Consider a square object lying motionless in a horizontal plane. Three forces ar
ID: 2194976 • Letter: C
Question
Consider a square object lying motionless in a horizontal plane. Three forces are applied to this square. The first two forces are of equal magnitude but pointing in different directions and applied at different points: Force 1 is applied to the northwest corner and pointing west. Force 2 is applied to the southwest corner and pointing south. Is it possible to apply one final force vector in order to preserve static equilibrium? If yes, specify the vector's magnitude and direction, as well as where it is to be applied. If it is not possible, explain why not.Explanation / Answer
+x = right
+y = up
Sum of the forces in the x direction = 0 = F1x + F2x + F3x
F1x = -27.5 N * sin(30)
F2x = 11.2 N
F3x = -F1x - F2x = 27.5 N * sin(30) - 11.2 N = 2.55
Sum of the forces in the y direction = 0 = F1y + F2y + F3y
F1y = 27.5 N * cos(30)
F2y = 0
F3y = -F1y - F2y = -23.8 N
|F| = sqrt(F3x^2 + F3y^2) = 24.0 N
Direction = atan(F3y/F3x) = -83.9 degrees measure from the +x axis, so -96.1 measured from the -x axis.