A shaft 2 m long rotates at 1500 rpm between bearings. The bearings experience forces of 5 kN and 3 kN acting in the same plane.
The shaft 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.

Now through a process of elimination I know what the masses m1 and m2 should be and also what these forces f1 and f2 are.

My problem is taking the 2 m shaft and how I got to these values without just plugging in different numbers till I got there. I think it must be something about the summing of the moments from the right hand or left hand end. Hope someone can point me in the correct direction with this.....

Think it maybe something like F1+F2 = 2 ( Ra-Rb ) But not sure and if this was correct how I narrow it down to solve for one of the variables it being either F1 or F2. Could really be doing with a shove in the right direction with this as I have been at it for days!!

Surely someone must have a clue for this???

How about this as a solution and see if anyone agrees......
Ra=5kN Rb=3kN
Ra + (Rb) = F1+F2
Ra = F1+F2-Rb
Leaves
5kN=F1+F2-3kN
F1-F2=5kN-3kN
F1=F2+2kN
Now
F1+F2 must equal 8kN
So F1=6kn
F2=2kN
Would this be an acceptable route??

I looked at your problem and didn't get it. Could you upload a sketch of what's going on? These arms the balancing weights are mounted on stick out of the shaft 100 mm?

The image shows the dimensions please not though f1 and f2 are downwards though and the reactions upwards.
The shaft is 2 metres long.
Bearing reaction A is place at 0 metres and is 5kN upwards
Bearing reaction B is placed at 2meres and is 3kN upwards
There are point masses downwards at 0.5 metres from the left hand end and a point mass at 1.5 metres also downwards
Hope this helps....

The image shows the dimensions please not though f1 and f2 are downwards though and the reactions upwards.

huh?
What about the "radius arms" that are 100mm each. They won't usually give dimensions to something in a problem unless it's important.

The balancing weights are placed 180 deg from the bearing reactions? Don't these weights rotate with the shaft? How is that 180 deg maintained?

Still not getting it.

I used the radius arms to find the masses once I worked out what the forces of F1 and F2 were. My problem was justifying how I got the forces F1 and F2 rather than just plugging in various different values so the radius arms are irrelevant for what I am trying achieve. This below is what I am trying to solve then from there I solve for the masses using the radius arms....
Ra=5kN Rb=3kN
Ra + (Rb) = F1+F2
Ra = F1+F2-Rb
Leaves
5kN=F1+F2-3kN
F1-F2=5kN-3kN
F1=F2+2kN
Now
F1+F2 must equal 8kN
So F1=6kn
F2=2kN

Originally Posted by Jock

F1-F2=5kN-3kN

where does this come from?

This is where I need the help Iam not sure how to get the F2 = 2kn and F1 = 6kn as required. Any ideas????
I thought by moving the F1+F2 from the right hand side of the equation to left I could change them from addition to subtraction

It appears that you have determined forces F1 and F2 that cause reactions Ra and Rb for a weightless horizontal stationary beam. You've assumed the shaft is not rotating at all - it's a beam with the two masses F1 and F2 and two simple supports at Ra and Rb. So by setting sum of vertical forces = 0 and sum of moment about either end = 0 you can determine that F1= 6N and F2=2 N.

However, for a rotating shaft you want to balance the centripetal forces of the two masses with the bearing reaction forces, which gives:

$\displaystyle \sum F = 0 = R_a+R_b - m_1 \omega^2 R - m_2 \omega^2 R$

$\displaystyle \sum T_a = 0 = 0.5 (m_1 \omega^2 R) + 1.5(m_2 \omega^2 R) - 2(R_b)$

Now solve for m_1 and m_2.

Surely you need to know the force if you don't know the mass though??
I follow the rest of the answer though

Originally Posted by Jock

Surely you need to know the force if you don't know the mass though??

I have no idea what you mean. From your previous work you know that $\displaystyle F_1= 6N$; this force equals $\displaystyle m_1\omega^2R$, and $\displaystyle F_2= 3N = m_2 \omega^2 R$

I only got the F1 and F2 values through crunching numbers for a day then in turn I got the masses I needed . I need to find a better route to getting to that point rather than telling someone it was through trial and error

As I indicated back in post #11: there are two equations in two unknowns:

Sum of vertical forces = 0: $\displaystyle \sum F = 0 = R_a + R_b -(F1 + F_2)$

Sum of moments about one end or the other = 0. Let's take moments about point A: $\displaystyle \sum T_a = 0 = 0.5F1 + 1.5F_2- 2R_b$

Now you can solve for F1 and F2, and you should find F1=6N and F2=2N.

BTW, what class are you doing this for? Using this method is introductory mechanics....