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Statics: Force Equilibrium

Statics: Force Equilibrium. The condition of equilibrium How to solve Example Whiteboards (Demo: Force scales, masses). F 2. F 3. F 1. Statics – acceleration = 0 Force Equilibrium - <The sum of all forces > = 0. Adding the three forces tip to tail:. They add to zero. How to solve:

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Statics: Force Equilibrium

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

  2. F2 F3 F1 Statics – acceleration = 0 Force Equilibrium - <The sum of all forces > = 0 Adding the three forces tip to tail: They add to zero

  3. How to solve: • Net force in the x dir. = 0 • Net force in the y dir. = 0 • Step By Step: • Draw Picture • Calculate weights • Express/calculate components • Set up a <sum of all forces> = 0 equation for x and another for the y direction • Do math.

  4. In the y direction: The weight of the mass is down (-): wt = mg = (5.0 kg)(9.8 N/kg) = -49 N Find the tension in the lines: Since the tensions are equal, they both have identical upward (+) components: Tsin(18o) + Tsin(18o) = 2Tsin(18o) T T 18o 5.0 kg • Example 1 - One unknown • Break forces in to components • Express the unknown forces as components • Set up a <sum of all forces> = 0 equation for x and another for the y direction • Do math. And now our vertical expression becomes: -49 N + 2Tsin(18o) = 0 which is solvable: T = (49 N)/(2sin(18o)) = 79.28 N (getting a car un-stuck)

  5. Whiteboards: Force Equilibrium - one unknown 1A | 1B | 1C | 2A | 2B | 2C TOC

  6. y Fx Step 1 - Set up your horizontal equation. Call the horizontal component of the unknown force Fx, Solve for Fx B = 14 N A = 23 N 56o 29o x  Find F, and  F Ax = (23 N)cos(29o) = 20.12 N x, Bx = -(14 N)cos(56o) = -7.83 N X direction: 20.12 N - 7.83 N + Fx = 0, Fx = -12.29 N W 20.12 N - 7.83 N + Fx = 0, Fx = -12.29 N

  7. y Step 2 - Set up your vertical equation. Call the vertical component of the unknown force Fy , Solve for Fy B = 14 N A = 23 N 56o 29o x  Find F, and  F Ay = (23 N)sin(29o) = 11.15 N , By = (14 N)sin(56o) = 11.61 N Y direction: 11.15 N + 11.61 N + Fy = 0, Fy = -22.76 N W 11.15 N + 11.61 N + Fy = 0, Fy = -22.76 N

  8. y B = 14 N A = 23 N 56o 29o x  Find F, and  F Step 3 - Solve for F, and  Fx = -12.29 N Fy = -22.76 N Magnitude = ((-12.29 N)2 + (-22.76 N)2) = 25.9 N = 26 N Angle = Tan-1(22.76/12.29) = 62o W 26 N, 62o

  9. Step 1 - Set up the horizontal equation, using T as the tension in the cables: T T 25.0o 25.0o Find the horizontal force acting on the walls 15.0 kg The two cables have identical horizontal components Tcos(25o) that cancel out, one to the right, and one to the left: Tcos(25o) - Tcos(25o) = 0 W Tcos(25o) - Tcos(25o) = 0

  10. Step 2 - Set up the vertical equation, using T as the tension in the cables: T T 25.0o 25.0o Find the horizontal force acting on the walls 15.0 kg Vertical: The weight of the 15 kg mass is down (-) (15 kg)(9.8 N/kg) = 147 N. Since the tensions are equal, they both have identical upward (+) components: Tsin(25o) + Tsin(25o) = 2Tsin(25o) 2Tsin(25o) - 147 N = 0 W 2Tsin(25o) - 147 N = 0

  11. T T 25.0o 25.0o Find the horizontal force acting on the walls 15.0 kg Step 3 - Solve for the answer 2Tsin(25o) - 147 N = 0, so T = (147 N)/(2Tsin(25o)) = 173.92 N The wall experiences the horizontal component of this:Tcos(25o) = (173.92 N)cos(25o) = 157.62 N = 158 N W 158 N

