DESIGN OF MACHINERY SOLUTION MANUAL 12-4-1! PROBLEM 12-4Statement:A three-bladed ceiling fan has 1.5 ft by 0.25 ft equispaced rectangular blades that normally weigh 2 lb each. Manufacturing tolerances will cause the blade weight to vary up to plus or minus 5%. The mounting accuracy of the blades will vary the location of the CG versus the spin axis by plus or minus 10% of the blades' diameters. Calculate the weight of the largest steel counterweight needed at a 2-in radius to statically balance the worst-case blade assembly.Given:Blade dimensions:Length:lb1.5 ft.Width: wb0.25 ft.Nominal weight: Wbnom2 lbf.Manufacturing tolerances: Weight tw0.05 CG offset tCG0.10Assumptions:1. The blades are held in place by a bracket such that their base is 6 in from the center of rotation making the tip 24 in from the center. Thus, the blade sweep diameter is 48 in.2. There is one heavy (maximum weight) blade at 0 deg and two light (minimum weight) ones at 120 and 240 deg, respectively. Thus,W11 twWbnom.W12.100 lbf=r115 in.θ10 deg.W21 twWbnom.W21.900 lbf=r215 in.θ2120 deg.W31 twWbnom.W31.900 lbf=r315 in.θ3240 deg.Solution:See Mathcad file P1204.1. There are two factors to be taken into account, the variation in blade weight and the error or eccentricity in the location of the global CG. The variation in blade weight about its spin axis will be considered first.2. Resolve the position vectors into xy components in the arbitrary coordinate system associated with the freeze- frame position of the linkage chosen for analysis.R1xr1cosθ1.R1x15.000 in=R1yr1sinθ1.R1y0.000 in=R2xr2cosθ2.R2x7.500 in=R2yr2sinθ2.R2y12.990 in=R3xr3cosθ3.R3x7.500 in=R3yr3sinθ3.R3y12.990 in=3. Solve equation 12.2c for the mass-radius product components.mRbxW1R1x.W2R2x.W3R3x.gmRbx3.000 in lb.=mRbyW1R1y.W2R2y.W3R3y.gmRby0.000 in lb.=4. Solve equations 12.2d and 12.2e for the position angle and mass-radius product required.θbatan2 mRbxmRby,θb180.000 deg=mRbbmRbx2mRby2mRbb3.000 in lb.=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 12-4-25. Now, account for the fact that the blades' spin axis can be eccentric from their CG.Maximum eccentricity:re48 in.tCG.re4.800 in=mR product due to eccentricity:mRbeW1W2W3gre.mRbe28.320 in lb.=6. Add the two mR products and divide by the 2-in radius specified for the counterweight to find the maximum weight required.Rcw2.0 in.mRbmRbbmRbemRb31.320 in lb.=mcwmRbRcwmcw15.66 lb=Note that 90% of the counterweight is required to balance the eccentricity. The manufacturer would be well advised to try to control this variation more tightly.2nd Edition,