Drawbridge - % p_o_e> + p_e_t> = p_o_t> so...

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Sheet1 Page 1 % Drawbridge problem % Replacement and zero set (i.e., statics) concepts % n1> to the left, n2> downward, and n3> = n1> x n2> points o,m,e,t % o is base, m is middle, e is end, and t is top of the drawbridge constants p,q,r,s constants H,L,theta newtonian n overwrite on % Define unit vector directions c> = n1> b> = n2> d> = n3> p_o_e> = l*sin(theta)*n1>-l*cos(theta)*n2> p_o_t> = -H*n2>
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Unformatted text preview: % p_o_e> + p_e_t> = p_o_t> so p_e_t> = p_o_t> - p_o_e> p_e_t> = p_o_t>-p_o_e> a> = unitvec(p_e_t>) % Define applied forces P> = -p*n2> Q> = q*n1> R> = r*n2> S> = s*a> % Part a) Calculate replacement using point o F> = P>+Q>+R>+S> p_o_m> = 0.5*l*sin(theta)*n1>-0.5*l*cos(theta)*n2> T> = cross(p_o_m>,R>)+cross(p_o_e>,S>) % Part b) Determine function f f = dot(T>,n3>)...
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This note was uploaded on 06/13/2011 for the course EGM 3400 taught by Professor Matthews during the Spring '08 term at University of Florida.

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