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FRQ: Spring Force and Static Friction — The Shower-Curtain Tension Rod

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  CHALLENGE  |  10 points  |  25 min

▤ Scenario

A spring-loaded tension rod holds a shower curtain across the opening of a tiled shower. The rod consists of two metal tubes, one of which slides inside the other; an ideal spring inside the tubes that pushes them apart; and a rubber cap on each end. To install the rod, the owner pushes the tubes together to shorten it, holds it horizontally between two parallel vertical walls, and lets go, so that the spring presses the rubber caps against the walls, as shown in Figure 1.

Figure 1 — The tension rod installed between two tiled walls; the spring is shown cut away.

The spring constant of the spring is k = 1500 N/m. Installed in this shower, the spring is compressed by Δx = 4.0 cm from its relaxed length. The coefficient of static friction between each rubber cap and the tiled wall is μ_s = 0.40. The rod and the curtain together have a mass of 3.0 kg, distributed symmetrically about the midpoint of the rod, and any curtain hung in this question hangs the same way. The curtain hangs freely from rings on the rod, touching nothing else, and exerts only vertical forces on the rod.

The two tubes slide against each other with negligible friction, and the deformation of the rubber caps and the walls is negligible. The rod remains horizontal and at rest, and all forces lie in the vertical plane that contains the rod. Take the bathroom as an inertial reference frame, with upward and toward the right-hand wall as positive, and use g = 10 N/kg.

✎ Free Response Questions

Part A

i.    Consider the rod — both tubes, the spring and the caps — and the curtain together as a single object, installed as described in the Scenario.

The following rectangle represents the rod and curtain. On the rectangle, draw and label the forces (not components) exerted on the rod and curtain. Draw each force as a distinct arrow starting on, and pointing away from, the point at which the force is exerted.

Figure 2 — The rectangle representing the rod and curtain together.

ii.   Suppose the curtain is replaced by a heavier one, hung in the same way. Derive an expression for the maximum total mass M_max of the rod and curtain for which the rod can remain at rest in this shower without slipping. Express your final answer in terms of k, Δx, μ_s, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

iii.  The owner wants to hang a heavier curtain, so that the rod and new curtain together would have a mass of 6.0 kg. By turning an adjusting collar, the owner can change the rod’s relaxed length, and so change how much the spring is compressed when the rod is installed in the same shower. The spring constant does not change. Calculate the least compression of the spring for which the rod and new curtain would remain at rest without slipping.

Part B

The owner keeps the original curtain and moves the rod, without adjusting it, to a second shower that is identical to the first except that its walls are 2.0 cm closer together, so that the spring is compressed by 6.0 cm. The rod and curtain again remain at rest, with their mass distributed symmetrically about the midpoint of the rod.

Let f_1 be the magnitude of the static friction force exerted on the rod by one wall of the first shower, and let f_2 be the magnitude of the static friction force exerted on the rod by one wall of the second shower.

Indicate whether f_2 is greater than, less than, or equal to f_1 by writing one of the following.

•   f_2 > f_1

•   f_2 < f_1

•   f_2 = f_1

Justify your response.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A (i) — Model Answer

Five forces are exerted on the rod and curtain. Earth exerts the gravitational force F_g = (3.0 kg)(10 N/kg) = 30 N, directed downward, drawn from the middle of the rectangle. Each wall exerts two forces on the rubber cap it touches: a normal force, perpendicular to the wall and so horizontal, directed away from that wall toward the rod’s midpoint; and a static friction force along the wall, directed upward, because without it the rod would slide down.

The spring is inside the rod: it pushes the left tube to the left and the right tube to the right. Both pushes act between two parts of the object, so they are internal forces and get no arrows.

Figure 3 — Forces on the rod and curtain, drawn to one scale; the values are found in A (ii) and B.

Scoring (2 points):

▸ A1 — 1 point: For drawing a horizontal normal force at each end of the rectangle, each starting at that end and directed away from the wall that exerts it, toward the rod’s midpoint, as distinct labelled arrows, with no other horizontal force.

▸ A2 — 1 point: For drawing an upward static friction force at each end and a downward gravitational force starting on the rectangle’s vertical center line, as distinct labelled arrows, with no other vertical force.

Scoring Note: Any label that names the force is accepted — F_N, N or normal force; f, f_s or friction; F_g, Mg, mg or weight — provided the labels tell the normal forces apart from the friction forces. Separate downward arrows for the rod’s weight and the curtain’s weight, each starting on the rectangle’s vertical center line, are accepted in place of one gravitational force. An arrow labelled as a spring force or a tension, whether drawn in place of a normal force or beside it, is extraneous: the spring’s pushes act between parts of the object, and nothing outside the object pulls on the rod along its length. A single slanted arrow at an end, standing for the normal and friction forces together, does not show them as distinct forces. Arrow lengths are not scored, and the response is not required to name the object exerting each force.

