ela-2017:RST.9-10.7ela-2017:RST.9-10.8ela-2017:RST.9-10.9ela-2017:RST.9-10.10
ela-2017:RST.9-10.7 Translate quantitative or technical information expressed in words in a text into visual form (e.g., a table or chart) and translate information expressed visually or mathematically (e.g., in an equation) into words.
ela-2017:RST.9-10.8 Assess the extent to which the reasoning and evidence in a text support the author's claim or a recommendation for solving a scientific or technical problem.
ela-2017:RST.9-10.9 Compare and contrast findings presented in a text to those from other sources (including their own experiments), noting when the findings support or contradict previous explanations or accounts.
ela-2017:RST.9-10.10 By the end of grade 10, read, comprehend, and respond to science/technical texts in the grades 9-10 text complexity band independently and proficiently.
Learning goal: Translate quantitative or technical information expressed in words in a text into visual form, such as a table or chart, and translate information expressed visually or mathematically, such as in an equation, into words; assess the extent to which the reasoning and evidence in a text support the author's claim or a recommendation for solving a scientific or technical problem; compare and contrast findings presented in a text to those from other sources, including your own experiments, noting when the findings support or contradict previous explanations or accounts; and read, comprehend and respond to science and technical texts in the grades 9-10 text complexity band independently and proficiently.
The texts. This lesson uses the original friction text from the first science lesson (with its hypothetical class results) and two new texts written for this lesson.
Original technical memo — Fixing Wheel Slip on the Competition Ramp (a fictional school robotics team).
To: Team members. From: Drive subteam. Problem: Our robot, mass 2.0 kg, with rubber wheels and all four wheels driven, spins its wheels and slides back on the practice ramp, which is built like the competition ramp: 20° of painted plywood. Recommendation: bolt 1.0 kg of steel plate to the chassis. Evidence: (1) At last month's pushing challenge, adding weight stopped our wheels from spinning when we pushed the 3 kg box. (2) Heavier robots have more grip, because friction is proportional to normal force. (3) We measured the friction between our tires and the painted plywood by locking the wheels and tilting a board until the robot slid; it slid at 16.7°, giving μ_s = tan 16.7° ≈ 0.30. Conclusion: more weight means more friction, so the robot will climb.
Original lab report excerpt — Group 4's area trial (fictional). "Narrow side down, 0.500 kg, we measured 1.2, 1.2 and 1.3 N, much less than the 1.5 N we got on the large face. Friction therefore depends on contact area, which contradicts the textbook. Note: we had sanded the narrow side smooth first so the scale would not catch."
Translating words into a visual. Section 5 of the friction text reports its results in sentences. Numbers in sentences are hard to compare; a table lines them up, and a graph shows the pattern. Translating means preserving every quantity with its unit and condition. Table: normal force F_N in newtons (4.9, 9.8, 14.7) against mean kinetic friction f_k in newtons (1.5, 2.9, 4.4), with a column for μ_k = f_k ÷ F_N (0.306, 0.296, 0.299). Graph: F_N on the horizontal axis, because it is the variable the procedure changes, and f_k on the vertical axis, because it is measured. The three points fall close to a straight line through the origin. A straight line through the origin is what "proportional" looks like, and its slope is the ratio f_k ÷ F_N, so the slope of the best line is μ_k ≈ 0.30.
Translating an equation or a visual into words. The equation f_k = μ_k F_N says, in words: "the kinetic friction force equals the coefficient of kinetic friction multiplied by the normal force." A fuller translation adds what the equation implies and its conditions: "for a given pair of dry, rigid surfaces, doubling the normal force doubles the kinetic friction; the coefficient is fixed by the surfaces; the relationship is approximate." The graph translates as: "As the normal force rises from 4.9 N to 14.7 N, the kinetic friction rises from 1.5 N to 4.4 N along a nearly straight line through zero, with slope about 0.30." Good translations name the variables, the direction and form of the relationship, and the key numbers.
Assessing reasoning and evidence for a recommendation. A recommendation for solving a technical problem makes a prediction: if we do this, the problem will be solved. Assess it by asking whether each piece of evidence is relevant to this problem, and whether the reasoning connects the evidence to the prediction without a hidden gap.
Reading the ramp mathematics. θ (theta) is the ramp's angle above horizontal. In a right triangle drawn with the ramp as its slanted side, sin θ is the opposite side divided by the slanted side, cos θ the adjacent side divided by the slanted side, and tan θ the opposite side divided by the adjacent side, which also equals sin θ ÷ cos θ. The robot's weight mg points straight down; on a ramp it splits into a part along the ramp, mg sin θ, which pulls the robot downhill, and a part into the ramp, mg cos θ, which the normal force balances. Use a calculator in degree mode: tan 20° ≈ 0.364. A check without trigonometry: the team's own robot slid at 16.7°, so any ramp steeper than 16.7° is too steep for those tires.
