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

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

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Solution Steps

Step 1: Resolve the Forces into Components
  • Resolve the 25 lb force into its x and y components:

    • F1x=25cos(60)=25×0.5=12.5 F_{1x} = 25 \cos(60^\circ) = 25 \times 0.5 = 12.5 lb
    • F1y=25sin(60)=25×0.866=21.65 F_{1y} = 25 \sin(60^\circ) = 25 \times 0.866 = 21.65 lb
  • Resolve the 45 lb force into its x and y components:

    • F2x=45cos(10)=45×0.985=44.325 F_{2x} = 45 \cos(10^\circ) = 45 \times 0.985 = 44.325 lb
    • F2y=45sin(10)=45×0.174=7.83 F_{2y} = 45 \sin(10^\circ) = 45 \times 0.174 = 7.83 lb
Step 2: Sum the Components
  • Sum the x-components:

    • FRx=F1x+F2x=12.5+44.325=56.825 F_{Rx} = F_{1x} + F_{2x} = 12.5 + 44.325 = 56.825 lb
  • Sum the y-components:

    • FRy=F1y+F2y=21.65+7.83=29.48 F_{Ry} = F_{1y} + F_{2y} = 21.65 + 7.83 = 29.48 lb
Step 3: Calculate the Magnitude and Direction of the Resultant Force
  • Magnitude of the resultant force:

    • FR=FRx2+FRy2=56.8252+29.482=3228.36+869.71=4098.07=63.99 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):

    • θ=tan1(FRyFRx)=tan1(29.4856.825)=tan1(0.519)=27.33 \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

  • Magnitude: 63.99 lb
  • Direction: 27.33° above the x-axis
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