1. The problem statement, all variables and given/known data A. A shaft 2m long rotates at 1500 revs min-1 between bearings as shown. The bearings forces of 5kN and 3kN acting in the same plane as shown. A single mass is to be used to balance the shaft, so that the reactions are zero. The mass is to be placed at a radius of 200mm from the shaft centre, 180o from the direction of the bearing reactions. Determine the size and position (a and b) of the mass to be used. B. The shaft in part A is to be balanced using two mases (m1 and m2) placed 0.5m and 1.5m from end A and 180o from the direction of the bearing reactions, each on radius arms 100mm long. Calculate the sizes of m1 and m2. 2. Relevant equations F=m*r*ω2 3. The attempt at a solution A. ω=1500 revs min-1=1500*(2π/60)=50π rad s-1 To balance the shaft the bearing reactions must equal to centrifugal force of the mass. RA+RB=F F=5kN+3kN=8kN F=m*r*ω2 8000=m*0.2*50π2 4934.8022m=8000 m=1.6211kg Calculating moment about RB RA*2-8*b=0 10-8b=0 8b=10 b=1.25m a+b=2m a=2-1.25=075m B. F1+F2=8kN Taking moment about RB RA*2-F1*1.5-F2*0.5=0 10-1.5*F1-0.5*F2=0 1.5*F1+0.5*F2=10 F1+F2=8 1.5*F1+0.5*F2=10 Solving equation F1=8-F2 1.5*(8-F2)+0.5*F2=10 12-1.5*F2+0.5*F2=10 F2=2kN F1+2=8 F1=6kN F1=m1*r*ω2 6000=m1*0.1*50π2 2467.4011m1=6000 m1=2.4317kg F2=m2*r*ω2 2000=m2*0.1*50π2 2467.4011m2=2000 m2=0.8106kg Did I got it right? I'm not sure where I went wrong, but I suspect the m1 should be smaller then m2 as RA is bigger then RB. Any tips?