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183_notes:examples:calculating_the_force_due_to_a_stretched_spring [2014/07/22 02:31] – caballero | 183_notes:examples:calculating_the_force_due_to_a_stretched_spring [2014/07/22 04:39] – pwirving | ||
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=== Approximations & Assumptions === | === Approximations & Assumptions === | ||
- | * Origin is at chamber wall $\langle 0,0,0 \rangle$ | + | * Origin is at chamber wall $\langle 0,0,0 \rangle\,m$ |
* Assume no forces due to drag or to friction | * Assume no forces due to drag or to friction | ||
| | ||
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=== Representations === | === Representations === | ||
- | $ {\vec F_{spring}} = -k_ss\hat{L}$ | + | $ {\vec F_{spring}} = -k_s\vec{s}$ |
- | | + | $ |\vec{s}| = |L - L_0|$ |
{{183_notes: | {{183_notes: | ||
+ | <WRAP todo> | ||
==== Solution ==== | ==== Solution ==== | ||
- | $\vec{L} = \langle 0.38,0,0 \rangle m - \langle 0,0,0 \rangle m = \langle 0.38, | + | To determine the spring force, you will need to compute: |
+ | $$ {\vec F_{spring}} = -k_s\vec{s} | ||
- | $|\vec{L}| = 0.38m$ | + | You will start be determining the position vector ($\vec{L}$) of the mass and the length of the position vector ($|\vec{L}|$), |
+ | | ||
- | $\hat{L} = \dfrac{(0.38, | + | $$|\vec{L}| = 0.38m$$ |
- | $ s = 0.38m - 0.20m = 0.18m$ | + | These can be used to compute the unit (direction) vector for the stretch ($\hat{s}$), which is in the same direction as the position vector: |
+ | | ||
- | $\vec{F} = -(8N/m)(0.18m)(1,0,0) = \langle 1.44,0,0 \rangle N/m$ | + | You can then compute the magnitude of the stretch |
+ | $$ |\vec{s}| | ||
+ | Finally, you can compute the force: | ||
+ | $$\vec{F} = -k_s|\vec{s}|\hat{s} = -(8N/ | ||
+ | which points to the left. That is consistent with the diagram above. | ||