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Statics: Equilibrant The condition of equilibrium How to solve Example Whiteboards (Demo: Force scales, masses)

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Presentation on theme: "Statics: Equilibrant The condition of equilibrium How to solve Example Whiteboards (Demo: Force scales, masses)"— Presentation transcript:

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2 Statics: Equilibrant The condition of equilibrium How to solve Example Whiteboards (Demo: Force scales, masses)

3 Statics – acceleration = 0 Force Equilibrium - = 0 F1F1 F2F2 F3F3 Adding the three forces tip to tail: They add to zero

4 How to solve: Net force in the x dir. = 0 Net force in the y dir. = 0 Step By Step: 1.Draw Picture 2.Calculate weights 3.Express/calculate components 4.Set up a = 0 equation for x and another for the y direction 5.Do math.

5 Find F, and  such that the system will be in equilibrium (This force is called the equilibrant) W A = 23 N B = 14 N F 29 o 56 o y x  Example: x y A B F Sum0 0

6 W A = 23 N B = 14 N F 29 o 56 o y x  Example: x y A20.12 11.15 B-7.83 11.61 F Sum0 0 Find F, and  such that the system will be in equilibrium (This force is called the equilibrant)

7 W A = 23 N B = 14 N F 29 o 56 o y x  Example: x y A20.12 11.15 B-7.83 11.61 F-12.29 -22.76 Sum0 0 Find F, and  such that the system will be in equilibrium (This force is called the equilibrant)

8 W A = 23 N B = 14 N F 29 o 56 o y x  Example: x y A20.12 11.15 B-7.83 11.61 F-12.29 -22.76 Sum0 0 -22.76 -12.29 Mag = √(22.76 2 +12.29 2 ) ≈ 26 N  = Atan(22.76/12.29) ≈ 62 o Trig angle = 180+62 = 242 o 

9 Whiteboards: Equilibrant 1 | 2 TOC

10 Find the equilibrant for the forces indicated. Express as a magnitude and a trig angle 22.3 N at 64.5 o W A = 15.0 N B = 35.0 N 23.0 o 42.0 o y x x y A13.81 5.86 B-23.42 -26.01 Equil.9.61 20.15 Sum0 0 Mag = √(9.61 2 +20.15 2 ) ≈ 22.32 N  = Atan(20.15/9.61) ≈ 64.50 o 

11 Find the equilibrant for the forces indicated. Express as a magnitude and a trig angle 19.6 N at 24.5 o W A = 18.0 N B = 29.0 N 17.0 o 28.0 o y x x y A5.26 17.21 B-25.61 -13.61 C2.49 -11.74 Equil17.85 8.14 Sum0 0 Mag = √(17.85 2 +8.14 2 ) ≈ 19.6 N  = Atan(8.14/17.85) ≈ 24.5 o C = 12.0 N 12.0 o 


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