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## Homework Statement

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, 180

^{o}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 (m

_{1}and m

_{2}) placed 0.5m and 1.5m from end A and 180

^{o}from the direction of the bearing reactions, each on radius arms 100mm long. Calculate the sizes of m

_{1}and m

_{2}.

## Homework Equations

F=m*r*ω

^{2}

## 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.

R

_{A}+R

_{B}=F

F=5kN+3kN=8kN

F=m*r*ω

^{2}

8000=m*0.2*50π

^{2}

4934.8022m=8000

m=1.6211kg

Calculating moment about R

_{B}

R

_{A}*2-8*b=0

10-8b=0

8b=10

b=1.25m

a+b=2m

a=2-1.25=075m

B.

F

_{1}+F

_{2}=8kN

Taking moment about R

_{B}

R

_{A}*2-F

_{1}*1.5-F

_{2}*0.5=0

10-1.5*F

_{1}-0.5*F

_{2}=0

1.5*F

_{1}+0.5*F

_{2}=10

F

_{1}+F

_{2}=8

1.5*F

_{1}+0.5*F

_{2}=10

Solving equation

F

_{1}=8-F

_{2}

1.5*(8-F

_{2})+0.5*F

_{2}=10

12-1.5*F

_{2}+0.5*F

_{2}=10

F

_{2}=2kN

F

_{1}+2=8

F

_{1}=6kN

F

_{1}=m

_{1}*r*ω

^{2}

6000=m

_{1}*0.1*50π

^{2}

2467.4011m

_{1}=6000

m

_{1}=2.4317kg

F

_{2}=m

_{2}*r*ω

^{2}

2000=m

_{2}*0.1*50π

^{2}

2467.4011m

_{2}=2000

m

_{2}=0.8106kg

Did I got it right?

I'm not sure where I went wrong, but I suspect the m

_{1}should be smaller then m

_{2}as R

_{A}is bigger then R

_{B}. Any tips?