Spring forces

Every spring has a natural length, the length it takes when nothing pulls or pushes on it. Move it away from that length and it resists. Stretch it and it pulls back in; compress it and it pushes back out. The further you move it, the harder it resists, and for an ideal spring that relationship is a straight line.

That is Hooke's law: $F_s = -k\cdot\Delta x$. Here $\Delta x$ is the displacement from the natural length, $k$ is the spring constant in newtons per meter, and the minus sign sets the direction. A larger $k$ means a stiffer spring, so the same stretch produces a larger force. The size of the spring force is $k\cdot\Delta x$; the direction is what the minus sign carries.

Two questions run through the topic: how big is the spring force, and which way does it point? The size is easy, $k$ times the displacement. The direction is where the points are lost, because the spring force never points the way you moved the spring. It points back toward where the spring wants to be (§2).

A spring also rarely acts on its own. Hang a mass from it or connect it between two carts and the resting position is set by the whole force balance, not by the spring alone (§4). And when two springs are combined, the system gets stiffer or softer depending on how they are joined (§5).

The most common spring mistake is to write Hooke's law as $F_s = k\cdot\Delta x$ and lose the minus sign. The magnitude comes out right, but the direction comes out backward, as if a stretched spring pushed outward along the pull. It does the opposite.

The minus sign makes the spring force a restoring force: it always points opposite the displacement, back toward the natural length. Stretch the spring to the right ($\Delta x > 0$) and the force points left, back toward the natural length. Compress it to the left ($\Delta x < 0$) and the force points right, back toward the natural length. Force and displacement always have opposite signs.

Graph the spring force against $\Delta x$ and this shows up clearly. The line runs through the origin with a negative slope of $-k$: when $\Delta x$ is positive, $F_s$ is negative, and the other way around. A line with positive slope, where force and displacement share a sign, is the dropped-minus-sign error drawn out.

Keeping the minus sign is not just bookkeeping. The restoring direction is exactly what makes a displaced spring snap back, overshoot, and return again, which is the oscillation you meet when springs come back in Unit 7. Drop the sign and that whole behavior disappears.

Setup. A block is attached to a spring with $k = 200$ N/m. Take rightward as positive, with the spring's natural length at the origin. (a) The block is pulled $0.10$ m to the right. (b) Instead it is pushed $0.05$ m to the left. Find the spring force in each case, magnitude and direction.

(a) Stretched right, $\Delta x = +0.10$ m. Use the signed law: $F_s = -k\cdot\Delta x = -200 \cdot (+0.10) = -20$ N. The magnitude is $20$ N and the sign is negative, so the force points left, back toward the natural length. It is not $+20$ N along the stretch.

(b) Compressed left, $\Delta x = -0.05$ m. Now $F_s = -k\cdot\Delta x = -200 \cdot (-0.05) = +10$ N. The magnitude is $10$ N and the sign is positive, so the force points right, again back toward the natural length. The two minus signs, one from the law and one from the displacement, multiply to a positive force.

The trap to avoid. In part (a) the answer is $20$ N pointing left, not $+20$ N pointing right along the pull. The magnitude $k\cdot\Delta x$ is only half the answer. The minus sign decides the direction, and in both cases it sends the force back toward the natural length.

Hooke's law gives the spring force at any displacement, but it does not by itself say where the spring will sit. The resting point comes from the full force balance, with the spring as one force among several.

A hanging mass. Hang a mass $m$ from a spring and it stretches until the spring's upward pull balances gravity: $k\cdot\Delta x = mg$, so the stretch is $\Delta x = mg/k$. The new equilibrium is below the natural length, and how far below depends on $m$, $g$, and $k$ together. The spring force there is not zero; it is $mg$, exactly enough to hold the weight.

Two connected carts. Put two carts on a frictionless track joined by a light spring and push only on cart $A$ with a force $F$. It is tempting to treat the pair as one particle: the center of mass accelerates at $a_{cm} = F/(m_A + m_B)$. That is true for the system as a whole, but it does not give either cart's individual motion. Cart $B$ feels only the spring force, so its acceleration is set by that spring force and its own mass, not by the combined mass.

The rule: use the center-of-mass result for whole-system quantities, and a separate free-body diagram for any single piece. Mixing them, by reading the system acceleration as one cart's acceleration, is the trap. The center-of-mass picture comes from Topic 2.1; momentum, which makes the connected-cart analysis even cleaner, arrives later in the course.

Two springs can be combined into one effective spring, and the rule depends on how they are joined. The catch is surprising: joining two springs end to end makes the pair softer, not stiffer, and that is exactly where the rules get swapped.

Parallel. Springs side by side share the load. They both stretch by the same amount, their forces add, and the combination is stiffer: $k_{eq} = k_1 + k_2$. The effective constant is always larger than either spring alone.

Series. Springs joined end to end share the stretch. The same force runs through both, each one stretches, and the stretches add, so the chain is softer: $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$. The effective constant is always smaller than the smaller spring.

The fastest check is physical, not algebraic. A series pair is a longer, floppier chain that is easier to stretch, so its constant must fall below the smallest spring. A parallel pair pulls together and resists more, so its constant must rise above the largest. Two identical springs of constant $k$ give $k/2$ in series and $2k$ in parallel. If your series answer is bigger than the springs, or your parallel answer is smaller, you have swapped the rules.

Setup. Two springs have $k_1 = 300$ N/m and $k_2 = 600$ N/m. (a) Find the effective constant in series and in parallel. (b) A $3$ kg mass hangs from the series combination. How far does it stretch? Use $g = 10$ N/kg.

(a) Series. $\frac{1}{k_{eq}} = \frac{1}{300} + \frac{1}{600} = \frac{2}{600} + \frac{1}{600} = \frac{3}{600}$, so $k_{eq} = 200$ N/m. That is below $300$, the smaller spring, just as a softer chain should be. Parallel. $k_{eq} = 300 + 600 = 900$ N/m, above $600$, the larger spring.

(b) Stretch of the series pair. The hanging mass settles where the spring pull balances gravity: $k_{eq}\cdot\Delta x = mg$, so $\Delta x = mg/k_{eq} = (3 \cdot 10)/200 = 0.15$ m.

The trap to avoid. Adding the constants for the series pair would give $900$ N/m, the parallel answer, and a stretch four and a half times too small. The reciprocal sum is for series, the plain sum is for parallel, and the sanity check, series softer than the smallest, catches the slip every time.