# Stretch length of horizontal and vertical springs

If the spring in Figure A is stretched a distance d, how far will the spring in Figure B stretch? The spring constants are the same.

The answer is "by half". I don't get it, to me it's the same.

• Possible duplicates: physics.stackexchange.com/q/41291/2451 , physics.stackexchange.com/q/45247/2451 and links therein. – Qmechanic Jan 12 '15 at 9:56
• By half. Replace one of the weights with a connection to "solid ground". What changes? – Hot Licks Jan 12 '15 at 20:17
• Yeah, that's a duplicate, but I far prefer this simple question with an obvious picture. +1 – Almo Jan 13 '15 at 8:51
• @Mörre - How can the weight on one side "know" whether the thing on the other side is a weight or a connection to solid ground? – Hot Licks Jan 14 '15 at 12:23

Think on the equilibrium position:

1. in figure A you have a force $mg$ exerted at the lower end and an identical force pushing the opposite direction exerted by the wall (if this force didn't exist, the spring with mass attached would just fall down by gravity).
2. In figure B you have a force $mg/2$ exerted on the right end of the spring and an identical force of $mg/2$ on the other end.

Start with figure A and ask what is the force exerted on the mass $m$ by the end of the spring? We know the mass $m$ is stationary, so the net force on it is zero, and we know there is a gravitational force downwards of $mg$. So the force exerted on the mass by the spring is an upwards force of $mg$ and therefore the tension in the spring must be $mg$.

Now take figure B and apply the same reasoning to either of the masses on the ends of the spring. The same argument tells us that the tension in the spring is only half as great. And that's why the spring extension is only half as big.

I suspect the confusion arises because in figure A it's tempting to ignore the upper end of the spring i.e. the joint where the upper end of the spring is attached. However the joint must be exerting an upwards force on the spring of $mg$ to balance out the downwards force of $mg$ exerted by the mass $m$. So in effect the spring has two blocks of mass $m$, one at each end, with one mass pulling down and the other pulling up. Now compare to figure B, which has a mass of $m/2$ at each end, and it should be obvious why the extension is only half as great in B.

• This picture is misleading. The spring on the left is equivalent to a spring with double the length on the right. This is seen if you treat the fixed wall on the left as a mirror. – ja72 Jan 26 '15 at 18:47

The easiest way for me to understand this is as follows:

Consider a "free body diagram" of each spring alone.

In case one, a force $M*G$ pulls down on the spring. An equal and opposite force pulls up on the spring at the ceiling to satisfy the spring's equilibrium (it's not accelerating). Therefore, the force everywhere within the spring is $M*G$. Because of this we end up with $M*G=k*X$.

In case two, a force pulls on the left of the spring and a force pulls on the right of the spring. The left force must also support the weight $\frac{m*g}{2}$. The right force must support the right weight in the same fashion. Therefore, the spring is in equilibrium by these two opposing forces. Everywhere within the spring the force felt is $\frac{m*g}{2}$. Because of this we end up with $\frac{m*g}{2}= k*X$

## protected by Qmechanic♦Jan 15 '15 at 20:39

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).