⚠︎ Common error (partial credit): Draws the five forces correctly but adds a pair of horizontal arrows labelled as spring forces or tensions, pointing outward toward the walls — earns 1 of 2 points, A2. The spring is part of the rod-and-curtain object, so its pushes on the two tubes are internal, and nothing outside the object pulls on the rod along its length.

⚠︎ Common error (partial credit): Draws the normal forces pointing toward the walls, as though each wall pulled on the rod, but draws the friction and gravitational forces correctly — earns 1 of 2 points, A2. A wall can only push, so each normal force points away from the wall that exerts it.

⚠︎ Common error (partial credit): Draws both normal forces correctly but draws the friction forces pointing downward, or draws no friction forces — earns 1 of 2 points, A1. Only static friction from the walls can hold the rod up.

✗ Common error (no credit): Draws the gravitational force downward and a single upward normal force, as if the rod rested on a floor, with no horizontal forces and no friction — earns 0 points. The walls are vertical, so their normal forces are horizontal, and only static friction holds the rod up.

Part A (ii) — Model Answer

Start with one tube. Horizontally, only two forces are exerted on the right-hand tube: the spring pushes it toward the right-hand wall, and that wall pushes it back with the normal force F_N. An ideal spring has negligible mass, so it pushes on each of its ends with a force of the same magnitude, kΔx, given by Hooke’s law. The tube is at rest, so Newton’s second law with zero acceleration gives, toward the right-hand wall:

ΣF_x = kΔx − F_N = 0,  so  F_N = kΔx

Figure 4 — Horizontal forces on the right-hand tube: the wall’s push balances the spring’s.

The same argument applies to the left-hand tube, so each wall presses on the rod with a normal force of magnitude kΔx, whatever the load. Now take the rod and curtain together. Vertically, the upward static friction forces from the two walls balance the downward gravitational force, and because the rod and curtain are symmetric about the midpoint, each wall supplies half:

f_L + f_R − Mg = 0,  so  f = Mg/2 at each wall

Static friction can be no larger than μ_s·F_N, so the rod holds without slipping only while

Mg/2 ≤ μ_s·kΔx
M_max = 2μ_s·kΔx/g

For this rod, M_max = 2(0.40)(1500 N/m)(0.040 m)/(10 N/kg) = 4.8 kg, well above the 3.0 kg it carries. The normal force from each wall is set by the spring’s compression, not by the weight of the load.

Scoring (3 points):

▸ A3 — 1 point: For a multistep derivation that includes Newton’s second law with zero acceleration (the condition for translational equilibrium) or the condition for rotational equilibrium, applied to the rod and curtain or to any part of it. This point is earned for the correct starting principle alone, even if the steps that follow are incorrect, but at least one step must follow; a single equation on its own, or a final expression for M_max on its own, is not a multistep derivation.

▸ A4 — 1 point: For indicating that each wall’s normal force has magnitude kΔx, the magnitude of the spring’s push on each tube, so that the largest static friction force at each wall is μ_s·kΔx.

▸ A5 — 1 point: For a correct expression, M_max = 2μ_s·kΔx/g, or equivalent. Where A4 was not earned, an expression for M_max in terms of k, Δx, μ_s and g that follows correctly from the response’s own normal force also earns this point; any other error prevents it.

Scoring Note: A correct, isolated, final expression for M_max earns points A4 and A5.

Scoring Note: A3 and A4 carry no consistency credit. The equal split of the load between the two walls may be stated by symmetry; an argument from rotational equilibrium is also accepted, and neither justification is required.

⚠︎ Common error (partial credit): Splits the spring’s push between the two walls, so that each wall presses with kΔx/2, and obtains M_max = μ_s·kΔx/g — earns 2 of 3 points, A3 and A5. An ideal spring pushes on each of its ends with the full kΔx, and each tube is balanced by its own wall.

⚠︎ Common error (partial credit): Uses the correct normal force but lets a single wall hold the whole load, writing Mg = μ_s·kΔx, and obtains M_max = μ_s·kΔx/g — earns 2 of 3 points, A3 and A4. Both walls exert friction on the rod, and by symmetry each supports half the load.

