So the title could be Finding Equilibrium in a Mass Spring System with Torque

[Dren]

1.

Homework Equations

// Tension of the spring
// Where sinL between `bottom right` of the spring and the bar
Ft = sinL * kx

// Where x0 is the initial length of the spring
Ft = sinL * k(x - x0)

// Force of gravity
Fg = -mg

// Find x0
// sqr = square root
x0 = sqr(100^2 + 100^2)
// Converted to meters
x0 = 1.421

Ft + Fg = 0
sinL * k(x - x0) - mg = 0
sinL * k(x - x0) = mg
1/1.421 * 1000(x - 1.421) = 10 * 9.8
(x - 1.421) = 98 / 1/1.421 * 1000
x - 1.421 = 98 / 703.73
x = 1.5602579540448752

// Converted to centimeters and rounded
x = 156cm

But for some reason this doesn't seem to be the right answer. ;(

If someone could help me would be really appreciated.

Thanks.

 

[billy_joule]

x0 = 1.421

x = 1.5602579540448752

Does a spring get longer or shorter when you pull on it's ends?

 

[Dren]

Does a spring get longer or shorter when you pull on it's ends?

I think longer, isn't it?

 

[Dren]

Does a spring get longer or shorter when you pull on it's ends?

I meant x to be bigger than x0 after the rigid bar has been starching the spring.

 

[billy_joule]

I think longer, isn't it?

Yes.

I meant x to be bigger than x0 after the rigid bar has been starching the spring.

The diagram shows the spring already stretched, with length $\sqrt2~ m$ as you've found.
If the bar were removed, the spring would contract to its relaxed length. which must be $< \sqrt2~ m$

 

[Dren]

Yes.The diagram shows the spring already stretched, with length $\sqrt2~ m$ as you've found.
If the bar were removed, the spring would contract to its relaxed length. which must be $< \sqrt2~ m$

I see, so I guess I was trying to solve the wrong problem from the beginning, so the question is: find the resting length if the bar is removed?

I this case I have to subtract the length caused by the bar on the spring, so that we'd find the original length of the spring without the bard acting on it?

BTW: Sorry for my silly questions.

 

[Dren]

I see, so I guess I was trying to solve the wrong problem from the beginning, so the question is: find the resting length if the bar is removed?

I this case I have to subtract the length caused by the bar on the spring, so that we'd find the original length of the spring without the bard acting on it?

BTW: Sorry for my silly questions.

Updated my comment.

 

[billy_joule]

I see, so I guess I was trying to solve the wrong problem from the beginning, so the question is: find the resting length if the bar is removed?

I this case I have to subtract the length caused by the bar on the spring, so that we'd find the original length of the spring without the bard acting on it?

Yes,that's one way of saying it.

Ft + Fg = 0

This isn't right.
There are three forces acting on the bar - it's weight force, the springs force, and a reaction force where it's connected to the wall.

The simplest way to find T is take moments (torques) about the bar-wall joint, that way we don't need to consider the wall reaction.

 

[Dren]

Yes,that's one way of saying it.This isn't right.
There are three forces acting on the bar - it's weight force, the springs force, and a reaction force where it's connected to the wall.

The simplest way to find T is take moments (torques) about the bar-wall joint, that way we don't need to consider the wall reaction.

Does the following sound reasonable:$m = 10kg$
$g = 9.8m/s^2$
$k = 1000N/m$

$x0 = \sqrt{2}$
$theta = 45°$

$T = k * x$

$Ft = L * T * sin(theta)$
$Fg = L/2 * m * g$

$L * T * sin(theta) = L/2 * mg$
$2L * T * sin(theta) = mg$
$T = 1/2 * mg / sin(theta)$

$k * x = 1/2 * mg / sin(theta)$
$x = (1/2 * mg / sin(theta)) / k$
$x = (1/2 * 98 / \sqrt{2}/2) / 1000$

$x = 98 / \sqrt{2} * 1000$
$x ~= 0.06929m$

// Now finally subtract from the original length
$diff = x0 - x$
$diff = \sqrt{2} - 0.06929$
$diff = 1.3449170978168133914017m$
$diff = 1.3449170978168133914017 * 100$

// Rounded
$diff = 134cm$

 

[Dren]

I think it seems to be ok.

 

[billy_joule]

Yes, looks good.
Except here you call moments forces;

$Ft = L * T * sin(theta)$
$Fg = L/2 * m * g$