Replace the force system acting on the frame by a resultant force and couple moment at point A.
Take F_1 = 8kN , F_2 = 4kN , F_3 = 3kN.
Determine the magnitude of the resultant force.
Determine the angle between the resultant force and the x axis.
Determine the moment of the resultant force about the point A.
Set origin and positive directions
origin at A
Up and right are positive directions
ccw moment is positive
Fr = resultant force at A
Sum forces in the x direction
Frx = 8(4/5) – 3(5/13)
Frx = 5.246
Sum forces in the y direction
Fry = – F1(3/5) – F2 – F3(12/13)
Fry = – 8(3/5) – 4 – 3(12/13)
Fry = – 11.569
Fr = √(5.246² + 11.569²)
Fr = 12.7 kN
direction
tanθ = -11.569 / 5.246
θ = -65.6°
Sum moments about A
M = F1(3/5)(4) – F1(4/5)(5) – F2(1) – F3(12/13)(2) + F3(5/13)(5)
M = 8(3/5)(4) – 8(4/5)(5) – 4(1) – 3(12/13)(2) + 3(5/13)(5)
M = 96/5 – 160/5 – 4 – 72/13 + 75/13
M = -16.569 kN•m
M = 16.569 kN•m clockwise
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