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Statics: Force Equilibrium

Statics: Force Equilibrium. The condition of equilibrium How to solve Example Whiteboards. 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

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Statics: Force Equilibrium

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  1. Statics: Force Equilibrium The condition of equilibrium How to solve Example Whiteboards

  2. 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.

  3. In the x direction: Tcos(18o) – Tcos(18o) = 0 Not very helpful 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 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)

  4. Whiteboards: Force Equilibrium - one unknown 1 TOC

  5. Step 1 - Set up the horizontal equation, using T as the tension in the cables: T T 25.0o 25.0o What is the tension in the two cables? 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

  6. Step 2 - Set up the vertical equation, using T as the tension in the cables: T T 25.0o 25.0o What is the tension in the two cables? 15.0 kg Vertical: The weight of the 15 kg mass is down (-) (15 kg)(9.81 N/kg) = 147.15 N. Since the tensions are equal, they both have identical upward (+) components: Tsin(25o) + Tsin(25o) = 2Tsin(25o) 2Tsin(25o) – 147.15 N = 0 W 2Tsin(25o) – 147.15 N = 0

  7. T T 25.0o 25.0o What is the tension in the two cables? 15.0 kg Step 3 - Solve for the answer 2Tsin(25o) – 147.15 N = 0, so T = (147.15 N)/(2sin(25o)) = 174.093 N ≈ 174 N W 174 N

  8. In the y direction: T1 = (12.5 kg)(9.81 N/kg) = 122.625 N (down) Find the tensions T1, T2, and T3 T2 has an upward component: T2 sin(180-40o) = T2 sin(140o) T2 T3 40o T3 also has an upward component: T3 sin(20o) 20o T1 12.5 kg So our expression becomes: T2 sin(140o) + T3 sin(20o) - 122.625 N = 0 (Making up positive)

  9. In the x direction: T2 has an leftward (-) component: T2 cos(140o) 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: T2 cos(140o) + T3 cos(20o) = 0 (One is +, the other -)

  10. Now it’s MATH time!!!!! • Two equations, two unknowns: • T2 cos(140o) + T3 cos(20o) = 0 • T2 sin(140o) + T3 sin(20o) - 122.625 N = 0 • Re-Write them like this: • cos(140o) T2 + cos(20o) T3 = 0 • sin(140o) T2 + sin(20o) T3 = 122.625 N • Matrices: • A B • [ cos(140o) , cos(20o)] [T2] = [0 ] • [ sin(140o) ,sin(20o)] [T3] = [122.625 N] • The answer is [A]-1[B] • T2 = 133.06 • T3 = 108.47 • (Demo with calculator)

  11. Or you can substitute Two equations, two unknowns: T3 cos(20o) - T2 cos(40o) = 0 and T2 sin(40o) + T3 sin(20o) - 122.625 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) - 122.625 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!!!

  12. Whiteboards: Two Unknowns 1 | 2 TOC

  13. 24.0o 62.0o P Q Trig Angles P = 62o Q = 180-24 = 156o 17.0 kg Weight of mass: (17.0 kg)(9.81 N/kg) 166.77 N Force Equations: Pcos(62o) + Qcos(156o) = 0 Psin(62o) + Qsin(156o) = 166.77 Solutions P = 152.7 N Q = 78.5 N TOC P = 152.7 N, Q = 78.5 N

  14. y Q = ? P ? 61o 31o x 81o Find P and Q 34.0 N Step 1 - Set up the horizontal equation (34.0 N)cos(180+81o) = -5.319 N, Pcos(31o), Qcos(180+61o): -5.319 N + Pcos(31o) + Qcos(180-61o) = 0 W -5.319 N + Pcos(31o) + Qcos(180-61o) = 0

  15. y Q = ? P ? 61o 31o x 81o Find P and Q 34.0 N Step 2 - Set up the vertical equation (34.0 N)sin(180+81o) = -33.581 N, Psin(31o), +Qsin(180-61o): -33.581 N + Psin(31o) + Qsin(180-61o) = 0 W -33.581 N + Psin(31o) + Qsin(180-61o) = 0

  16. Step 3 - Do Math: -33.581 N + Psin(31o) + Qsin(180-61o) = 0 -5.319 N + Pcos(31o) + Qcos(180-61o) = 0 Substitution: -33.581 N + Psin(31o) + Qsin(180-61o) = 0, P = (33.581 N-Qsin(180-61o))/sin(31o) -5.319 N + Pcos(31o) + Qcos(180-61o) = 0, substituting: -5.319 N + {(33.581 N+Qsin(180-61o))/sin(31o)}cos(31o) + Qcos(180-61o) = 0 -5.319 N + (33.581 N)/tan(31o) + Qsin(180-61o)/tan(31o) - Qcos(180-61o) = 0 Q = 26.061 = 26 N, P = 21 N • Matrices: • Psin(31o) + Qsin(180-61o) = 33.581 N • Pcos(31o) + Qcos(180-61o) = 5.319 N • J K • [sin(31o) , sin(180-61o)] [P] = [33.581 N] • [cos(31o) ,cos(180-61o)] [Q] = [5.319 N ] • Answer matrix will be [J]-1[K] W Q = 26 N, P = 21 N

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