# Mass Spring System and Torque

1. Oct 8, 2016

### Dren

1.

2. Relevant 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.

2. Oct 8, 2016

### billy_joule

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

3. Oct 8, 2016

### Dren

I think longer, isn't it?

4. Oct 8, 2016

### Dren

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

5. Oct 8, 2016

### billy_joule

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 $< \sqrt2~ m$

6. Oct 8, 2016

### 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.

Last edited: Oct 8, 2016
7. Oct 8, 2016

### Dren

Updated my comment.

8. Oct 8, 2016

### billy_joule

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.

9. Oct 9, 2016

### Dren

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
$diff = 134cm$

10. Oct 10, 2016

### Dren

I think it seems to be ok.

Last edited: Oct 10, 2016
11. Oct 10, 2016

### billy_joule

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