Please show work please print clear Find the resultant force

Please show work please print clear.

Find the resultant force(magnitude and angle) from adding F1+F2+F3 where F1=140 Ibs at 60 degrees F2=150Ibs at 110 degrees F3=120 Ibs at 280 degrees 75 degrees 79 Ibs 169 Ibs 38 degrees 43 degrees 149 lbs

Solution

The x & y components of the given forces are x₁ = 140cos 60, x₂ = 150cos 110 and x₃ = 120cos 280 ;

Y₁   = 140sin 60, y₂ = 150sin 110 and y₃ = 120sin 280. The x and y components of the resultant force R are The x and y components of the resultant force R are as under.

x = x₁ + x₂ + x₃ = 140cos 60 + 150cos 110 + 120cos 280 = 140 (0.5)+ 150(- 0.34) + 120 (0.17) = 70 – 51 + 20.4 = 39.4

Y = y₁ + y₂ + y₃ = 140sin 60 + 150sin 110 + 120 sin280 = 140(0.87) + 150(0.94) + 120 (- 0.98) = 121.8 + 141 – 117.6 = 145.2.

Therefore, R =   (x² + y²) = [ ( 39.4)² + (145.2)²] = ( 1552.36 + 21083.04) = 22635.4 = 150.45 (approximately). Also if is the angle of R, then

tan = y/x = 145.2/ 39.4 = 3.685. Therefore,      = tan^-1 3.685 = 74.82 degrees. The magnitude and the angle of the resultant force are 150.45 and 74.82 degrees respectively. However, if we have to choose, oone of the given options, then magnitude= 149 lbs and angle = 75 degrees will be chosen.

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