The memo's claim is that adding 1.0 kg will let the robot climb. Evidence (3) is a sound measurement and is relevant: μ_s ≈ 0.30 for tire on painted plywood. Evidence (2) is partly correct: friction is proportional to normal force in the simple model. But the reasoning leaves out something decisive: on a slope, weight does two jobs. The component of weight pulling the robot down the ramp is mg sin θ; the normal force is mg cos θ, so the maximum static friction the tires can supply is μ_s mg cos θ. (A driven wheel that grips without spinning pushes on the ramp through static friction, even though the robot is moving, because the patch of tread touching the ramp is not sliding over it.) To climb without slipping, the tires must supply at least mg sin θ. Both sides contain m, so mass cancels: the robot slips whenever tan θ is greater than μ_s, whatever it weighs. Here tan 20° ≈ 0.364, larger than 0.30, so in this model the robot slips at 20° with or without the steel plate. The memo's own measurement shows this: the robot with locked wheels slid at 16.7°, and 20° is steeper.
Evidence (1) is relevant to the general claim that added weight can increase grip, but it is not sufficient evidence for the ramp recommendation, because it does not transfer. In the pushing challenge the resisting force was the box's friction, which does not grow when the robot gets heavier; extra robot weight raised the grip without raising the load. On a ramp, the load the tires fight is the robot's own weight, so the two situations differ in exactly the relation that matters.
Verdict: the evidence supports the claim that heavier robots have more available friction, but it does not support the recommendation, because the reasoning ignores that the downhill force grows by the same factor. Better recommendations follow from the same equation: raise μ_s above tan 20° ≈ 0.364 with softer or grippier tread or cleaner surfaces; or, only if the team controls the ramp and the competition rules allow it, lower the angle below about 16.7°. The model has limits: rubber tires do not follow the simple friction model exactly, and the calculation assumes all four wheels stay in contact and are driven. Meeting the condition also removes only the traction limit: to climb, the motors must supply enough torque, applied smoothly enough that the wheels do not spin, and starting or speeding up on the ramp needs some margin above 0.364. A locked-wheel tilt test shows only whether the robot can hold; a powered climb on the real ramp should settle the matter.
Comparing findings with other sources and your own experiment. Findings from one source support a previous explanation when they agree with what it predicts, and contradict it when they cannot be reconciled. The friction text's finding that turning the block on its side changed friction by less than trial variation supports the widely taught explanation that, for dry rigid surfaces, sliding friction does not depend noticeably on apparent contact area. Group 4's finding appears to contradict it: 1.2 N on the narrow side against 1.5 N on the face. But their note reveals that they sanded the narrow side first, changing the surface condition, which the text says determines μ_k. Two variables changed at once, so their result cannot separate area from smoothness; it does not contradict the explanation, it fails to test it. Your own trial from the first science lesson is a third source: compare your μ_k values with 0.30 and your area result with both.
Reading grades 9-10 science texts independently. At this level texts mix prose, symbols, equations, tables and qualifications. A proficient reader identifies the question and purpose first, translates every equation into words, turns numbers in sentences into a table, marks conditions ("dry," "approximately," "at low speeds") and asks what evidence supports each claim.
A misconception. A finding that disagrees with a textbook is not automatically a discovery or an error. First check whether the two findings measured the same thing under the same conditions.
Worked words-to-table translation: Section 5 sentence: "with 1.000 kg (F_N = 9.8 N), trials 2.9, 3.0 and 2.8 N, mean 2.9 N." Table row: total mass 1.000 kg; F_N 9.8 N; trial 1, 2.9 N; trial 2, 3.0 N; trial 3, 2.8 N; mean f_k (2.9 + 3.0 + 2.8) ÷ 3 = 2.9 N; μ_k 2.9 ÷ 9.8 ≈ 0.296. Every number from the sentence appears once, with its unit, and the computed columns show their arithmetic.
Worked equation-to-words translation: The memo uses the condition for slipping, tan θ > μ_s. In words: "the robot slides back when the tangent of the ramp angle is greater than the coefficient of static friction between tires and ramp." Check with numbers: at 20°, tan 20° ≈ 0.364, greater than 0.30, so it slips; at 15°, tan 15° ≈ 0.268, less than 0.30, so in this model it holds. The angle at which tan θ = 0.30 is about 16.7°, which matches the memo's own tilt test. Mass does not appear anywhere in the condition, and saying so in words is the key point of the translation.
Worked comparison with a previous explanation: Suppose a website for bicycle riders (an invented example for this lesson) states that wider tires give more grip. This seems to contradict section 2's claim that friction does not depend noticeably on contact area. Compare conditions: section 2 limits its claim to "dry, rigid surfaces" and names rubber as an exception. A rubber tire deforms around the road surface, so the simple model is not expected to hold for it. The two sources describe different conditions, so the website does not contradict the text; the text's own limits already allow for it. Neither source is shown to be wrong.
Translation set (representation): (1) Turn section 5 of the friction text into a complete table with means and μ_k for every row, including the narrow-side trial. (2) Plot f_k against F_N and draw the best straight line through the origin; state its slope. (3) Translate f_k = μ_k F_N, W = Fd and tan θ > μ_s into sentences that name every variable and one condition for each.
