Question

Four masses \(A, B, C\) and Das shown below are to be completely balanced.

$$ \begin{array}{c|c|c|c|c} \hline & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\ \hline \text { Mass (kg) } & 10 & 4 & - & 5 \\ \hline \text { Radius (mm) } & 125 & 150 & 100 & 200 \\ \hline \end{array} $$

Find the followings:

(a) The angle between the masses \(\mathrm{A}\) and \(\mathrm{C}\)

(b) The axial distance between the planes of rotation of \(\mathrm{B}\) and \(\mathrm{C}\)

(c) The magnitude of mass C

Answer 1

Sol Given Data :- mass Radius 10 125 4 150 (6.299) 5. 100 200 O 125 mm lio Q 149.869 120mm for complete balancing the the mass system all the forces & couples should be balanced Efn=0 Efy=0 EMR.P=0 7300X110) 200mm ici (C) Emiriconosco MATA ces out moriscos 110° + mere case tmodo cs 240 = 0 1Q X:125 + 46.151134202) + mare codi + 54.2% (-0.5)=0 mcrc cort = -0.544788 z mirisino =0 lox 125 sino t48.15 sinto the resino +58.2 sonayo no 22 - © mero sino = 0.3022 Squaring & adding equ + ① mcrccoshe thino = 120.544700? + (0.3022} me x1 = 0.622 me = 6.299 kg) 19 Now from equ o 6.299X-) coso e = -0.544788 Oc = 149.0697

(b) for the calculation of distance B to let B toc = x couple about refernce plane say (B) should be zero - MANA (AB) = mercat moro (2+ (D) 10 X •125 X [ 10 X 170 6.3xxn to 5 X : 2 X Cat :55) 0.75 = 0.55 +1.632 n = 0.12269 m = 122.7mm - Hence ü 122.7 mma Bc distance @ Rip tntt 550mmy 6oomm (122.4mmi)