Mechanical principles rotating systems

AI Thread Summary
A 2 m long shaft rotating at 1500 RPM experiences forces of 5 kN and 3 kN, requiring a balancing mass to achieve zero reactions at the bearings. The mass is to be positioned 200 mm from the shaft center, 180° opposite the bearing reactions. The discussion focuses on calculating the size and position of this mass and introduces a second scenario with two masses placed at different distances from one end of the shaft. Users express confusion about using simultaneous equations to determine the masses and seek guidance on their calculations. Clarification on a missing figure is also requested to assist in solving the problem effectively.
Mitch1
Messages
55
Reaction score
1

Homework Statement


A shaft 2 m long rotates at 1500 revs min–1 between bearings as
shown in FIGURE 2. The bearings experience forces of 5 kN and
3 kN 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 200 mm from the shaft centre, 180° from the
direction of the bearing reactions. Determine the size and position (a
and b) of the mass to be used.

HNCPic1.jpg


(b) The shaft in part (a) is to be balanced using two masses (m1 and m2)
placed 0.5 m and 1.5 m from end A and 180° from the direction of
the bearing reactions, each on radius arms 100 mm long. Calculate
the sizes of m1 and m2.

Homework Equations


F=mrω^2

up forces must equal down to be in equilibrium

The Attempt at a Solution


It is B) that I am confused about

Do you need to use some sort of simultaneous equation to work out any of the two masses ? I have tried using the balancing equating but I can't seem to get the correct answer
 
Physics news on Phys.org
Sorry, the attached figure doesn't appear in your post. :frown:

If you have an attempted solution to the problem, please post it.
 
5000N x 2m - ((m1 x .1m x157^2)x1.5) - ((m2 x .1m x 157^2)x0.5m) =0
10,000 - (m1 x 3697.35) + (m2x1232.45)=0
10000/(3697.35+1232.45)=m1+m2
2.028= m1+m2

Pretty sure this is wrong to be honest but not sure where to go from here or what approach I should be making

Thanks
 
Mitch1 said:
5000N x 2m - ((m1 x .1m x157^2)x1.5) - ((m2 x .1m x 157^2)x0.5m) =0
10,000 - (m1 x 3697.35) + (m2x1232.45)=0
10000/(3697.35+1232.45)=m1+m2
2.028= m1+m2

Pretty sure this is wrong to be honest but not sure where to go from here or what approach I should be making

Thanks
Any word on the figure missing from the OP? That would be a big help to anyone trying to guide you.
 
Sorry yes there you go
 

Attachments

  • image.jpg
    image.jpg
    34.2 KB · Views: 432
This is a better view:

image-shaft.jpg

Along with your calculations:
Mitch1 said:
5000N x 2m - ((m1 x .1m x157^2)x1.5) - ((m2 x .1m x 157^2)x0.5m) =0
10,000 - (m1 x 3697.35) + (m2x1232.45)=0
10000/(3697.35+1232.45)=m1+m2
2.028= m1+m2

Pretty sure this is wrong to be honest but not sure where to go from here or what approach I should be making

Thanks
 
No problem, do u think that is on the right lines or way off?
 
Mitch1 said:
No problem, do u think that is on the right lines or way off?
I haven't had a chance to look at it. If someone else has any suggestions, please feel free to dive in. :smile:
 
No problem cheers
 
Back
Top