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

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

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.

Example 1 - One unknown 1.Break forces in to components 2.Express the unknown forces as components 3.Set up a = 0 equation for x and another for the y direction 4.Do math. 5.0 kg Find the tension in the lines: 18 o In the y direction: The weight of the mass is down (-): wt = mg = (5.0 kg)(9.8 N/kg) = -49 N Since the tensions are equal, they both have identical upward (+) components: Tsin(18 o ) + Tsin(18 o ) = 2Tsin(18 o ) T T And now our vertical expression becomes: -49 N + 2Tsin(18 o ) = 0 which is solvable: T = (49 N)/(2sin(18 o )) = N (getting a car un-stuck)

Whiteboards: Force Equilibrium - one unknown 1A1A | 1B | 1C | 2A | 2B | 2C1B1C2A2B2C TOC

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

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

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

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

2Tsin(25 o ) N = 0 W 15.0 kg 25.0 o Find the horizontal force acting on the walls 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(25 o ) + Tsin(25 o ) = 2Tsin(25 o ) 2Tsin(25 o ) N = 0 T T Step 2 - Set up the vertical equation, using T as the tension in the cables:

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

Step By Step: - Two unknowns 1.Break all forces into components 2.Express the unknown forces as components 3.Set up a = 0 equation for x and another for the y direction 4.Do math kg 40 o 20 o T1T1 T2T2 T3T3 Find the tensions T 1, T 2, and T 3 In the y direction: T 1 = (12.5 kg)(9.8 N/kg) = N (down) T 2 has an upward component: T 2 sin(40 o ) T 3 also has an upward component: T 3 sin(20 o ) So our expression becomes: T 2 sin(40 o ) + T 3 sin(20 o ) N = 0 (Making up positive)

Step By Step: 1.Take all the given forces and break them into components 2.Express the unknown forces as components 3.Set up a = 0 equation for x and another for the y direction 4.Do math kg 40 o 20 o T1T1 T2T2 T3T3 Find the tensions T 1, T 2, and T 3 So our expression becomes: T 3 cos(20 o ) - T 2 cos(40 o ) = 0 (Making right positive) In the x direction: T 2 has an leftward (-) component: T 2 cos(40 o ) T 3 has an rightward (+) component: T 3 cos(20 o )

Now it’s MATH time!!!!! Two equations, two unknowns: T 3 cos(20 o ) - T 2 cos(40 o ) = 0 T 2 sin(40 o ) + T 3 sin(20 o ) N = 0 Re-Write them like this: -cos(40 o ) T 2 + cos(20 o ) T 3 = 0 sin(40 o ) T 2 + sin(20 o ) T 3 = N Matrices: A B [-cos(40 o ), cos(20 o )] [T 2 ] = [0 ] [ sin(40 o ), sin(20 o )] [T 3 ] = [122.5 N] The answer is [A] -1 [B]

Or you can substitute Two equations, two unknowns: T 3 cos(20 o ) - T 2 cos(40 o ) = 0 and T 2 sin(40 o ) + T 3 sin(20 o ) N = 0 From the first equation T 3 cos(20 o ) = T 2 cos(40 o ), then T 3 = T 2 cos(40 o )/cos(20 o ) Plug this into the second equation: T 2 sin(40 o ) + ( T 2 cos(40 o )/cos(20 o ) ) sin(20 o ) N = 0 Solve: T 2 sin(40 o ) + T 2 cos(40 o ) tan(20 o ) = N T 2 ( sin(40 o ) + cos(40 o ) tan(20 o ) ) = N T 2 = (122.5 N) / ( sin(40 o ) + cos(40 o ) tan(20 o ) ) = 133 N and T 3 = ( N) cos(40 o )/cos(20 o ) = 108 N Wow!!!

Whiteboards: Two Unknowns 11 | 2 | 323 TOC

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

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

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