University Physics I · topic UPI-12.04
Beams, supports, and reactions
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.
Beams, Supports & Reactions: predict a finite change
A support reaction is the force or couple a support exerts on a structure, with one component for each motion the support prevents: one for a roller or a cable, two for a pin, three for a fixed end. Use the supplied topic model Y_out = 6*x*sqrt(d*d+h*h)/(d*h), where Y_out is CABLE TENSION T (kN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. h is Anchor height h of D above A, measured in m. Keep d (Cable attachment E at distance d) = 4 m; x (Load position x from A) = 2.75 m fixed. For this exercise use only 1.2125 m ≤ h ≤ 2.7875 m 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 h = 1.559 m and 1.7795 m. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δh across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CABLE TENSION T against h
Predictions from the supplied Beams, Supports & Reactions 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): CABLE TENSION T / kN
Horizontal axis (x): h: Anchor height h of D above A / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| h: Anchor height h of D above A / m | CABLE TENSION T / kN |
|---|---|
| 1.2125 | 14.219705444097219 |
| 1.6062499999999997 | 11.069655964070389 |
| 1.9999999999999998 | 9.223780407186634 |
| 2.39375 | 8.032956498181582 |
| 2.7874999999999996 | 7.2148132649102985 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of h separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At h = 1.559 m, substitute into Y_out = 6*x*sqrt(d*d+h*h)/(d*h): Y₀ = 11.3592 kN.
- At h = 1.7795 m, the same fixed-parameter model gives Y₁ = 10.1484 kN.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.21073 kN; the input change is Δh = 0.2205 m.
- Average sensitivity is ΔY_out/Δh = -5.49085 kN per m. 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 = 11.3592 kN, Y₁ = 10.1484 kN; ΔY_out = -1.21073 kN; average sensitivity = -5.49085 kN per m.
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.
Beams, Supports & Reactions: infer an input with uncertainty
A support reaction is the force or couple a support exerts on a structure, with one component for each motion the support prevents: one for a roller or a cable, two for a pin, three for a fixed end. Use the supplied topic model Y_out = 6*x*sqrt(d*d+h*h)/(d*h), where Y_out is CABLE TENSION T (kN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. h is Anchor height h of D above A, measured in m. Keep d (Cable attachment E at distance d) = 4 m; x (Load position x from A) = 2.75 m fixed. For this exercise use only 1.2125 m ≤ h ≤ 2.7875 m 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 = 8.6405371 kN, with a bounded reading uncertainty ±0.21386 kN. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown h by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 m, 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.
CABLE TENSION T against h
Predictions from the supplied Beams, Supports & Reactions 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): CABLE TENSION T / kN
Horizontal axis (x): h: Anchor height h of D above A / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| h: Anchor height h of D above A / m | CABLE TENSION T / kN |
|---|---|
| 1.2125 | 14.219705444097219 |
| 1.6062499999999997 | 11.069655964070389 |
| 1.9999999999999998 | 9.223780407186634 |
| 2.39375 | 8.032956498181582 |
| 2.7874999999999996 | 7.2148132649102985 |
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(h) = 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(1.2125) = 14.2197 kN and Y_out(2.7875) = 7.21481 kN. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives h_line = 1.2125 + (8.64054 − 14.2197)(2.7875 − 1.2125)/(7.21481 − 14.2197) = 2.46694 m. 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 h = 2.17325 m; substituting back gives Y_out = 8.6405371 kN.
- The interpolation error in the input is |h_line − h| = 0.293686 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.4266771 kN and 8.8543971 kN. Sort the roots even if the response decreases: 2.10598 m ≤ h ≤ 2.24551 m.
- The lower and upper input deviations are 0.0672722 m and 0.0722557 m. 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 h ≈ 2.17325 m; compatible interval [2.10598, 2.24551] m. Endpoint interpolation gives 2.46694 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.