- #1
notabigfan
- 12
- 0
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.
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.