Using the data below: Do results from the first experiment support the validity of the condition of rotational equilibrium? justify your answer

Report on Laboratory Experiment "Rotational Equilibrium" . 06065 (9.8)= 100 + 16. 5 = 116.5 DATA TABLES 0.499 - 0.3 = 0.19 9 0.6517 Part I. Testing the condition of rotational equilibrium. 16.5 +50 = 46-5 0.85 - 0.40 . 1165 ( 9.8 ) 16.5 Xu (117) = 0.499 = 0. 351 MIT(g) = 100 g + Meth M2T(g) = 50g + Meth 1 . 1417 MIto MIT MIT X1 M270 (8) (8) X2 (kg) (m) (8) 0. 499-30 700 (8) (kg ) (m) 114:5 0 . 1165 46 . 5 10.06 65 Old5(-351) 0.30 50 2 . 194 F1 = MIT(kg) x8 (111/52) 0.85 di = Xo - X1 F2 = MIT (kg) x8(171/52) d2 = X2 - Xo 023- (N) (m) 1. 1417( 199) (N). (m) 1. 1417 0. 227 0. 199 0. 6517 0. 351 Tcow = 71 = FI(N) di(m) Tow = 12 = F2 (N) dz(m) 0.727- 0.023 (N .+1) 2 O. 227 0. 023 = 0. 204 % difference between tow and Tew 2 10.2 2, 102 (100) Moth: Mass of clamp plus hanger Xo: Position of pivot point di: lever arm of force F1 X1: Position of 100 g mass d2: lever arm of force F2 X2: Position of 50 g mass Part II. Determination of the mass of an unknown object. Table 2 499- 300 Mith(kg ) = 0.0165 Xo (11) = 0.499 0. 75-0,409 Mu-measured X1 di = Xo - Xi Mk X2 d2 = X2 - Xo = . laq ( kg ) (m) (m1) ( kg ) (m) (m) 0.065 0.300 0 . 199 0.1 0.75 0 . 251 Mu-calculated (kg) using equation (3) 0. 120 % difference between Mu-calculated and Mu-measured 3 .2 Meth: Mass of clamp plus hanger Xo: Position of pivot point MA: Mass of known object X1: Position of unknown mass Mu-calculated: Mass unknown calculated X2: Position of known mass Mu-measured: Mass unknown measured d1: lever arm of force F1 d2: lever arm of force F2 QUESTION: Do results from the first experiment support the validity of the condition of rotational equilibrium? Justify your