University Physics I · topic UPI-05.03
Pulley and Atwood-type systems
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Pulley and Atwood-type systems: predict a finite change
An Atwood-type system is a set of bodies joined by ideal ropes over pulleys, solved by writing Newton's second law for each body and closing the equations with one length constraint per rope. Use the supplied topic model Y_out = mc*ml*g*(1+n)/(ml+n*n*mc), where Y_out is TENSION PER SEGMENT T (N). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; g = 9.81 m s⁻². ml is Load mass M, measured in kg. Keep n (Segments holding the load n) = 2; mc (Counterweight mass m) = 4 kg fixed. For this exercise use only 6.125 kg ≤ ml ≤ 21.875 kg on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at ml = 9.59 kg and 11.795 kg. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δml across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
TENSION PER SEGMENT T against ml
Predictions from the supplied Pulley and Atwood-type systems model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): TENSION PER SEGMENT T / N
Horizontal axis (x): ml: Load mass M / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ml: Load mass M / kg | TENSION PER SEGMENT T / N |
|---|---|
| 6.125 | 32.58915254237288 |
| 10.0625 | 45.450647482014396 |
| 14 | 54.936 |
| 17.9375 | 62.220331491712706 |
| 21.875 | 67.99009900990099 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of ml separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At ml = 9.59 kg, substitute into Y_out = mc*ml*g*(1+n)/(ml+n*n*mc): Y₀ = 44.1162 N.
- At ml = 11.795 kg, the same fixed-parameter model gives Y₁ = 49.9553 N.
- Signed change: ΔY_out = Y₁ − Y₀ = 5.83905 N; the input change is Δml = 2.205 kg.
- Average sensitivity is ΔY_out/Δml = 2.64809 N per kg. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 44.1162 N, Y₁ = 49.9553 N; ΔY_out = 5.83905 N; average sensitivity = 2.64809 N per kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Pulley and Atwood-type systems: infer an input with uncertainty
An Atwood-type system is a set of bodies joined by ideal ropes over pulleys, solved by writing Newton's second law for each body and closing the equations with one length constraint per rope. Use the supplied topic model Y_out = mc*ml*g*(1+n)/(ml+n*n*mc), where Y_out is TENSION PER SEGMENT T (N). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; g = 9.81 m s⁻². ml is Load mass M, measured in kg. Keep n (Segments holding the load n) = 2; mc (Counterweight mass m) = 4 kg fixed. For this exercise use only 6.125 kg ≤ ml ≤ 21.875 kg on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 58.363819 N, with a bounded reading uncertainty ±1.4439 N. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ml by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.001575 kg, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
TENSION PER SEGMENT T against ml
Predictions from the supplied Pulley and Atwood-type systems model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): TENSION PER SEGMENT T / N
Horizontal axis (x): ml: Load mass M / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ml: Load mass M / kg | TENSION PER SEGMENT T / N |
|---|---|
| 6.125 | 32.58915254237288 |
| 10.0625 | 45.450647482014396 |
| 14 | 54.936 |
| 17.9375 | 62.220331491712706 |
| 21.875 | 67.99009900990099 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(ml) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(6.125) = 32.5892 N and Y_out(21.875) = 67.9901 N. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives ml_line = 6.125 + (58.3638 − 32.5892)(21.875 − 6.125)/(67.9901 − 32.5892) = 17.5922 kg. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives ml = 15.7325 kg; substituting back gives Y_out = 58.363819 N.
- The interpolation error in the input is |ml_line − ml| = 1.85974 kg. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 56.919919 N and 59.807719 N. Sort the roots even if the response decreases: 14.9789 kg ≤ ml ≤ 16.5237 kg.
- The lower and upper input deviations are 0.753594 kg and 0.791172 kg. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer ml ≈ 15.7325 kg; compatible interval [14.9789, 16.5237] kg. Endpoint interpolation gives 17.5922 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.