Static Equilibrium & Elasticity

A rigid body is in equilibrium when the external forces on it add to zero and their torques about a point add to zero, so neither its centre of mass nor its rotation accelerates; static equilibrium also requires it to start at rest. Use the supplied topic model Y_out = fa+fb-600, where Y_out is NET FORCE ΣF_y (N). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. fa is Lift F₁ at end A, measured in N. Keep fb (Lift F₂ at end B) = 300 N; p (Axis position p from A) = 1 m fixed. For this exercise use only 189.75 N ≤ fa ≤ 410.25 N 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 = 24.255 N, with a bounded reading uncertainty ±12.899 N. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown fa by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.02205 N, 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.

The center of gravity is the point where a body's entire weight can be taken to act when computing torques; in a uniform gravitational field it coincides with the center of mass, the mass-weighted average position r_cm = Σ mᵢrᵢ / M. Use the supplied topic model Y_out = deg(atan(0.3*(20+load)/(15+load*shelf))), where Y_out is TIPPING ANGLE θ_c (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; deg converts radians to degrees; unmarked trigonometric functions use radians. load is Load mass m, measured in kg. Keep tilt (Tilt about the edge θ) = 15 °; shelf (Height of the load's center) = 1.2 m fixed. For this exercise use only 6.375 kg ≤ load ≤ 53.625 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 = 16.135698 °, with a bounded reading uncertainty ±0.08713 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown load by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.004725 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.

A couple is a pair of equal and opposite forces whose lines of action are a perpendicular distance d apart; it exerts no net force, and its moment C = F d is the same about every point. Use the supplied topic model Y_out = f*(d/2+p)+f*(d/2-p), where Y_out is NET M₁ + M₂ (N m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Force magnitude F, measured in N. Keep d (Separation of the lines d) = 0.5 m; p (Position of P from the midpoint) = -0.8 m fixed. For this exercise use only 13.1875 N ≤ f ≤ 36.8125 N 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 = 13.799375 N m, with a bounded reading uncertainty ±0.69103 N m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0023625 N, 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.

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.