cables_vector_soln%20rev%20class%2021[1]

cables_vector_soln%20rev%20class%2021[1] - e e k A A A A T...

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x y z A (4,-3,6) B(-4,-6,4) C (0,5,0) W = 3000 lb Determine the tensions in the cables via a direct vector solution x y z A (4,-3,6) B(-4,-6,4) C (0,5,0) W = 3000 lb TA T B T C O O free body diagram This example problem can be solved by writing the equilibrium equations in terms of the x, y, z components and solving the three simultaneous equations for TA , TB , TC . But there is a more direct vector way to solve this problem. 0 0 A A B B C C T T T W = + + - = F e e e k
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x y z A (4,-3,6) B(-4,-6,4) C (0,5,0) W = 3000 lb TA T B T C O free body diagram define the vector cross products these are three vectors (not unit vectors) that are orthogonal to the unit vectors in their definitions A B θ n A x B =ABsin θ n n is a unit vector perpendicular to A , B B C A C A B = × = × = × r e e s e e t e e
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If we dot this equilibrium equation with r similarly dotting the equilibrium equation with s and t gives This is done most effectively with vectors in MATLAB: since 0 A A B B C C T T T W + + - = e
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Unformatted text preview: e e k A A A A T W W T = = r e r k r k r e B B C C W T W T = = s k s e t k t e ( 29 ( 29 B B C B C B C C = = = = r e e e e r e e e e >> OA =[4 -3 6]; >> ea=OA/norm(OA); >> OB = [-4 -6 4]; >> eb =OB/norm(OB); >> OC = [ 0 5 0]; >> ec=OC/norm(OC); >> W = 3000*[ 0 0 1]; >> r =cross(eb, ec); >> s =cross(ea, ec); >> t =cross(ea, eb); >> TA = dot(W, r)/dot(r, ea) TA = 2.3431e+003 >> TB=dot(W, s)/dot(s, eb) TB = 2.4739e+003 EDU>> TC=dot(W, t)/dot(t, ec) TC = x y z A (4,-3,6) B(-4,-6,4) C (0,5,0) W = 3000 lb TA T B T C O define W k define unit vectors define r,s,t solve for tensions TA = 2343 lb TC = 2700 lb TB = 2474 lb A A W T = r k r e B B C C W T W T = = s k s e t k t e B C A C A B = = = r e e s e e t e e...
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cables_vector_soln%20rev%20class%2021[1] - e e k A A A A T...

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