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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 04:55] (current) – pwirving
@@ Line 16: / Line 16: @@
 === Approximations & Assumptions ===
   * Origin is at chamber wall $\langle 0,0,0 \rangle\,m$
   * Assume no forces due to drag or to friction
@@ Line 23: / Line 22: @@
 === Representations ===
- $ {\vec F_{spring}} = -k_ss\hat{L}$
+ $ {\vec F_{spring}} = -k_s\vec{s}$
- $ s = |\vec L| - L_0$
+ $ |\vec{s}| = |L - L_0|$
+{{183_notes:spring237.jpg}}
+{{183_notes:spring_235.jpg}}
-{{183_notes:spring_force_jpeg.jpg}}
 ==== Solution ====
- $\vec{L} = \langle 0.38,0,0 \rangle m - \langle 0,0,0 \rangle m = \langle 0.38,0,0 \rangle m$
+To determine the spring force, you will need to compute:
+$$  {\vec F_{spring}} = -k_s\vec{s} = -k_s|\vec{s}|\hat{s}$$
- $|\vec{L}| = 0.38m$
+You will start be determining the position vector ($\vec{L}$) of the mass and the length of the position vector ($|\vec{L}|$),
+ $$\vec{L} = \langle 0.38,0,0 \rangle m - \langle 0,0,0 \rangle m = \langle 0.38,0,0 \rangle m$$
- $\hat{L} = \dfrac{(0.38,0,0)}{0.38} = \langle 1,0,0 \rangle m$
+ $$|\vec{L}| = 0.38m$$
- $ 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:
+ $$\hat{s} = \hat{L} = \dfrac{\langle 0.38,0,0\rangle}{0.38} = \langle 1,0,0 \rangle$$
- $\vec{F} = -(8N/m)(0.18m)(1,0,0) = \langle 1.44,0,0 \rangle N/m$
+You can then compute the magnitude of the stretch $(|\vec{s}|)$:
+ $$ |\vec{s}| = |L - L_0| = 0.38m - 0.20m = 0.18m$$
+Finally, you can compute the force:
+$$\vec{F} = -k_s|\vec{s}|\hat{s} = -(8N/m)(0.18m)\langle 1,0,0\rangle = \langle -1.44,0,0 \rangle\,N$$
+which points to the left. That is consistent with the diagram above.