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Two forces P and Q pass through a point A which is 4 m to the right of and 3 m above a moment center O


Two forces P and Q pass through a point A which is 4 m to the right of and 3 m above a moment center O. Force P is 890 N directed up to the right at 30° with the horizontal and force Q is 445 N directed up to the left at 60° with the horizontal. Determine the moment of the resultant of these two forces with respect to O. 

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Answer #1

a)

Qx = -Qcos(60) = -222.5N

Qy = Qsin(60) = 385.38N

b)

Px = Pcos(30) = 770.76N

Py = Psin(30) = 445N

c)

Resultant Force F

Fx = Px + Qx = 548.26N

Fy = Py + Qy =830.38N

F = 995.4N theta = 56.56 Degress

d)

Moment M is in z direction(r X F) and anticlockwise is positive

(Mx = moment due to force Fx)

Mx = -Fx*3 = -1644.78 Nm

My = Fy*4 = 3321.52 Nm

Total Moment = Mx + My = 1676.74 Nm

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