# Balancing a plank on springs (Torques)

1. Apr 9, 2014

### notabigfan

Hello, I would like some help with the following question. I highly appreciate your help ! I got an answer that doesn't make sense so I know there must be a concept that I don't get.

There is a complicated piece of machinery sitting on a plank of metal which
is 2 m long. The metal and machinery have a combined mass of 430 kg. The center
of mass is approximately 1.2 m from the left side of the plank. We need to attach 2
springs to the plank to lift it off the ground, but it needs to be level. One spring has
a spring constant of 2200 N/m and is attached to the right edge of the plank. If the
other spring has a spring constant of 1500 N/m, where should we put it to make sure
everything is level?

So what I did is that I wanted to balance torques and forces.

Forces:

mg= Fs1 + Fs2
mg= 2200Δx + 1500 Δx ( Since it is level, then I assumed that delta x is the same)
mg = 3700 Δx
430*9.8= 3700 Δx
Δx = 1.4

Then, I started balancing torques.
I have set the left end to be my pivot.
mg(1.2)= Fs2y ( Where y is the distance of the second spring from the left end)
mg(1.2) = KΔx y
430*9.8*1.2 = 1500*1.4 y

y = 2.4 , which is obviously impossible since the length of the plank is just 2 metres.

2. Apr 9, 2014

### haruspex

What about the spring at the right-hand end?

3. Apr 9, 2014

### BvU

Hello small fan,

Could you please use the template ? There's good reasons for it being required in PF. Read the guidelines if you don't believe me.

I don't understand how you can find Δx = 1.4 from 430*9.8= 3700 Δx ?

Your torque balancing makes a strange impression on me.
Why is Fs1 missing ?
Did you make a drawing ?

Ha Haru! Please take over, bedtime for me!

4. Apr 9, 2014

### notabigfan

Can't I just use one pivot ?

Hey ! I was wondering why do I need to consider both ends? The object has one pivot.

5. Apr 9, 2014

### notabigfan

Hey !

I thought I am following it. I am sorry if I didn't. Ok .. I multiply 430 by 9.8 and divide the product by 3700.
Δx is the compression of the springs.

6. Apr 9, 2014

### notabigfan

Also, Fs1 is missing because it passes through the pivot. So the torque it's producing is equal to zero.

If you want to show me a better approach please do. Thanks :)

7. Apr 9, 2014

### haruspex

Not according to what you posted. You have one spring at the right hand end, one spring at some unknown distance from the left hand end, and a pivot at the left hand end.

8. Apr 9, 2014

### notabigfan

My bad ... got it. Thanks :)