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Problem 3 - 25 points Initials: An old single-leaf drawbridge is made of a thin rectangular board of mass M, width a board (into the page) and length b which can cables pivot at one end. When it is not in use, the span of the bridge is held at a 60 angle from the horizontal direction. The two M ideal cables that hold it at its farther end 300 b are anchored to the ground on either side of the leaf so that they form a 30 angle with the horizontal direction, as shown in hinge the figure. The acceleration due to gravity has canal magnitude g. a- Draw the free body diagram of the rectangular board. Each force should be properly labeled and represented where it applies. Let FH and OH be the unknown magnitude and direction of the net hinge force, measured from the horizontal direction. b- Set up all the scalar equations expressing the static equilibrium of the drawbridge.Initials: elainal c- Determine the magnitude T of the tension force exerted by each of the two cables on the board. ehsan vilestents eskillers if andlod cubsage d- Determine the magnitude FH of the net hinge force exerted on the board and the angle OH that the force forms with respect to the horizontal direction. boley slugns ins (0=1) 0 nodeng Telugus and at enondibroo forthat or enfantsis(I edInitials: ing to the 139 ~4 Initials: Simple Marmonk Physics 8A Final Equation Sheet 7 (t ) = x(t) i + y(t)j AT = 12-11 =4xi+ 4yj Newton's 1s' Law AT B = constant = > F = 0 VAV = At - = VAU-x1 + VAD-y) B = lim 47 = vxi + vyj Newton's 2nd Law At - 0 4t [F = ma aAv = = QAV-xi + aAV-yl a = lim 40 = Newton's 3rd Law At to at = axi + ay ) FMe/You = -Fyou/Me B (t ) = vo + alt - tol Tension: r (t) = ro + volt - to] + salt - to]2 T Ax = X2 -X1 Ax X2 - X1 VAV-x At t2 - t1 Normal Force Ax dx Vx = lim - N At-0 At dt Avx Vx2 - Vx1 aAV-x - At Gravity t2 - t1 AVX dux Fo = mg ax = lim - At-0 At dt Ux (t) = Vox + axlt - tol Static Friction fs S HSN x (t) = xo + Vox[t - tol+ 5 ax[t - to]2 v2 = Vox + 2ax[x - xol Kinetic Friction Ay = y2 - y1 fx = HKN Ay _ yz - y1 VAV-y At tz - t1 Spring Force Ay dy Fap = - kx (x measured from relaxed position) Vy = lim At-0 4t dt AVy Vy2 - Vy1 Uniform Circular Motion aAv-y At t2- 1 a = Avy dvy r ay = lim At-0 At dt v = rw Vy (t) = Voy + dy[t - tol 2 TT 1 T = W y(t) = yo + Voy[t - to] + 5ay[t - to]2 vy = Voy vay + 2ayly - yolInitials: Fext = 0 = AP = 0 Work and Energy 2mz B mi - miz unit - Uzi W = F . di mi + mz my + mz 2m1 m2 - m1 W = FL cose (constant angle and magnitude) V26 mi + mz m + mz W. = -mgAh (height axis pointing upward) WSp = - =(X3 - XA) (x measured from relaxed position) K = =mv Rotational Kinematics Wnet = > Wi = 4K U + K =E e (t) = angular position AE = WNC 40 = 02 - 01 Ug = mgh (height axis pointing upward) 40 02 - 01 WAD Usp = = kx2 (x measured from relaxed position) At t2- t1 40 do w = lim - Universal Gravitation At-0 At dt AW W 2 - W 1 GMm aAV =- At t2 FG = 12 Aw dw GM a = lim g = At-0 At dt GMm w (t ) = wo+ alt- to] UG = r 0 (t) = 0 0 + wolt - to] += alt- to]2 GM W2 = wo+ 2a[0 - Oo] Vorb = s = ro (arc length) v = rw 2TY 3/2 atan = ra T = 12 2 VGM arad = -= rw2 Rotational Dynamics 2GM Vesc = 1 = Et=1 miri (point masses) Ip = ICM + Md2(parallel axis theorem) Linear Momentum It| = rF sin 0 p = mv Text = la P = P L = rp sino; L = Iw (@=angular velocity) Fext = dp W = | Tdo JO1 Krot = =102Initials: nitials: Cu. Cp and Y = Cp/Cy for an Ideal Gas Simple Harmonic Oscillator dzx 2+ w 2 x = 0 1. For a monoatomic gas, d=3 dt2 x(t) = A cos(wt + ) Cy = 38, Cp = 38 and Y = = 1.67 w = 2nf = ~ (@=angular frequency) 2. For a diatomic gas, d=5 Mass-spring: w = Cy = SR, Cp = 12 and Y = = = 1.4 3. For a triatomic or polyatomic gas, d=6 Simple pendulum: w = Cy = 3R, Cp = 4R and y = = = 1.33 Physical pendulum: w = mgLem Quadratic Equation Torsional pendulum: w = ax2 + bx + c = 0 has the solutions 20 (-b + vb2 - 4ac Fluids Derivatives P = V; P = perp Pgauge = Pabs - Patm d(xn) - = nxn-1 P = Po + pgh dx FB = PgVsub d(cos (ax + b)) -a sin(ax + b) Qm = PVA = pQv = const. dx 1' P + pgy + 5 puz = const. Lengths, areas and volumes Circumference of circle: 2nR Thermodynamics Area of disk: TR Surface area of sphere: 4TR PV = nRT = NKBT Lateral area of cylinder: 2TRh Eint = nCUT Volume of cylinder: TRZh AEint = Q - W R = NAKB; Co = AR Volume of sphere: 4TR3/3 2Cp Cp = CD + R ; Y = Cy Trigonometry e =- Wnet = 1 - Lout (efficiency) sin (90) = cos (09) = - cos (180) = 1 Qin Qin sin (09) = cos (909) = sin (180) = 0 TL eideal = 1 - TH cos (60) = sin (30) = 1/2 sin (60) = cos (309) = V3/2 sin (450) = cos (450) = 12/2 W Isobaric CPRAT PAV cos (180- 0) = - cose CVRAP P=const. Isochoric V= const. sin (180- 0) = sine Isothermal nRT In ( vo) nRT In T=const. cos (90 + 0) = - cos (90 - 0) = - sine Adiabatic 0 - # ( P, VJ - POVO) PV=const. sin (90 + 0) = sin (90 - 0) = cose cos2 0 + sin- 0 = 1Initials: Table of rotational inertias Axis Axis Axis Solid cylinder Hoop about Annular cylinder (or disk) about central axis (or ring) about central axis central axis 1=MR2 ( @ ) 1 = +M(R, 2 + R2? ) ( b ) 1 = 1MR (c) Axis Axis Axis Solid cylinder Thin rod about Solid spherea (or disk) about axis through center bout any central diameter perpendicular to diameter length 2R 1=!MR + $ ML2 (d ) 1= AML? (e) 1=3MR (D) Axis Axis Axis Thin spherical shella Hoop about any bout any R diameter Slab about 2 F perpendicular diameter axis through center 1 = AMR (8) 1= ;MR? (h ) 1 = 12 M(2 + 6? ) (1 )