CELE Engineering Mechanics — Frames, Machines and CablesMisconception Buster
If you have been missing Frames, Machines and Cables questions on your CELE mocks, the cause is almost always a misconception. This page lists the ones Professional Regulation Commission (PRC) — Board of Civil Engineering exploits most often in the CELE Engineering Mechanics subtest and shows how to correct them before exam day.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Engineering Mechanics subtest is marked as "Core" in the official pattern, and Frames, Machines and Cables appears in position 4th of 8 in the CELE Engineering Mechanics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Frames, Machines and Cables - Misconception Buster
In the PRC Civil Engineer Licensure Examination, Frames, Machines and Cables is one of the most mark-rich yet conceptually misunderstood topics in Engineering Mechanics. Many examinees carry incorrect mental models from their undergraduate years — applying truss analysis rules to multi-force members, misidentifying two-force members, forgetting Newton's Third Law at pins, or confusing parabolic and catenary cable shapes. These are not minor slips; they are systematic errors that cause examinees to lose 5–10 points per sitting. This guide exposes the 11 most dangerous misconceptions, explains exactly why smart students fall for them, and provides trap questions that mirror the style of actual board exam items. Study each misconception until the correct reasoning feels automatic.
Summary
The 11 misconceptions in this guide cluster around four root causes that you must eliminate before the board exam: (1) CLASSIFICATION ERRORS — confusing two-force and multi-force members is the single most expensive mistake; always run the three-point checklist (two loading points, pin endpoints, negligible weight) before deciding. (2) FBD DISCIPLINE — the whole-frame FBD gives external reactions only; dismembering is mandatory for internal pin forces; Newton's Third Law means pin force arrows must be reversed on adjacent FBDs. (3) CABLE PHYSICS — H is constant (vertical loads only); maximum tension is at the steepest point (supports, not midspan); tension at the lowest point equals H, never zero; parabola applies to horizontal-projection loads, catenary to arc-length loads. (4) FORMULA SCOPE — H = wL²/8d applies ONLY to symmetric cables with equal support heights; for unequal heights, set up boundary conditions from the general parabola equation. Mastering these four areas eliminates the most common sources of partial and total mark losses in the Mechanics portion of the CE board examination.
Misconceptions
All members in a frame or machine carry only axial (two-force) loads, just like truss members.
Tags
- conceptual_gap
- common_error
- truss_vs_frame_confusion
Topic
Frames — Member Classification
Severity
critical
Exam Impact
If you treat a multi-force member as two-force, you assign the wrong direction to internal forces, set up the wrong equilibrium equations, and get completely wrong pin reactions — every computed value will be incorrect.
The Reality
A frame contains at least one multi-force member — a member loaded at more than two points, or at two points that are not both pins, or that carries an applied moment. Such a member develops both axial force AND shear AND bending moment. You cannot assign its force as simply 'along the member.' The very definition of a frame (versus a truss) is the presence of multi-force members.
Trap Question
Question
A frame member PQ has pins at P and Q and also carries a 20 kN transverse load applied at its midpoint. What is the direction of the resultant pin force at P?
Explanation
A two-force member requires loading at EXACTLY two points and no applied moments — PQ has a third load point at its midpoint, making it a multi-force member. The pin force direction is determined by equilibrium, not by geometry of the member axis.
Wrong Answer
The force at P acts along the line PQ because PQ is pinned at both ends.
Correct Answer
The direction cannot be assumed along PQ. PQ is a multi-force member; the pin at P has two unknown components Px and Py that must be solved from equilibrium equations.
Misconception Id
M1
Correct Vs Incorrect
Correct Approach
First check: is member BC loaded only at its two endpoints B and C, and at no other point, and is it free of applied moments? If YES, it is a two-force member and the force is along BC. If NO (e.g., there is an external load at a midpoint D, or a moment applied somewhere along BC), it is a multi-force member — treat it with unknown Bx, By components at each pin.
Incorrect Approach
Member BC is straight and has pins at both B and C, so the force in BC must act along BC at the same angle as the member. I will use this direction directly in equilibrium.
Why Students Believe It
Students learn truss analysis first, where every member is a two-force member by definition. When they encounter frames, they carry the same mental template and assume any straight bar still carries only tension or compression along its axis.
When you dismember a frame and draw FBDs, the pin forces on the two members sharing a pin act in the same direction.
Tags
- common_error
- newtons_third_law
- FBD_error
Topic
Frames — FBD Dismembering
Severity
critical
Exam Impact
Both equilibrium equations become wrong. The computed pin forces have the wrong sign, yielding answers that are equal in magnitude but wrong in sign — a deceptively close incorrect answer that costs full marks.
