University Physics I · topic UPI-15.05
Buoyancy and Archimedes' principle
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.
Buoyancy & Archimedes' Principle: predict a finite change
The buoyant force is the net upward force that a fluid at rest exerts on a body immersed in it, equal in magnitude to the weight of the fluid the body displaces. Use the supplied topic model Y_out = 100*min(rb/rf,1), where Y_out is SUBMERGED FRACTION (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. rf is Fluid density ρ_f, measured in kg m⁻³. Keep rb (Block density ρ_b) = 600 kg m⁻³ fixed. For this exercise use only 847.813 kg m⁻³ ≤ rf ≤ 1202.19 kg m⁻³ 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 rf = 925.775 kg m⁻³ and 975.388 kg m⁻³. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SUBMERGED FRACTION against rf
Predictions from the supplied Buoyancy & Archimedes' Principle 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): SUBMERGED FRACTION / %
Horizontal axis (x): rf: Fluid density ρ_f / kg m⁻³
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rf: Fluid density ρ_f / kg m⁻³ | SUBMERGED FRACTION / % |
|---|---|
| 847.8125 | 70.77036490969407 |
| 936.40625 | 64.07475387952611 |
| 1025 | 58.536585365853654 |
| 1113.59375 | 53.879612740283434 |
| 1202.1875 | 49.90902001559657 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rf separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rf = 925.775 kg m⁻³, substitute into Y_out = 100*min(rb/rf,1): Y₀ = 64.8106 %.
- At rf = 975.388 kg m⁻³, the same fixed-parameter model gives Y₁ = 61.514 %.
- Signed change: ΔY_out = Y₁ − Y₀ = -3.29655 %; the input change is Δrf = 49.6125 kg m⁻³.
- Average sensitivity is ΔY_out/Δrf = -0.066446 % per kg m⁻³. Its negative sign means that this response decreases 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 = 64.8106 %, Y₁ = 61.514 %; ΔY_out = -3.29655 %; average sensitivity = -0.066446 % per kg m⁻³.
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.
Buoyancy & Archimedes' Principle: infer an input with uncertainty
The buoyant force is the net upward force that a fluid at rest exerts on a body immersed in it, equal in magnitude to the weight of the fluid the body displaces. Use the supplied topic model Y_out = 100*min(rb/rf,1), where Y_out is SUBMERGED FRACTION (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. rf is Fluid density ρ_f, measured in kg m⁻³. Keep rb (Block density ρ_b) = 600 kg m⁻³ fixed. For this exercise use only 847.813 kg m⁻³ ≤ rf ≤ 1202.19 kg m⁻³ 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 = 56.391971 %, with a bounded reading uncertainty ±0.97244 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rf by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0354375 kg m⁻³, 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.
SUBMERGED FRACTION against rf
Predictions from the supplied Buoyancy & Archimedes' Principle 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): SUBMERGED FRACTION / %
Horizontal axis (x): rf: Fluid density ρ_f / kg m⁻³
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rf: Fluid density ρ_f / kg m⁻³ | SUBMERGED FRACTION / % |
|---|---|
| 847.8125 | 70.77036490969407 |
| 936.40625 | 64.07475387952611 |
| 1025 | 58.536585365853654 |
| 1113.59375 | 53.879612740283434 |
| 1202.1875 | 49.90902001559657 |
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(rf) = 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(847.813) = 70.7704 % and Y_out(1202.19) = 49.909 %. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rf_line = 847.813 + (56.392 − 70.7704)(1202.19 − 847.813)/(49.909 − 70.7704) = 1092.06 kg m⁻³. 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 rf = 1063.98 kg m⁻³; substituting back gives Y_out = 56.391971 %.
- The interpolation error in the input is |rf_line − rf| = 28.0793 kg m⁻³. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 55.419531 % and 57.364411 %. Sort the roots even if the response decreases: 1045.94 kg m⁻³ ≤ rf ≤ 1082.65 kg m⁻³.
- The lower and upper input deviations are 18.0366 kg m⁻³ and 18.6696 kg m⁻³. 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 rf ≈ 1063.98 kg m⁻³; compatible interval [1045.94, 1082.65] kg m⁻³. Endpoint interpolation gives 1092.06 kg m⁻³.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.