A three-bladed ceiling fan has 1.5 ft by 0.25 ft equispaced rectangular blades t

Question

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. I need the solution solved step by step please.

Explanation / Answer

Given:

Blade dimensions:

Length: l = 1.5 ft & Width: w = 0.25 ft .

Nominal weight: W = 2 lbf .

Manufacturing tolerances: Weight tw = 0.05

CG offset tcg = 0.10

Assumptions:

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,

W1 = ( 1 + tw)*W

W1 = 2.100 lbf , r1 = 15.in, 1 = 0.deg

W2 = ( 1 - tw)*W

W2 = 1.900 lbf , r2 = 15.in, 2 = 120.deg

W3 = ( 1 - tw)*W

W3 = 1.900 lbf , r3 = 15.in, 3 = 240.deg

Now,

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 freezeframe position of the linkage chosen for analysis.

R1x = r1*cos(1) , R1x = 15.000 in , R1y = r1*sin(1) , R1y = 0.000 in

R2x = r2*cos(2) , R2x = - 7.500 in , R2y = r2*sin(2) , R2y = 12.990 in

R3x = r3*cos(3) , R3x = - 7.500 in , R3y = r3*sin(3) , R3y = - 12.990 in

3). Solve for the mass-radius product components.

m*Rbx = [- W1*R1x - W2*R2x - W3*R3x]/g   m*Rbx = - 3.000 in lb.

m*Rby = [- W1*R1y - W2*R2y - W3*R3y]/g   m*Rbx = 0.000 in lb.

4.) Solve equations for the position angle and mass-radius product required.

b = atan2(m*Rbx , m*Rby) , b = 180.000 deg

m*Rbb = [ m*Rbx2 + m*Rby2]1/2 , m*Rbb = 3.000 in lb

5.) Now, account for the fact that the blades' spin axis can be eccentric from their CG.

Maximum eccentricity: Re = 48.in* tcg . Re = 4.800 in

mR product due to eccentricity:

m*Rbe = (W1 + W2 + W3)*Re/g , m*Rbe = 28.320 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.

Rcw 2.0.in

m*Rb = m*Rbb + m*Rbe , m*Rb = 31.320 in lb.

Mcw = m*Rb/Rcw , Mcw = 15.66 lb

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