The Reality
Newton's Third Law: the force that member A exerts on member B through a shared pin is equal in magnitude and opposite in direction to the force that member B exerts on member A. When you draw FBDs of dismembered parts, the pin force arrows MUST point in opposite directions on the two FBDs. Violating this is the single most common error in frame analysis.
Trap Question
Question
In a frame analysis, you find that the pin at joint B exerts a force of 30 kN upward on member AB. What is the force that pin B exerts on member BC at the same joint?
Explanation
Newton's Third Law applies at every dismembered joint. The force on member AB from pin B is +30 kN (upward). The reaction on member BC from the same pin B is −30 kN (downward). Drawing both arrows in the same direction violates Newton's Third Law and is the most frequent dismembering error.
Wrong Answer
30 kN upward on member BC, because the pin force is the same at joint B.
Correct Answer
30 kN downward on member BC.
Misconception Id
M2
Correct Vs Incorrect
Correct Approach
If pin C pushes member AB upward with 15 kN (+y on AB's FBD), then by Newton's Third Law, member AB pushes member BC downward with 15 kN (−y on BC's FBD). Always reverse the arrow when transferring a pin force to the other member's FBD.
Incorrect Approach
I found that pin C pushes member AB upward with 15 kN. So on member BC's FBD, I will also draw pin C pushing upward with 15 kN.
Why Students Believe It
Students think of the pin as 'pushing both members the same way,' similar to how a single external reaction pushes the whole structure. They forget that they are now looking at the internal interaction between members.
The horizontal tension component H in a cable varies along the cable span.
Tags
- conceptual_gap
- formula_confusion
- constant_H
Topic
Cables — Concentrated Loads
Severity
critical
Exam Impact
If you allow H to vary, you get a different H at every segment, making the problem unsolvable — or you solve it with a wrong H and get wrong tensions and wrong cable geometry throughout.
The Reality
For any cable in equilibrium under vertical loads only (no horizontal distributed loads), the horizontal component of tension H is CONSTANT throughout the entire cable. This is a direct consequence of horizontal equilibrium: if there are no horizontal external loads between any two points on the cable, then ΣFx = 0 demands that H_left = H_right. This fact is the foundation of all cable calculations.
Trap Question
Question
A cable carries three vertical concentrated loads. The tension in the leftmost segment is 45 kN at an angle of 20° to the horizontal. What is the horizontal component of tension in the rightmost segment (at a steeper angle of 35°)?
Explanation
With only vertical loads, H is constant throughout. H = 45 cos 20° = 42.3 kN in every segment. The tension in the right segment is T_right = 42.3/cos 35° = 51.6 kN — higher T but same H.
Wrong Answer
H_right is larger because the right segment is steeper, so H_right ≠ H_left.
Correct Answer
H_right = H_left = 45 cos 20° = 42.3 kN
Misconception Id
M3
Correct Vs Incorrect
Correct Approach
T is larger at the right support because the vertical component Vy is larger there (steeper angle). But H = T cos θ is the SAME everywhere. The increase in T is entirely due to the increase in the vertical component, not H. Use H = constant and T = H/cos θ at any point.
Incorrect Approach
The cable is steeper at the right support, so H must be larger there because the tension T is larger there.
Why Students Believe It
Tension clearly changes magnitude along the cable (steeper = higher tension), so students assume both components change. They also confuse this with beam shear diagrams where internal forces vary.
A cable under its own self-weight forms a parabola.
Tags
- formula_confusion
- conceptual_gap
- catenary_vs_parabola
Topic
Cables — Catenary vs Parabola
Severity
major
Exam Impact
Using parabola formulas for a catenary problem gives wrong H, wrong T_max, and wrong cable length. Board exam items sometimes explicitly state 'the cable supports its own weight' as a cue for catenary.
The Reality
The parabola applies ONLY when the load is uniformly distributed per unit HORIZONTAL length (e.g., the deck load of a suspension bridge). When the load is the cable's own self-weight — distributed per unit LENGTH OF CABLE — the true shape is a CATENARY (described by a hyperbolic cosine function y = a cosh(x/a)). For shallow sags (sag/span < 1/10), the parabola approximates the catenary with small error, which is why the approximation works in most exam problems. However, for deeply sagging cables or when the problem specifically states 'uniform load along cable length,' you must use catenary equations.
Trap Question
Question
A cable hangs freely under its own weight between two towers of equal height 50 m apart with a sag of 5 m. Which shape does the cable take?
Explanation
Self-weight is distributed per unit of cable arc length, which produces a catenary. The parabola arises from load per unit horizontal projection. Here d/L = 0.10 is at the threshold — the parabola approximation is marginal. If the exam problem specifies 'own weight' without shallow sag, the catenary is technically correct.
Wrong Answer
Parabola, using H = wL²/8d.
Correct Answer
Catenary (hyperbolic cosine). However, since d/L = 5/50 = 0.10, the parabola is an acceptable approximation for engineering calculations.
