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Am I missing something? enter image description here

It seems obvious to me that at $+A$ and $-A$, the spring has restorative forces equal in magnitude but opposite in direction. But since gravity is always pulling it down, the spring at position $-A$ must have less net force acting on it. But my book says that at both positions, the spring has its maximum $\sum F = ma$. How does this make sense?

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    $\begingroup$ I think you are correct. $\endgroup$ – paparazzo Jul 29 '15 at 20:09
  • $\begingroup$ Give more text. What is lower case a. A is upper case in the diagram. And the equation does not even include gravity. If that is a horizontal spring is a is acceleration then that would be true. $\endgroup$ – paparazzo Jul 29 '15 at 21:31
  • $\begingroup$ Even though the question is not detailed it is well posed, the equation $\sum F = ma$ does not need to include gravity since it is implicit in $F$ and $a$ is simply the acceleration. $\endgroup$ – Andre Maizel Jul 29 '15 at 22:38
  • $\begingroup$ I don't ignore gravity. As said before, it is implicit in $\sum F$ since it means the sum over all forces, which in this case are gravity and spring tension. $\endgroup$ – Andre Maizel Jul 29 '15 at 23:14
  • $\begingroup$ What do you mean by "force acting on it?" do you mean the net force acting on the spring by the top attachment point, weight of the hanging mass, and gravity on the spring mass? Or do you mean the force the spring exerts on the hanging mass? $\endgroup$ – Bill N Jul 30 '15 at 20:47
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The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a $\Delta x$ amount with $F_{spring}(0) = k\Delta x$.

So the spring force on point A is a little smaller than in point -A, since $ F_{spring}(A) = -k(A-\Delta x)$ and $ F_{spring}(-A) = k(A+\Delta x)$ so it compensates the "extra" force.

You have to notice that in this equilibrium position

$F_{spring} - mg = 0$ ,

so

$F_{spring} = k\Delta x = mg$

with

$\Delta x = mg/k$.

Substituting in

$ F_{net}(A) = F_{spring}(A) - mg = -k(A-\Delta x) - mg = -k(A - \frac{mg}{k}) - mg = -kA $

the same hold for the -A position

$ F_{net}(-A) = F_{spring}(-A) - mg = -k(-A-\Delta x) - mg = -k(-A-\frac{mg}{k}) - mg = kA $

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  • $\begingroup$ Why spring force equation always has a negative sign? $\endgroup$ – most venerable sir Jul 29 '15 at 23:04
  • $\begingroup$ If I am not mistaken, in your second to last equation, you want to add "mg" instead of subtracting b/c the two forces are in the same direction. $\endgroup$ – most venerable sir Jul 29 '15 at 23:06
  • $\begingroup$ Because it the force always act in the opposite direction of the compression/distension. If $\delta x$ is the deformation of the spring(with the same reference system as in the figure) we have for $ \delta x < 0$ a force that points in the upward direction, and if $\delta x > 0$ we have a force that points downwards. Note that in this case $\delta x < 0$ means elongation and $\delta x > 0$ means compression. $\endgroup$ – Andre Maizel Jul 29 '15 at 23:07
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    $\begingroup$ I always subtract mg because the reference system is such that $mg$ is negative. Notice that the net force is negative and points downwards, which means that the spring is pushing the particle down. $\endgroup$ – Andre Maizel Jul 29 '15 at 23:12
  • $\begingroup$ it is funny that my answer is basically the same as yours, your answer is +5 and mine is -1. $\endgroup$ – sintetico Jul 30 '15 at 10:28
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In accordance with Hooke's law the force is linear with distance. Incorporating gravity only means that the equillibirum position of the spring has changed, the "zero" around which it oscillates. The gravitational pull is already compensated by the spring. Thus the magnitude of the force is euqal at $-A$ and $+A$.

Edit: When the gravitational pull on the mass on the spring is considered, the spring elongates. This results in a new equillibirum position $x'_0 = x_0 + \Delta x = x_0 + \frac{m g}{k}$. Since the force is always (in Hooke's regieme) linear with distance, you can just neglect the gravitational force since it is compensated by the spring. It is a simple superposition of forces.

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  • $\begingroup$ Is your first sentence the reason why "incorporating gravity" will change the equilibrium position? Can you be more detailed? $\endgroup$ – most venerable sir Jul 29 '15 at 22:56
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The answer is no. The net force at the maximum elongation points has the same magnitude. This is because the rest point of the spring is modified by the gravitational weight of the mass. The mass oscillates around this new rest point, and at the points of maximum amplitude the net force is the same. One can tell that the gravitational force can be "integrated out".

To see this in more details, consider the net force on the mass, which in this case is the sum of the elastic force and the gravitational force (which is constant): $$ F=mg- k (x-x_0) $$ where $x_0$ is the rest position of the spring (without the mass). Now you can calculate the rest position $x_0'$ of the spring including the effect of gravitation, which is defined as the point where $F=0$. So you obtain that at the position $x=x_0'$ you have $$ 0=mg- k (x_0'-x_0)\Rightarrow x_0'=x_0+mg/k $$ Now, what happens if one changes coordinates? $$ F=mg- k (x-x_0)=mg- k [x-(x_0'-mg/k)]=mg- k [x-x_0']-mg=-k(x-x_0') $$ which means that, if one considers the displacement with respect to the new rest point $x_0'$, the force is simply given by $$ F=-k(x-x_0') $$ Now it is easy to see that at the two points of maximum elongation the net force is the same in magnitude, but opposite in sign.

Edit: Consider the maximum elongation $A$, which is measured with respect to the new rest point $x_0'$. One can reverse the transformation I'v done and obtain that the net force at $x-x'_0=\pm A$ (i.e., $x=x'_0\pm A$) is $$ F=mg- k (x'_0\pm A-x_0)=\mp k(x-x_0') $$ Note that the net force is the same in magnitude, but the elongation of the spring with respect to the original rest point $x_0$ (without considering the mass $m$) is not the same but it is $(x'_0-x_0\pm A)$.

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  • $\begingroup$ I think you are missing acceleration. At the to of bungee jump I feel weightless and at the bottom I fell heavy. $\endgroup$ – paparazzo Jul 29 '15 at 21:22
  • $\begingroup$ gravitational acceleration is still there. $\endgroup$ – sintetico Jul 29 '15 at 21:26
  • $\begingroup$ Not gravity. Acceleration from forces the the spring. $\endgroup$ – paparazzo Jul 29 '15 at 21:28
  • $\begingroup$ So, feeling "heavy" or "light" has of course a lot to do with acceleration, but it is not a measure of acceleration. $\endgroup$ – sintetico Jul 29 '15 at 22:24
  • $\begingroup$ What? Gravity does not change. Not possible for net force to be the same and elastic force to change. My mass does not change. If I feel no force then my net acceleration is zero. How hard is my bungee example? $\endgroup$ – paparazzo Jul 29 '15 at 22:49

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