• Create Account

Banner advertising on our site currently available from just \$5!

### #ActualBuckeye

Posted 11 February 2014 - 08:10 AM

Couple questions:

1. Is figure 1 the initial condition? That is, are all the masses all and all the spring forces are zero?

I would like calculate the new position of m2 and m3 mass after that user move the m4 mass.

2. If you're only interested in calculating the final position of m4, you don't need to consider damping coefficients. Damping is only a function of velocity. When all the masses are at rest there is no damping. The mass positions will all be colinear (all in a line).

This is just a quick derivation, so double check the algebra!

The letter L represents the distance between two masses
and assumes each spring was at it's rest position before m4 was moved

and ALL k's and distances are > 0.

Forces on m2 must sum to 0 since masses are at rest:
k12 * L12 = k24 * L24, so
1) L12 / L24 = k24 / k12

Forces on m3:
k13 * L13 = k34 * L34
2) L13 / L34 = k34 / k13

Distances:
You know L14 from the final position of m4, so:

L12 + L24 = L14
L13 + L34 = L14

L12 = L14 - L24
1) becomes ( L14 - L24 ) / L24 = k24 / k12
or L14 / L24 - 1 = k24 / k12
L14 / L24 = k24 / k12 + 1
L24 = L14 / ( k24 / k12 + 1 ) [everything on the right is known]

Similarly for L13 / L14

### #3Buckeye

Posted 11 February 2014 - 08:00 AM

Couple questions:

1. Is figure 1 the initial condition? That is, are all the masses all and all the spring forces are zero?

I would like calculate the new position of m2 and m3 mass after that user move the m4 mass.

2. If you're only interested in calculating the final position of m4, you don't need to consider damping coefficients. Damping is only a function of velocity. When all the masses are at rest there is no damping. The mass positions will all be colinear (all in a line).

This is just a quick derivation, so double check the algebra!

The letter L represents the distance between two masses
and assumes each spring was at it's rest position before m4 was moved.

Forces on m2 must sum to 0 since masses are at rest:
k12 * L12 = k24 * L24, so
1) L12 / L24 = k24 / k12

Forces on m3:
k13 * L13 = k34 * L34
2) L13 / L34 = k34 / k13

Distances:
You know L14 from the final position of m4, so:

L12 + L24 = L14
L13 + L34 = L14

L12 = L14 - L24
1) becomes ( L14 - L24 ) / L24 = k24 / k12
or L14 / L24 - 1 = k24 / k12
L14 / L24 = k24 / k12 + 1
L24 = L14 / ( k24 / k12 + 1 ) [everything on the right is known]

Similarly for L13 / L14

### #2Buckeye

Posted 11 February 2014 - 07:59 AM

Couple questions:

1. Is figure 1 the initial condition? That is, are all the masses all and all the spring forces are zero?

I would like calculate the new position of m2 and m3 mass after that user move the m4 mass.

2. If you're only interested in calculating the final position of m4, you don't need to consider damping coefficients. Damping is only a function of velocity. When all the masses are at rest there is no damping. The mass positions will all be colinear (all in a line).

This is just a quick derivation, so double check the algebra!

The letter L represents the distance between two masses
and assumes each spring was at it's rest position before m4 was moved.

Forces on m2 must sum to 0 since masses are at rest:
k12 * L12 = k24 * L24, so
1) L12 / L24 = k24 / k12

Forces on m3:
k13 * L13 = k34 * L34
2) L13 / L34 = k34 / k13

Distances:
You know L14 from the final position of m4, so:

L12 + L24 = L14
L13 + L34 = L14

L12 = L14 - L24
1) becomes ( L14 - L24 ) / L24 = k24 / k12
or L14 / L24 - 1 = k24 / k12
L14 / L24 = k24 / k12 + 1
L24 = L14 / ( k24 / k12 + 1 )

Similarly for L13 / L14

### #1Buckeye

Posted 11 February 2014 - 07:56 AM

Couple questions:

1. Is figure 1 the initial condition? That is, are all the masses all and all the spring forces are zero?

I would like calculate the new position of m2 and m3 mass after that user move the m4 mass.

2. If you're only interested in calculating the final position of m4, you don't need to consider damping coefficients. Damping is only a function of velocity. When all the masses are at rest there is no damping. The mass positions will all be colinear (all in a line).

The letter L represents the distance between two masses
and assumes each spring was at it's rest position before m4 was moved.

Forces on m2 must sum to 0 since masses are at rest:
k12 * L12 = k24 * L24, so
1) L12 / L24 = k24 / k12

Forces on m3:
k13 * L13 = k34 * L34
2) L13 / L34 = k34 / k13

Distances:
You know L14 from the final position of m4, so:

L12 + L24 = L14
L13 + L34 = L14

L12 = L14 - L24
1) becomes ( L14 - L24 ) / L24 = k24 / k12
or L14 / L24 - 1 = k24 / k12
L14 / L24 = k24 / k12 + 1
L24 = L14 / ( k24 / k12 + 1 )

Similarly for L13 / L14

PARTNERS