Misconception Id
M4
Correct Vs Incorrect
Correct Approach
If w is weight per unit horizontal distance → parabola (H = wL²/8d). If w is weight per unit cable length (or the problem says 'own weight') → catenary. For shallow sag (d/L ≤ 0.1), parabola is still acceptable as approximation.
Incorrect Approach
The cable hangs under its own weight, so I apply H = wL²/8d where w is the cable's unit weight.
Why Students Believe It
The most commonly taught cable formula (H = wL²/8d) gives a parabola. Students generalize this to ALL cable loading situations, including self-weight.
Maximum cable tension occurs at the lowest point (midspan) of the cable.
Tags
- common_error
- formula_confusion
- maximum_tension_location
Topic
Cables — Maximum Tension
Severity
major
Exam Impact
The board exam frequently asks for T_max. Answering with T at midspan gives T_min instead — completely opposite of what is asked.
The Reality
Cable tension T = √(H² + V²). At midspan of a symmetrically loaded parabolic cable, the vertical component V = 0, so T_midspan = H (minimum tension). At the SUPPORTS, V = wL/2 (maximum), so T_support = √(H² + (wL/2)²) — this is the MAXIMUM tension. The cable is steepest at the supports; steeper angle means larger vertical component means larger total tension.
Trap Question
Question
A suspension cable spans 80 m with a sag of 8 m and carries a uniform horizontal load of 5 kN/m. What is the maximum tension in the cable?
Explanation
H = 500 kN is the tension at midspan (minimum), not maximum. T_max is at the supports where both H and the vertical component contribute. Always use T_max = √(H² + V²) evaluated at the support.
Wrong Answer
T_max = H = wL²/8d = 5(80²)/(8×8) = 500 kN (at midspan)
Correct Answer
H = 5(80²)/(8×8) = 500 kN; V_support = wL/2 = 5(80)/2 = 200 kN; T_max = √(500² + 200²) = 538.5 kN (at supports)
Misconception Id
M5
Correct Vs Incorrect
Correct Approach
T is maximum at the supports where the cable is steepest. T_max = √(H² + (wL/2)²) for a symmetric parabolic cable. T_min = H at midspan.
Incorrect Approach
T_max is at midspan because the cable has maximum sag there. T_max = H = wL²/8d.
Why Students Believe It
The midspan is where the cable sags the most, which students intuitively link to maximum stress. Also, many beam problems have maximum bending moment at midspan, reinforcing this idea by analogy.
You can analyze the entire frame as a single rigid body to find all internal member forces directly.
Tags
- procedural_error
- common_error
- FBD_error
Topic
Frames — Analysis Procedure
Severity
major
Exam Impact
Students who skip dismembering cannot solve for internal pin forces at all. They often guess or leave these blank, losing all points associated with the internal analysis.
The Reality
Treating the whole frame as one FBD gives you the EXTERNAL reactions (at the supports). This is Step 1 and is necessary. But to find the forces between members (pin forces at internal joints), you MUST dismember the frame and analyze each member separately. The whole-frame FBD has no information about how forces are distributed internally between parts.
Trap Question
Question
A two-member frame has a pin support at A, a roller at B, and an internal pin connecting the two members at C. After finding the support reactions using the whole-frame FBD, how do you find the force at the internal pin C?
Explanation
The whole-frame FBD only has external forces — pin C is internal and invisible to the whole-frame analysis. Dismembering exposes the internal pin forces as external forces on each separate member FBD.
Wrong Answer
Substitute the support reactions back into the whole-frame equilibrium equations to solve for pin C forces.
Correct Answer
Dismember the frame at pin C. Draw a FBD of one member (e.g., the upper member) with the known support reactions and unknown Cx, Cy components at pin C, then apply ΣFx = 0, ΣFy = 0, ΣM = 0 to that member.
Misconception Id
M6
Correct Vs Incorrect
Correct Approach
Step 1: FBD of whole frame → find external support reactions. Step 2: Dismember → separate FBDs of each member → apply equilibrium to each → find internal pin forces. The whole-frame FBD cannot reveal internal pin forces.
Incorrect Approach
I will apply ΣM = 0 and ΣF = 0 to the whole frame to find all unknown forces including the pin at C between members AB and BC.
Why Students Believe It
For trusses and simple beams, treating the whole structure gives all required forces. Students try the same shortcut for frames, not realizing that treating the whole frame only gives EXTERNAL reactions, not internal pin forces between members.
Mechanical advantage in a machine is always greater than 1 (machines always amplify force).
Tags
- formula_confusion
- conceptual_gap
- mechanical_advantage
Topic
Machines — Mechanical Advantage
Severity
minor
Exam Impact
A board exam problem may describe a machine where the effort arm < load arm (e.g., tweezers-type grip), and ask for the output force. Students who assume MA > 1 will invert the ratio and get a wrong (amplified) force.
