Fr = F1 + F2 + F3 + F4 = 0
100N[37o] + 200N[270] + 250N[200] + F4=0
X = 100*Cos37+200*Cos270+250*Cos200 = -155 N.
Y = 100*sin37+200*sin270+250*sin200 =
-225 N.
Tan Ar = Y/X = -225/-155 = 1.45161
Ar = 55.44o = Reference angle.
A = 55.44 + 180 = = 235.4o
F1+F2+F3 = Y/sin A = -225/sin235.4 = 273.3 N.[235.4o] = -155 - 225i.
Fr = -155 - 225i + F4 = 0
F4 = 155 + 225i = 273.3 N[55.44o] = The
equilibrant.
Note: The equilibrant is equal in magnitude but opposite in direction from
the sum of F1,F2, and F3.
Three children are fighting over a toy. One child pulls on the toy with a force of 100N at an angle of 37° north of east. The second child pulls on the toy with a force of 200N south. The third child pulls with a force of 250N at an angle of 20° south of west. Watching all of this from nearby, you decide that you have had enough of the bickering, and so you go over and take hold of the toy in order to prevent the children from dragging it (and themselves) away. What equilibrant (magnitude and angle) should you apply to the toy to prevent it, the children, and you from being pulled away? Please help. I really don't understand this problem!
1 answer
1 answer