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Bridge Construction

Transcript: 1/3 of bridge opp 2 crossbar 30* Placement: center of luggage, allows for control of entire bag 4 dyson balls on bottom of luggage so it can easily move in every direction Luggage 62 cm tall, 25 cm 65 cm R = 5,833.33g d HOW 4.8N Compression bridge weight 3) 5.5N 4) 6N 6.4N 20N 2.5 N 30N E = Stress = F/A R = 4,333.33g R x 60 = 6,000 x 25 + 10,000 x 20 = 350,000g 19.3N 1) 6N 2) 5N Move in opposition of the resonance frequency in a structure with springs or pendulums Reduce overall movement of the bridge Measure angle = 37* 1/3 of bridge length 10N 2.1 N 12N 3.5N Filled in at end of each day F bridge length 5,083.33g + 5,833.33g + 5,083.33g = 16,000g Ft = 10N Fx = 6.4N Fy = 7.7N What would happen if there were 3 supporting pillars? 7.933 strips used @ 1,000 pounds/strip = 7,933.33 pounds = $13,090 1 1/2 W 20N Lab 3: Static Equilibrium Hyper Physics 40* spring scale 22 cm 50N Sum of Forces table 70 deg. Triangle = greatest ratio = strongest shape 2 Equilibrium 2 1) 7N 2) 6N 32.1N 2.2 N Shape A (L-Shape) = .075kg Shape B (Triangle) = .25kg Shape C (Square) = .15kg Shape D (Pentagon) = .1kg Shape E (Hexagon) = .075kg 20 cm Tsin(θ) Angle: 45 deg. Fx = 14.1N Fy = 14.1N 1 Macro Increase in tension force = Increase in both horizontal & vertical force Truck: 6,000kg Bridge: 10,000kg 50* stand adj Example: 15 cm 2(Ft)(sinθ)=Mg 2(2.1)(sinθ)=(.35)(9.81) 4.2(sinθ)=3.4335 sinθ=.8175 θ=54.8° Hexagon =.075kg Add girders or harmonic absorbers R + R =16,000 - 5,833.33 = 10,166.7/2 = 5,083.33g How does this relate to real life? How do the angles change when the separation distance changes? amount spring is stretched 2 45 cm 3 F 8N .36kg 1 COMPLETED 3 Ft and angle are inversely related As angle increases, force of tension decreases 3) 8N 4) 7N Fx 60m F = -kx 2(Ft)(sinθ)=Mg sinθ=Fy/Ft Ft=Mg/2(sinθ) Ft(sinθ)=Fy Fy=Mg 50 deg 2 BUT AT 45 degrees: Fx = Fy Try to keep structure stable while removing blocks Different shapes of different sizes are used to make it more difficult to determine which blocks to remove Important to think before making any moves - one wrong move could affect the entire structure 4.6N 4.5N Twister 2 F & x - direct relationship Questions Model 7N spring constant 45* Fy How can we apply these concepts to our bridge? Creating a New Problem the space figure (geometric 3-dimensional figure) is symmetrical balancing point is centered string supports figure in centered point string is ab R = 5,833.33g R = 4,333.33g R = 5,833.33g 2.3 N r Bridge: 172g = .172kg .172 x 9.8 = 1.69N Load: 15lbs/2.2 = 6.8kg 6.8 x 9.8 = 66.8N Model Ratio: 66.8N/1.69N = 39.5 Avg. Force (N) Build a cheap yet efficient bridge that allowed cars to safely pass over it Extra points for trucks Materials: go back to original shape when force is applied Certain point where the material will not go back to the original shape - will shortly break: Young's Modulus Tension Angle WHEN 6 + 8 = 10 100 = 100 Constant velocity, changing angle Measure hypothenuse = 10N g=9.8m/s^2 55 cm R = 5,083.33g R = 5,833.33g R = 5,083.33g 6N 5,833.33 + 4,333.33 + 5,833.33 = 16,000 sin70 = Fy/4.6N --- Fy = 4.3N cos70 = Fx/4.6 ---Fx = 1.6N Fx +Fy = Ft --- 4.3 +1.6 = 4.6 distance from end A R + R = 16,000-4,333.33 = 11,666.7/2 = 5,833.33g 7.7N 20m 19.3N Reaction at each support when truck is... spring force 15 cm Questions: Changing Angles 2 Ft How can we ensure the experimental values are correct? adj F = 6N ; F = 8N opp A Strain Balancing Cans 2 3N length Theoretical Values How can this be prevented? sin50 = Fy/7 --- Fy = 5.4N cos50 = Fx/7 --- Fx = 4.5N Fx + Fy = Ft --- 5.4 + 4.5 = 7 5000kg bridge weight Equations Model Ratio d Tacoma Narrows Bridge 2 sin60 = Fy/5.6 --- Fy = 4.8N cos60 = Fx/5.6 --- Fx = 2.8N Fx +Fy = Ft ---4.8 + 2.8 =5.6 Questions string Calculate θ for different Ft 60 deg. 8N MR = .313 adj string truck weight Fx decreases while Fy increases distance from end A 100g & 250g Put a ruler between two books. Then press down on the middle so that the "beam" bends. Questions: What must be happening to the molecules of the ruler? What are the micro and macro elements? Data Table θ 6N Experiment 1: Bending a Beam Triangle = .25kg Equilibrium Game 8 + 12 = 15 225 = 225 22.9N F = 10sin37 = 6.018N F = 10cos37 = 7.986N L-Shape = .075kg 25* 20N F = 8N ; F = 12N How does velocity affect our values? Why must it be constant? 4000kg 50* 75* Plan 5.6N 20N F 2 2 The Lever Law Resonance 70 deg Ft = 30N Fx = 19.3N Fy = 22.9N What would we do differently? Use a box to insert weights so that they are evenly distributed Add weights quickly so that they are not acting on bridge while waiting to add more Pythagorean Theorem 37* .72kg F Angle = 50 degrees 5.6N MR = .695 F = 20N spring scale What is the relationship between the angle and the force values? 2 60 deg Angle = Consistent opp Angle: 75 deg. Fx = 5.2N Fy = 19.3N 25 cm Ft truck weight 90 deg 22cm high - comfortable for avg. human to pull 14cm wide - average width of hand Mass of suspended weight (350 grams) Height of crossbar

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