6. A drawbridge design uses a beam with mass M and length L. The beam is hinged to a wall at its left end and supported

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6 A Drawbridge Design Uses A Beam With Mass M And Length L The Beam Is Hinged To A Wall At Its Left End And Supported 1 (34.58 KiB) Viewed 19 times

6 A Drawbridge Design Uses A Beam With Mass M And Length L The Beam Is Hinged To A Wall At Its Left End And Supported 2 (32.89 KiB) Viewed 19 times

6 A Drawbridge Design Uses A Beam With Mass M And Length L The Beam Is Hinged To A Wall At Its Left End And Supported 3 (36.94 KiB) Viewed 19 times

6. A drawbridge design uses a beam with mass M and length L. The beam is hinged to a wall at its left end and supported by a rope on its right end. The rope is perpendicular to the beam. See figure. Find the hinge force components and the magnitude of the tension force in the rope. You may assume the drawbridge is in static equilibrium. The knowns are M, L, g and e. (20 points) You must use a FBD to generate your equations. hinge beam center of mass wall rope

Static Equilibrium ΣΕ = 0 ΣΕ = 0 Σr=0 (0) = max Universal Gravity Gm,m, Simple Harmonic Motion T = 27, 2π =0A a = 00² A max F =KA= mamax max ןןז T=t|F||F|sin 0 T=xF, -yF, T>0=CCW T<0=CW k c (hyp) = mr² point particle. L=10 L = Lyst a (adj) ax² +bx+c=0 sin = opp/hyp=b/c 0= adj/hyp = a/c 8 tan - opp/adj-b/a c²=a² + b² b (opp) cos Quadratic Equation x= -b± √b² - 4ac 2a

Constant Acceleration Equations (1) x(1) = x₂ + vt + at² (2) v(t)=v₁ + at (3) v² = +2a(x-x) W=F,Ar +F,A, (constant force) W=F|AF|cos (constant force) p=mv System Momentum Conservation = mv,+ m₂v, m₁v, m₂v₂ Inelastic Collision V₁ = V₂/ Elastic Collision V2-V₁ = V₁₁-V₂ ΣF, = ma, ΣF, =ma, ΣF, = ma r f₁ = μ₁n Energy Equations _my² +mgy, + = kx² + [W_ == my² +mgy, +_ kc 2 10+mgyors + W = 10² -mgycs2 2 K=-=-mv² U₁ = mgy U₁ -kx² no slip condition: v= cor