'Reversing' a moments calculation

[Beamy]

[URL]http://homepage.ntlworld.com/russelliott/shifting-cofg-1.png[/URL]

A rigid weightless beam is supported on 3 springs, spaced as marked. The forces in the 3 springs are initially equal (at 10N). The position of the 30N balancing force F to keep the system in equilibrium is determined from taking moments about F:

f1x1 + f2x2 = f3x3
10(20 + x2) + 10x2 = 10(29 - x2) [substituting values, to isolate x2]
200 + 10x2 + 10x2 = 290 - 10x2 [expanding]
30x2 = 90
x2 = 3

So far so good.

Homework Statement

What I'm trying to determine is what happens to the spring forces f1, f2 and f3 if the position of the balancing force F is shifted, say a further short distance x4 to the right (and with F remaining at 30N):

[URL]http://homepage.ntlworld.com/russelliott/shifting-cofg-2.png[/URL]

Homework Equations

The Attempt at a Solution

From the second diagram I get a number of moment equations:

f1(23 + x4) + f2(3 + x4) = f3(26 - x4) [taking moments about F]
23f1 + f1x4 + 3f2 + f2x4 = 26f3 - f3x4 [expanded]
f1x4 + f2x4 + f3x4 = 26f3 - 23f1 - 3f2 [further expanded]
30x4 = 26f3 - 23f1 - 3f2 [given that f1 + f2 + f3 = F = 30]

30(23 + x4) = 20f2 + 49f3 [taking moments about f1]
30x4 = 20f2 + 49f3 - 690

20f1 + 30(3 + x4) = 29f3 [taking moments about f2]
30x4 = 29f3 - 20f1 - 90

49f1 + 29f2 = 30(26 - x4) [taking moments about f3]
30x4 = 780 - 49f1 - 20f2

but I can't isolate f1, f2 or f3. What mathematical technique do I need?

 

[member 553083]

To solve for the three unknown forces f1, f2, and f3, you will need to use a system of three equations with three unknowns. You can use substitution or elimination to solve the system of equations.