1. In the system shown, cable A has an Ultimate Stress of 201 MN/m². For this application the cable must have a factor of safety of 8. Find the required diameter of cable A. 60.0⁰ Cable A 95.0⁰ Cable B 755 kg
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Q: A 557-N force is applied to stretch a cable, 66-m long and has a cross-sectional area of 2.8 mm^2.…
A: Given that-Force, F=557 NLength, L=66marea, A=2.8 mm2Y=1×1011 N/m2
Q: 3. Using the Method of Sections determine the load in members BD, CD, and CE.
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Q: Part A As shown, a truss is loaded by the forces P=2.14 kN and P=0.860 kN and has the dimension a=…
A: Solution:-Given thatP1=2.14 kN & P2=0.860 kNa=2.9 m
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Q: Assume Young's modulus for bone is 1.50 x 1010 N/m2. The bone breaks if stress greater than o= 99260…
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Q: Assume Young's modulus for bone is 1.50 x 1010 N/m2. The bone breaks if stress greater than o= 45844…
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Q: beyond the maximu Fter some time?
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Q: Assume Young's modulus for bone is 1.50 x 1010 N/m. The bone breaks if stress greater than 0= 80179…
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Q: Load (N) Length (mm) 40 100 40.2 200 40.4 250 41 300 42 350 43 400 45 450 48 420 51 40
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Q: A 557-N force is applied to stretch a cable, 66-m long and has a cross-sectional area of 2.8 mm^2.…
A: Given that-Force, F=557 NLength, L=66marea, A=2.8 mm2Y=1×1011 N/m2We know that,Hook's law FA=Y×∆LL…
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A: Dear student The strain is defined as Change in length per unit original Length. Strain = ∆L/L
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Can you please redo using the force triangle method