University Physics I · topic UPI-05.04
Elevators and apparent weight
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.
Elevators & Apparent Weight: predict a finite change
Apparent weight is the normal force a support exerts on a body, N = m(g + a_y) with upward positive; it is the quantity a scale actually reads, and it equals the true weight mg only while the vertical acceleration is zero. Use the supplied topic model Y_out = m*max(g+a,0), where Y_out is NORMAL FORCE N (N). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; g = 9.81 m s⁻². m is Rider mass m, measured in kg. Keep a (Lift acceleration a_y (up positive)) = 1.5 m s⁻²; v (Lift velocity v_y (up positive)) = 2 m s⁻¹ fixed. For this exercise use only 48.5 kg ≤ m ≤ 111.5 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 m = 62.36 kg and 71.18 kg. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
NORMAL FORCE N against m
Predictions from the supplied Elevators & Apparent Weight 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): NORMAL FORCE N / N
Horizontal axis (x): m: Rider mass m / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rider mass m / kg | NORMAL FORCE N / N |
|---|---|
| 48.5 | 548.535 |
| 64.25 | 726.6675 |
| 80 | 904.8000000000001 |
| 95.75 | 1082.9325000000001 |
| 111.5 | 1261.065 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 62.36 kg, substitute into Y_out = m*max(g+a,0): Y₀ = 705.292 N.
- At m = 71.18 kg, the same fixed-parameter model gives Y₁ = 805.046 N.
- Signed change: ΔY_out = Y₁ − Y₀ = 99.7542 N; the input change is Δm = 8.82 kg.
- Average sensitivity is ΔY_out/Δm = 11.31 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 = 705.292 N, Y₁ = 805.046 N; ΔY_out = 99.7542 N; average sensitivity = 11.31 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.
Elevators & Apparent Weight: infer an input with uncertainty
Apparent weight is the normal force a support exerts on a body, N = m(g + a_y) with upward positive; it is the quantity a scale actually reads, and it equals the true weight mg only while the vertical acceleration is zero. Use the supplied topic model Y_out = m*max(g+a,0), where Y_out is NORMAL FORCE N (N). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; g = 9.81 m s⁻². m is Rider mass m, measured in kg. Keep a (Lift acceleration a_y (up positive)) = 1.5 m s⁻²; v (Lift velocity v_y (up positive)) = 2 m s⁻¹ fixed. For this exercise use only 48.5 kg ≤ m ≤ 111.5 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 = 983.1783 N, with a bounded reading uncertainty ±41.683 N. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0063 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.
NORMAL FORCE N against m
Predictions from the supplied Elevators & Apparent Weight 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): NORMAL FORCE N / N
Horizontal axis (x): m: Rider mass m / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rider mass m / kg | NORMAL FORCE N / N |
|---|---|
| 48.5 | 548.535 |
| 64.25 | 726.6675 |
| 80 | 904.8000000000001 |
| 95.75 | 1082.9325000000001 |
| 111.5 | 1261.065 |
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(m) = 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(48.5) = 548.535 N and Y_out(111.5) = 1261.07 N. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives m_line = 48.5 + (983.178 − 548.535)(111.5 − 48.5)/(1261.07 − 548.535) = 86.93 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 m = 86.93 kg; substituting back gives Y_out = 983.1783 N.
- The interpolation error in the input is |m_line − m| = 0 kg. This is within the requested numerical tolerance.
- Invert both limits Y_out = 941.4953 N and 1024.8613 N. Sort the roots even if the response decreases: 83.2445 kg ≤ m ≤ 90.6155 kg.
- The lower and upper input deviations are 3.6855 kg and 3.6855 kg. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer m ≈ 86.93 kg; compatible interval [83.2445, 90.6155] kg. Endpoint interpolation gives 86.93 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.