Question

Replace the force system shown in Figure 6 with a single force and couple system acting at point B.

Answer 1

`(In DeltaAGF)  b=tan^(-1)  (GD)/(DA) =  tan ^(-1)   3/4 = 36.87°`

`(In DeltaAEF) a = tan ^ (-1)  (AE)/(AF) = tan^(-1)  3/2 = 56.32°|`

`(InDeltaFHD)C = tan^(-1)   (DC)/(CH) = tan ^(-1)   6/2 =71.57° `

ΣFx = 1000cos(36.87) + 632cos(71.57) – 722cos(56.32) = 599.42
ΣFy = - 1000sin(36.87) + 632sin(71.57) – 722sin(56.32) = -601.23

`R = sqrt(Fx^2 +FY^2)`

`R = sqrt(599.42^2 +(-601.23)^2)`

`R = 848.99 N`

`Theta = tan^(-1)  ((Sigma Fy)/(SigmaFx))` 

`= tan^(-1)  ((-601.23)/(599.42))`

= -45.086⁰
`ΣM_(B^F) `= - [722cos(56.32)x3] + [1000cos(36.87)x6] – [632sin(71.57)x2]
`ΣM_(B^F)` = 2399.66 N-m

The magnitudes of the resultant force and couple at B are 848.99 N and 2399.66 N-m clockwise.