Questions: 1. Determine the magnitude, direction, and sense of the resultant for the 2 -force system shown.

Solution Steps

• Resolve the 25 lb force into its x and y components:

  • \( F_{1x} = 25 \cos(60^\circ) = 25 \times 0.5 = 12.5 \) lb
  • \( F_{1y} = 25 \sin(60^\circ) = 25 \times 0.866 = 21.65 \) lb

• Resolve the 45 lb force into its x and y components:

  • \( F_{2x} = 45 \cos(10^\circ) = 45 \times 0.985 = 44.325 \) lb
  • \( F_{2y} = 45 \sin(10^\circ) = 45 \times 0.174 = 7.83 \) lb

• Sum the x-components:

  • \( F_{Rx} = F_{1x} + F_{2x} = 12.5 + 44.325 = 56.825 \) lb

• Sum the y-components:

  • \( F_{Ry} = F_{1y} + F_{2y} = 21.65 + 7.83 = 29.48 \) lb

• Magnitude of the resultant force:

  • \( F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2} = \sqrt{56.825^2 + 29.48^2} = \sqrt{3228.36 + 869.71} = \sqrt{4098.07} = 63.99 \) lb

• Direction of the resultant force (angle with the x-axis):

  • \( \theta = \tan^{-1}\left(\frac{F_{Ry}}{F_{Rx}}\right) = \tan^{-1}\left(\frac{29.48}{56.825}\right) = \tan^{-1}(0.519) = 27.33^\circ \)

Final Answer