Problem #3: A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is

9 ft. The angular speed of the small pulley is 1750 rev/min as it delivers 2 hp. The service is such that a service factor K, of 1.25 is appropriate. A) Find Fc, Fi, (F₁), and F2, assuming operation at the maximum tension limit B) Find H, n, and belt length C) Find the dip Angle of contact and length of belt D-d Od = π-2 sin-¹ 2C 8p+2 sin-¹ 2C 1 L = √√4C²-(D-d)² + (Dºp+dºa) Surface velocity of pulley, V V = wr; nadu V = 12 Appendix A: Equations Weight of belt, w sin D-d 20 -√4C² -ID-87² sin D-d 20 w being rotational speed and r the radius n being rotational speed in rev/min, d in inches. V is obtained in ft/min #p/nWeight of belt, w w = 12ybt; y being weight density in lbf/in³, belt width b in inches, t belt thickness in inches. w is obtained in lbf/ft Centrifugal force, F. F = mr²w² Fc = (-²; w in lbf/ft, g in ft/s², V in ft/min. F, is obtained in lbf The tight-side tension (F₁) and loose-side tension (F₂) T F₁ = F₁+Fc+d F₂ = F₁+Fc-d F₂-Fe Allowable largest tension, (F₁). (F₁) a = bF₂CpCvi b is ten belt width, F the manufacturer's allowed tension, Cp the pulley correction factor,, C, the velocity correction factor, Transmitted torque, T T = (F₁-F₂) H W T is torque and d is diameter f is the friction coefficient and the contact angle T=-/nT = 63025 Initial tension, Fi F₁+F₂ 2 Tel-1 F₁ = H = Tw H = (F₁-F₂)V 33000 Transmitted horse power, H nfs = Allowable or design horse power, Ha Ha = Hnom Kendi The factor of safety, f=In Dip, H, in hp, n in rev/min. T is obtained in lbf.in F f is the friction coefficient and the contact angle H₂ Hnom K The level of friction development, 1₁ (F₁)a - Fc F₂-Fc dip = F₁ and F₂ in lbf, V in ft/min. H is obtained in hp c²w 96 Hrom is the nominal power, K, is the service factor, and no is the design factor C is the center-to-center distance in inches, w is the weight in lbf/in, F, in lbf