ME211solution_2 - The spring stretches x=0.15-0.08=0.07 m...

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The spring stretches x=0.15-0.08=0.07 m Applying the spring formula, we have 600(0.07) 42.0N sp Fk x == = The normal reaction B N can be obtained directly by summing moments about point A. 0; 42.0(0.05) (0.2) 0 B A MN =− = 10.5N B N = 0; 42.0 0 xx FA = 42.0N x A = 10.5 0 yy = 10.5N y A =
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Reaction force at point A can be obtained directly by summing moment about point B 0; (1.4) 60(9.81)(0.9) 0 By MA =− + = 379.39N y A = 0; 378.39 6(9.81) 2 0 yy FB + = 105.11N y B = 0; 0 xx == 105N B F =
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sin30 Ax A NN , cos30 Ay A cos30 x WW , y 0; B M = (5.25) (0.5) (3.5) (2.5) 0 Ay Ax y x W W +− = ( cos30 )(5.25) sin30 (0.5) 100sin30 (3.5) 100cos30 (2.5) 0 AA °+ ° ° ° = 81.6 lb A N = 0; 100cos30 81.6sin30 0 xx FB =− + ° ° = 45.8 lb x B = 0; 100sin30 81.6cos30 0 yy ° + ° = 20.67 lb y B 22 (45.8) ( 20.67) 50.2 lb B F =+ = N Ay B x B y W x y x N Ax W y 30° 30°
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0; A M = 3 () ( 2) ( ) 0 2 BC FL F L P L +− = 21 . 5 FF P += ------① From similar triangles 2 B C LL = ΔΔ = > 2 CB Δ=Δ = > 2 kk = = > 2 = ------② From ① and ② 0.3 B FP = 0.6 C = Deflection 0.6 C P x k =
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Due to the smooth pipe no frictions (tangential forces) exist 0; 45cos30 (36) 45sin30 (8) (20) (8tan30 ) 0 AC B MR R ° + ° = ------① 0; cos30 45 0 yC B FR R ° = ------② From
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This homework help was uploaded on 04/04/2008 for the course ME 211 taught by Professor Thouless during the Fall '08 term at University of Michigan.

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ME211solution_2 - The spring stretches x=0.15-0.08=0.07 m...

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