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--- 183_notes:examples:calculating_the_force_due_to_a_stretched_spring [2014/07/22 02:42] – caballero
+++ 183_notes:examples:calculating_the_force_due_to_a_stretched_spring [2014/07/22 04:50] – pwirving
@@ Line 27: / Line 27: @@
 {{183_notes:spring_force_jpeg.jpg}}
+{{183_notes:spring_235.jpg}}
 <WRAP todo>This FBD needs the forces of the Earth and the surface. It should also be abstract. For example, see the early fan cart problems.</WRAP>
 ==== Solution ====
-<WRAP todo>Needs more steps/explanatory words; Remember they are intro students</WRAP>
 To determine the spring force, you will need to compute:
@@ Line 42: / Line 42: @@
 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:
- $$\hat{s} = \hat{L} = \dfrac{(0.38,0,0)}{0.38} = \langle 1,0,0 \rangle m$$
+ $$\hat{s} = \hat{L} = \dfrac{\langle 0.38,0,0\rangle}{0.38} = \langle 1,0,0 \rangle$$
-You can then compute the magnitude of the stretch (|\vec{s}|):
+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:
-$$\vec{F} = -k_s|\vec{s}|\hat{s} = -(8N/m)(0.18m)(1,0,0) = \langle 1.44,0,0 \rangle\,N/m$$
+$$\vec{F} = -k_s|\vec{s}|\hat{s} = -(8N/m)(0.18m)\langle 1,0,0\rangle = \langle -1.44,0,0 \rangle\,N$$
+which points to the left. That is consistent with the diagram above.