • cfpops
    Lv 5
    1 decade ago

    The approach to this type of problem is to calculate the x and y components of each of separate forces, then sum all the x components to get a resultant x; then sum all of the y components to get a resultant y. Knowing x, y of the resultant force will allow you to calculate the hypoteneuse (magnitude of resulting force) and the angle with respect to the x-axis.

    F1 = 11.7 N at 72 deg

    F2 = 9.8 N at 131 deg

    F3 = 6.2 N at 211 deg

    F1x = -11.7/cos(72) = -37.86 N

    F1y = -11.7/sin(72) = -11.17 N

    F2x: 131 deg is 180-131 = 49 deg measured from -x axis

    F2x = +8.8/cos(49) = +5.77 N

    F2y = -8.8/sin(49) = -6.64 N

    F3x: 211 deg is 211-180 = 31 deg measured from -x axis

    F3x = +6.2/cos(31) = +5.31 N

    F3y = +6.2/sin(31) = +3.19 N

    Frx = F1x + F2x + F3x = -37.86 + 5.77 + 5.31 = -26.78 N

    Fry = F1y + F2y + F3y = -11.17 - 6.64 + 3.19 = -14.62 N

    Fa1: Resulting force angle with respect to -x axis:

    Fa1 = acrtan(14.62/26.78) = 28.63 deg

    Fa: Resulting force angle with respect to +x axis:

    Fa = 180 + 28.63 = 208.63 degrees

    Fr: Resulting force magnitude

    Fr = 14.62/sin(28.63) = 14.62/0.479

    Fr = 30.51 N

    Resulting force is 30.51 N away from the mass at an angle of 208.63 degrees from the positive x-axis.

    Hope this is correct, and hope it helps!

  • 1 decade ago

    Take the x-component of each force projected to the x-axis as Force(newtons) times cosine of the angle and add them together. Take the y-component as force times sine of the angle and add together the resultant vector [x,y] will have a magnitude of sqrt(x^2 + y^2) and angle arctan(y/x).

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