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
Posted: Sun May 22, 2022 1:18 pm
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