SE101A_HW1_FA2019_Sol.pdf - Homework 1 Solution SE 101A – Fall 2019 Q1 Solution As F = Gm1 m2 r2 and r = r1 r2 = 1m the attractive force equation

# SE101A_HW1_FA2019_Sol.pdf - Homework 1 Solution SE 101A –...

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Homework 1 Solution SE 101A – Fall 2019 Q1 Solution: As F = Gm 1 m 2 r 2 and r = r 1 + r 2 = 1 m , the attractive force equation between two spheres is constructed as, F 1 = 6 . 673 × 10 - 11 ( 250 )( 250 ) 1 2 = 4 . 17 × 10 - 6 N ( 4 pts ) It is given that spheres are identical. If we define the distance between earth and one sphere as X , the attractive force equation between one sphere and earth is constructed as, F 2 = Gm earth m 1 X 2 = 6 . 673 × 10 - 11 ( 5 . 976 × 10 24 )( 250 ) X 2 = 9 . 97 × 10 16 X 2 N ( 4 pts ) By equating F 1 to F 2 , X is computed as, X = s 9 . 97 × 10 16 4 . 17 × 10 - 6 = 1 . 546 × 10 11 m ( 2 pts ) 1
Homework 1 Solution SE 101A – Fall 2019 Q2 Solution: As the stress equation for eccentric loading of a short column is, σ = P A + Pe y I P is a force so its dimension is ML T 2 . A is a area so its dimension is L 2 . Therefore, if we consider the stress equation above, dimension of stress σ is computed as, ML T 2 L 2 = M T 2 L ( 5 pts ) For the second moment area I , since e and y are in lengths, dimension of Pe y is computed as, ML T 2 × L × L = ML 3 T 2 Therefore, in order to provide dimensional consistency between right hand side and left hand side of

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