| New Reply |
'Reversing' a moments calculation |
Share Thread | Thread Tools |
| Feb7-11, 01:51 PM | #1 |
|
|
'Reversing' a moments calculation![]() 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. 1. The problem statement, all variables and given/known data 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): ![]() 2. Relevant equations 3. 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? |
| New Reply |
| Thread Tools | |
Similar Threads for: 'Reversing' a moments calculation
|
||||
| Thread | Forum | Replies | ||
| Statistical moments and multipole moments | General Math | 0 | ||
| Moments - Principle of moments question | Introductory Physics Homework | 4 | ||
| help calculation torque, moments. | Precalculus Mathematics Homework | 8 | ||
| Calculation of Moments of Inertia | Introductory Physics Homework | 10 | ||
| nuclear magnetic moments and electric quadrupole moments | General Physics | 0 | ||