Balancing a plank on springs (Torques)

  • Thread starter notabigfan
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
haruspex
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mg(1.2)= Fs2y ( Where y is the distance of the second spring from the left end)
What about the spring at the right-hand end?
 
  • #3
BvU
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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
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Can't I just use one pivot ?

What about the spring at the right-hand end?
Hey ! I was wondering why do I need to consider both ends? The object has one pivot.
 
  • #5
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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!
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
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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!
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
haruspex
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Also, Fs1 is missing because it passes through the pivot.
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
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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.
My bad ... got it. Thanks :)
 

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