⚠︎ Common error (partial credit): Takes the normal force to be the load’s weight, F_N = Mg, as it would be for an object resting on a floor, so that M cancels and the response concludes that no load, or any load, can be held — earns 1 of 3 points, A3. The walls are vertical, so the normal forces are horizontal, and only the spring presses the caps against the walls.

Part A (iii) — Model Answer

From part A (ii), the rod holds a load of mass M without slipping as long as Mg/2 ≤ μ_s·kΔx. The least compression is the one that makes the two sides equal:

Δx_min = Mg / (2μ_s·k) = (6.0 kg)(10 N/kg) / [2(0.40)(1500 N/m)]
Δx_min = (60 N) / (1200 N/m) = 0.050 m

The spring must be compressed by at least 0.050 m, that is, 5.0 cm. At exactly that compression each wall pushes with 75 N and can supply at most 30 N of friction — exactly half of the 60 N load — so the rod would be on the verge of slipping. Any greater compression holds the heavier curtain with friction to spare.

Scoring (2 points):

▸ A6 — 1 point: For a relation that sets the largest total friction force the walls can supply, 2μ_s·kΔx, equal to the weight of the 6.0 kg load (or the equivalent relation for one wall, μ_s·kΔx = Mg/2), or that sets the response’s own part A (ii) expression equal to 6.0 kg.

▸ A7 — 1 point: For the value 0.050 m (5.0 cm), with units, or a value that follows correctly from the response’s own part A (ii) expression.

Scoring Note: A6 does not require the relation to be solved correctly. A response whose own part A (ii) expression leaves out friction, M_max = kΔx/g, and carries it through consistently to 0.040 m satisfies A6 and A7.

⚠︎ Common error (partial credit): Finds 0.050 m correctly but reports it as extra compression to add to the original 4.0 cm, recommending a total of 9.0 cm — earns 1 of 2 points, A6. The 0.050 m is the whole compression the spring needs, measured from its relaxed length.

✗ Common error (no credit): Having derived an expression with friction in part A (ii), abandons it here and sets the spring’s push equal to the load’s weight, kΔx = Mg, obtaining 0.040 m — earns 0 points. The spring’s push is horizontal and cannot hold up a vertical load; it matters because it sets the normal force, and friction does the holding.

Part B — Model Answer

f_2 = f_1

In both showers the rod and curtain are at rest, so the vertical forces on them balance: the friction forces from the two walls together support the 30 N gravitational force, and by symmetry each wall supplies half of it. Nothing about the load has changed, so f_2 = f_1 = (3.0 kg)(10 N/kg)/2 = 15 N.

What the larger compression does change is the normal force. By Hooke’s law the spring now pushes on each tube with kΔx = (1500 N/m)(0.060 m) = 90 N, so each wall, balancing that push, exerts a normal force of 90 N instead of 60 N, and the largest static friction force each wall could supply rises from μ_s·F_N = 24 N to 36 N. Static friction, however, takes only the value needed to keep the surfaces from slipping, so the higher limit does not raise the friction actually exerted. A larger normal force raises the maximum possible static friction, not the friction needed to hold an object at rest.

Figure 5 — The right-hand wall’s two forces in each shower: the limit rises, the friction does not.

Scoring (3 points):

▸ B1 — 1 point: For indicating f_2 = f_1.

▸ B2 — 1 point: For a justification that indicates that the rod and curtain are at rest in each shower, so the static friction from the walls must balance the gravitational force on them.

▸ B3 — 1 point: For a justification that indicates either that the gravitational force on the rod and curtain, and so the friction each wall must supply, is the same in both showers, or that the larger compression raises only the largest possible static friction force, μ_s·F_N, and not the static friction force actually exerted.

Scoring Note: B2 and B3 are content criteria: any wording with that content satisfies them, and numerical values are not required. Either branch satisfies B3. B2 and B3 may each be earned independently of the option indicated in B1 and of the response to part A.

⚠︎ Common error (partial credit): Indicates f_2 = f_1 because the rod and curtain weigh the same in both showers, without saying why the friction must match the weight — earns 2 of 3 points, B1 and B3.

⚠︎ Common error (partial credit): Indicates f_2 = f_1, arguing only that friction does not depend on how hard the surfaces are pressed together — earns 1 of 3 points, B1. The normal force does set the largest static friction a surface can supply; what it does not set is the friction a resting object needs.

✗ Common error (no credit): Indicates f_2 > f_1 because friction equals μ_s·F_N and the normal force rises from 60 N to 90 N, so the friction rises from 24 N to 36 N — earns 0 points. μ_s·F_N is the largest friction a wall could supply; a rod at rest needs only enough friction to balance its weight.

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