Question

(b) Figure 1 shows a system with TWO (2) weights on a rotating shaft. The given parameters for this system are: W = 66.7 N for 30° at a 101.6 mm radius and W = 89 N for 270° at a 152.4 mm radius. Rajah 1 menunjukkan satu sistem dengan DUA (2) berat pada aci yang berputar. Parameter yang diberikan untuk sistem ini adalah: W, = 66.7 N untuk 30° pada jejari 101.6 mm dan W2 = 89 N untuk 270ºpada jejari 152.4 mm.) (i) Determine the radii and angles of the balance weights needed to dynamically balance the system if the balance weight in plane 3 and 4 weights 66.7 N and 133.4 N respectively. (Tentukan jejari dan sudut imbangan berat yang diperlukan untuk mengimbangi sistem secara dinamik.jika berat imbangan pada satah 3 dan 4 adalah 66.7 N dan 133.4 N masing- masing (14 Marks/ Markah) (ii) Determine angular velocity to reduce the shaking forces and shaking moment if rotation is unbalanced at 100 rpm. (Tentukan halaju sudut untuk mengurangkan daya-daya gegaran dan momen gegaran jika putaran tidak seimbang pada 100 rpm.] (6 Marks/ Markah) W WI 228.6 mm -76.2 4 W: W Figure 1

Answer 1

m₂ % w²+ Mary w = m₂ 9₂ Wh + m 3 ?W? Sole Given Data 1 me= W, 6.8 kg My 101: 6 mm مه m2 = Wz = 9.03 kg M2 = 152.4 mm 8 m3 = Wa a 6.8 kg, my ha a 13.6 kg g tutt thu 228.6mm 76.2mn 3 Since the net force on the shaft must be equal to zero force on the disturbing must be equal to sum of contrifugal forces on the balancing acting therefore the centrifugal far fy = f + f₂ mat mury a m₂ 2 + m₂%

6.8X (101.6) + 2. 13.614 = 9.07 X( 1524) + 6.8295 13.694 6.893 = 691: 388 (i) In order moment acting on dynamic balance, must be equal to zero. Shaft Taking flane 3 reference plane, m₂ qw2 x(228.6) My My W²x (2286+76.2) m₂ r₂ X 228 6 = X 304.8 mary 9.07 x 152:4 X 22816 = 1366 X tu x 30' 8 .: Mu= 7.6. 23 mm 1 get from equation is 50.785 mm mass, "m3" be Oz and het Angle made by be by on respectively,

Then, Shorizontal forces = 0 3. m₂ l cos O₂ + m₂ r₂ cos O2+ m₂ h3 cos O₂ + my ly cosa = 0 (6.88 101.6X cos 30°) + (9.078 152.4x Cos 270° ) + (6 8x 90- 483 69 6 )+( 13+6 X 16 23 (6) 6 =0 998. 32 + 345 34 103037 1036.728 coy=0 li) 345.34 Cod Og + 1036. 728 Con By = -598.32 Also , & Vertical forces =0 Mal sin O + mar sin O2 + Mgra sin Oz + my ry sin Oy =0 6-8X 101.6 x sein 30' )+ ( 9.07 X 152.4 x Sis 270°) +(6-8 X 50.785 sin 03) +(13.676-23 sin Ou) = 0 345.34 sin Oz + 103 6.728 Sin Qy = 1036.8287 () lii) we get On solving equation (in and 03. 05.40, Ou = 135.80

(b) The Shaking forces and moment doesn't change ift the values of 16) Given that, N = 10 0 x 20 N 60 - ليا aux 100 10. 472 read's Completed