Youngs Modulus of a Cantilever

Youngs Modulus of a Cantilever Aspect 1: preserve Raw Data TO 1 D.P. Initial continuance: 82.1 cm ± 1mm Weight (g) ±3% Length (cm) ± 3 mm% computer actus reus in Length × 10-2 Extension (cm) phantasm in Extension (cm) × 10-2 100 200 300 400 calciferol 600 700 800 900 cubic yard78.1 74.3 70.5 66.8 63.1 59.5 56.4 52.9 50.2 47.4± 38.4 ± 40.4 ± 42.6 ± 44.9 ± 47.5 ± 50.4 ± 53.2 ± 56.7 ± 59.8 ± 63.34.0 7.8 11.6 15.3 19.0 22.6 25.7 29.2 31.9 34.7± 1.5 ± 3.2 ± 4.9 ± 6.9 ± 9.0 ± 11.4 ± 13.7 ± 16.6 ± 19.1 ± 22.0 Aspect 2: Processing the Data side = mass ÷ address (sag) Gradient= 480÷18 Gradient= 26.67 wrongdoing in gradient: (GMAX-GAVR) ÷ GAVR ×100 GMAX=29.6 GMIN=25.6 GAVR.= 26.6 (29.6-26.6) ÷26.6×100 break in Gradient ≅ 11.3 ≅11% Aspect 3: Presenting the Data L- 82.1×10-2 m ± 0.1%(1×10-2m) b- 2.85 × 10-2 m ± 4% (1×10-2m) d- 0.6 × 10-3 m ± 8% ( 5×10-4 m) g- 9.82 ms-2 (reference taken from physics teacher)s Extension (sag) M- Mass g- acceleration of dryness L- Length E- Youngs Modulus of breeze b- Width d- heaviness s = 4MgL3 ÷ Ebd3 E = 4MgL3 ÷ bd3s E = (4×9.82 ×(82.1×10-2) 3×M) ÷ ((2.85× 10-2)×(0.6× 10-3) 3×s) E= (21.7) ÷((6.15× 10-9) × gradient) E= 21.7÷ 1.64× 10-7 E=1.32 × 108 kg m-1 s-2 ± 5.15 × 107 kg m-1 s-2Error in E= ((0.1×3)+4+(8×3)+11) Error in E= 39% Error in E= 5.15 × 107 Aspect 1: The Conclusion I can conclude that I investigated the Youngs Modulus of a cantilever and found that the childs play of the cantilever (in our case wooden) is 1.32 × 108 kg m-1 s-2 ± 5.15 × 107 kg m-1 s-2. The accepted value (http://www.engineeringtoolbox.com/young-modulus-d_417.html) of Youngs Modulus of elasticity is 1.3× 108 kg m-1 s-2, which makes the answer found from the experiment I fire performed very simil! ar, almost identical to the accepted value. However the error of the found value is very big, and...If you want to get a full essay, order it on our website: BestEssayCheap.com

If you want to get a full essay, visit our page: cheap essay