  12. In the y direction: T1 = (12.5 kg)(9.8 N/kg) = 122.5 N (down) Find the tensions T1, T2, and T3 T2 has an upward component: T2 sin(40o) T2 T3 T3 also has an upward component: T3 sin(20o) 40o 20o T1 12.5 kg • Step By Step: - Two unknowns • Break all forces into components • Express the unknown forces as components • Set up a <sum of all forces> = 0 equation for x and another for the y direction • Do math. So our expression becomes: T2 sin(40o) + T3 sin(20o) - 122.5 N = 0 (Making up positive)

  13. In the x direction: T2 has an leftward (-) component: T2 cos(40o) Find the tensions T1, T2, and T3 T3 has an rightward (+) component: T3 cos(20o) T2 T3 40o 20o T1 12.5 kg • Step By Step: • Take all the given forces and break them into components • Express the unknown forces as components • Set up a <sum of all forces> = 0 equation for x and another for the y direction • Do math. So our expression becomes: T3 cos(20o) - T2 cos(40o) = 0 (Making right positive)

  14. Now it’s MATH time!!!!! • Two equations, two unknowns: • T3 cos(20o) - T2 cos(40o) = 0 • T2 sin(40o) + T3 sin(20o) - 122.5 N = 0 • Re-Write them like this: • -cos(40o) T2 + cos(20o) T3 = 0 • sin(40o) T2 + sin(20o) T3 = 122.5 N • Matrices: • A B • [-cos(40o) , cos(20o)] [T2] = [0 ] • [ sin(40o) ,sin(20o)] [T3] = [122.5 N] • The answer is [A]-1[B]

  15. Or you can substitute Two equations, two unknowns: T3 cos(20o) - T2 cos(40o) = 0 and T2 sin(40o) + T3 sin(20o) - 122.5 N = 0 From the first equation T3 cos(20o) = T2 cos(40o), then T3 = T2 cos(40o)/cos(20o) Plug this into the second equation: T2 sin(40o) + (T2 cos(40o)/cos(20o)) sin(20o) - 122.5 N = 0 Solve: T2 sin(40o) + T2 cos(40o) tan(20o) = 122.5 N T2(sin(40o) + cos(40o) tan(20o)) = 122.5 N T2 = (122.5 N) / (sin(40o) + cos(40o) tan(20o)) = 133 N and T3 = (132.92 N) cos(40o)/cos(20o) = 108 N Wow!!!

  16. Whiteboards: Two Unknowns 1 | 2 | 3 TOC

  17. y B = ? A ? 61o 31o x 81o Find A and B 34.0 N Step 1 - Set up the horizontal equation -(34.0 N)cos(81o) = -5.319 N, +Acos(31o), -Bcos(61o): -5.319 N + Acos(31o) - Bcos(61o) = 0 W -5.319 N + Acos(31o) - Bcos(61o) = 0

  18. y B = ? A ? 61o 31o x 81o Find A and B 34.0 N Step 2 - Set up the vertical equation -(34.0 N)sin(81o) = -33.581 N, +Asin(31o), +Bsin(61o): -33.581 N + Asin(31o) + Bsin(61o) = 0 W -33.581 N + Asin(31o) + Bsin(61o) = 0

  19. Step 3 - Do Math: -33.581 N + Asin(31o) + Bsin(61o) = 0 -5.319 N + Acos(31o) - Bcos(61o) = 0 Substitution: -33.581 N + Asin(31o) + Bsin(61o) = 0, A = (33.581 N-Bsin(61o))/sin(31o) -5.319 N + Acos(31o) - Bcos(61o) = 0, substituting: -5.319 N + {(33.581 N-Bsin(61o))/sin(31o)}cos(31o) - Bcos(61o) = 0 -5.319 N + (33.581 N)/tan(31o) - Bsin(61o)/tan(31o) - Bcos(61o) = 0 B = 26.061 = 26 N, A = 21 N • Matrices: • Asin(31o) + Bsin(61o) = 33.581 N • Acos(31o) - Bcos(61o) = 5.319 N • J K • [sin(31o) , sin(61o)] [A] = [33.581 N] • [cos(31o) ,cos(61o)] [B] = [5.319 N ] • Answer matrix will be [J]-1[K] W B = 26 N, A = 21 N

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