Independent performance: Write a 350- to 450-word assessment of the robotics memo addressed to the team. Success criteria: (1) the recommendation and each piece of evidence named separately; (2) for each piece of evidence, a judgment of whether it is accurate, whether it is relevant, and whether it is sufficient for the ramp problem; (3) the gap in the reasoning explained in words and with the numbers tan 20° ≈ 0.364 and μ_s ≈ 0.30; (4) at least one alternative recommendation that follows from the same evidence; (5) a comparison of the friction text's area finding with Group 4's and with your own trial, noting whether each supports or contradicts the previous explanation and why; (6) one limit of the simple friction model stated as the text states it. Then, working alone, read an unfamiliar grades 9-10 science or technical text that your teacher supplies (for example a textbook section or a product data sheet) and write a 150- to 200-word response that translates one of its tables, graphs or equations into words and judges how well its evidence supports one claim or recommendation.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Translate the graph of kinetic friction against normal force into words, including its slope and what the slope means.
As the normal force rises from 4.9 N to 14.7 N, the kinetic friction rises from 1.5 N to 4.4 N, and the points lie close to a straight line through the origin. A line through the origin means friction is proportional to normal force. Its slope, about 0.30, is the ratio f_k ÷ F_N, which is the coefficient of kinetic friction for these surfaces.
Translate the equation f_k = μ_k F_N into words and state what happens to f_k if F_N is tripled, with the condition under which that holds.
The kinetic friction force equals the coefficient of kinetic friction multiplied by the normal force. If F_N is tripled, f_k triples, because μ_k is fixed by the two surfaces. This holds approximately for dry, rigid surfaces at low speeds, as the text states; it is not exact for rubber or lubricated surfaces.
Assess the memo's evidence (1), about the pushing challenge. Is it accurate, and is it relevant to the ramp problem?
It may be accurate, since added weight can stop wheel spin when pushing a box, and it is relevant to the general claim that weight adds grip; but it is not sufficient for the ramp recommendation, because it does not transfer to the ramp. It bears on the general claim that weight adds grip, but it does not transfer to the ramp, because in the pushing challenge the resisting force came from the box and did not grow with the robot's weight, while on the ramp the force the tires must overcome is the robot's own weight component, mg sin θ, which grows in the same proportion as the available friction. An answer that says it is not relevant to the ramp problem, for this reason, is also correct: it says the same thing as 'relevant to the general claim but not sufficient for the ramp.'
Using the memo's own numbers, explain why adding 1.0 kg will not, in the simple model, stop the robot slipping on the 20° ramp.
The tires can supply at most μ_s mg cos θ up the ramp, and must supply at least mg sin θ. Mass appears in both, so it cancels, and the robot slips whenever tan θ > μ_s. tan 20° ≈ 0.364 is greater than 0.30, so the robot slips at any mass. The memo's tilt test, where the robot slid at 16.7°, already shows the ramp is too steep.
Suggest a recommendation that the memo's evidence does support, and explain why.
Raise μ_s above tan 20° ≈ 0.364 with softer or cleaner tread; or, only if the team controls the ramp and the rules allow it, lower the angle below about 16.7°. Both follow from tan θ > μ_s: the robot holds when the tangent of the angle is less than the coefficient, so either the coefficient must rise or the angle must fall. The recommendation should then be tested by a powered climb on the real ramp, because rubber tires do not follow the simple model exactly and because enough grip removes only the traction limit: the motors must still supply the torque to climb.
Does Group 4's narrow-side result contradict the explanation that sliding friction does not depend noticeably on apparent contact area? Explain.
No. Group 4 sanded the narrow side smooth before testing, which changed the surface condition as well as the area. The text says μ_k depends on the surfaces and their condition, so the lower reading (about 1.23 N against 1.5 N) could come from smoothness rather than area. Because two variables changed at once, their result cannot test the area explanation, so it neither supports nor contradicts it.
A student's own trial gives μ_k = 0.42 for a plastic block on a lab bench, not 0.30. Does this contradict the friction text? Explain what comparison is fair.
Not necessarily. The text's value of 0.30 is for one wood block on one wooden board, and the text says μ_k depends on the two materials and their condition. A different pair of surfaces is expected to have a different coefficient. A fair comparison tests whether the student's data also show friction proportional to normal force and no noticeable area effect; those are the text's general findings, while the number 0.30 belongs to its surfaces only.
Read the memo's evidence (3) the way a proficient reader of a science text would: state its purpose in the memo, turn its quantities into one table row with units and conditions, translate its equation into words, and name one condition that limits what it shows.
Purpose: it supplies a measured value of the grip between these tires and this surface, the number the recommendation should rest on. Table row: test, tilt test; surfaces, rubber tires on painted plywood; condition, wheels locked; angle when the robot slid, 16.7°; μ_s = tan 16.7° ≈ 0.30 (no unit). Equation in words: the coefficient of static friction equals the tangent of the angle at which the robot just begins to slide. Limiting conditions: it is a single test with locked wheels, so it measures grip without the wheels turning, and it holds only for these tires on this paint; a reader also notes that 0.30 is less than tan 20° ≈ 0.364, which the memo never compares.