The Reality
Mechanical advantage MA = (effort arm)/(load arm) = Load/Effort. MA can be less than 1, equal to 1, or greater than 1. When MA < 1, the machine sacrifices force to gain speed or range of motion (e.g., a fishing rod, a tweezers, a broom). Engineers often deliberately design MA < 1 machines. Conservation of energy still holds: what you gain in force, you lose in displacement.
Trap Question
Question
A lever-type mechanism has the effort applied 0.04 m from the pivot and the load 0.16 m from the pivot. A 200 N effort is applied. What is the output load?
Explanation
The effort arm (distance from pivot to effort) is 0.04 m; the load arm (distance from pivot to load) is 0.16 m. MA = 0.04/0.16 = 0.25. This is a speed-multiplying machine — it moves the load farther but with less force. Many students invert the ratio when MA < 1.
Wrong Answer
MA = 0.16/0.04 = 4, so Load = 4 × 200 = 800 N
Correct Answer
MA = effort arm/load arm = 0.04/0.16 = 0.25; Load = 0.25 × 200 = 50 N
Misconception Id
M7
Correct Vs Incorrect
Correct Approach
MA = effort arm/load arm. If effort arm = 0.05 m and load arm = 0.20 m, then MA = 0.05/0.20 = 0.25. Load = 0.25 × Effort. This machine trades force for reach (like a tongs or tweezers).
Incorrect Approach
MA must be ≥ 1 because machines always make work easier. So Load ≥ Effort always.
Why Students Believe It
Common machines discussed in textbooks (levers, pulleys, hydraulic jacks) often amplify force, so students overgeneralize that all machines have MA > 1.
Any member pinned at both ends is automatically a two-force member.
Tags
- conceptual_gap
- common_error
- two_force_member
Topic
Frames — Two-Force Member Identification
Severity
major
Exam Impact
Incorrectly calling a loaded member a two-force member means you assign only axial force to it, ignoring shear and moments — the entire FBD becomes wrong.
The Reality
A member is a two-force member if and only if: (1) it is loaded at EXACTLY TWO POINTS (the pin endpoints), (2) there are NO external forces or moments applied between those two endpoints, and (3) the weight of the member is negligible. Being pinned at both ends is a NECESSARY but NOT SUFFICIENT condition. If there is any transverse load, any applied moment, or even significant self-weight along the member's length, it is a multi-force member regardless of the pin configuration.
Trap Question
Question
A horizontal member GH is pinned at G and H. A 10 kN vertical load is applied at its midpoint. A student says GH is a two-force member because it has pins at both ends. Is the student correct?
Explanation
Pins at both ends is not enough. Two-force member requires that the ONLY loading is at the two pin endpoints and nowhere else. The midpoint load violates this — GH has three loading points and is therefore a multi-force member.
Wrong Answer
Yes, GH is a two-force member because both ends are pinned.
Correct Answer
No. GH is a multi-force member because there is a 10 kN load applied at its midpoint (a third loading point). The force at pin G is NOT directed along GH.
Misconception Id
M8
Correct Vs Incorrect
Correct Approach
Check: Are there any loads between C and D, any applied moments, or significant self-weight? If YES → multi-force member → use unknown Cx, Cy at C and Dx, Dy at D. If NO → two-force member → force is along CD.
Incorrect Approach
Member CD is pinned at C and D, therefore the force it exerts is directed along CD. I'll resolve it at angle θ_CD.
Why Students Believe It
The typical description of a two-force member is 'pinned at two ends' and students hear this as the sufficient condition, missing the crucial additional requirement of no intermediate loads.
The cable formula H = wL²/8d can be used directly when supports are at different heights.
Tags
- formula_confusion
- boundary_condition_error
- parabolic_cable
Topic
Cables — Unsymmetric Geometry
Severity
major
Exam Impact
Board exams sometimes describe cables between supports of unequal height. Misapplying H = wL²/8d gives a completely wrong H and T_max.
The Reality
The formula H = wL²/8d is derived for a SYMMETRIC parabolic cable — both supports at the same height, with the sag d measured from the support level to the lowest point at midspan. When supports are at DIFFERENT heights (unsymmetric cable), the lowest point is NOT at midspan, the sag is measured differently, and the formula does not apply directly. You must use the general parabolic equation y = (w/2H)x² and set up two boundary conditions (one at each support) to find H and the location of the lowest point.
Trap Question
Question
A cable supports a uniform horizontal load of 3 kN/m over a horizontal span of 24 m. The left support is 4 m higher than the right support. The sag below the left support is 2 m. Can you use H = wL²/8d directly?
Explanation
H = wL²/8d requires symmetric geometry. Unequal support heights break this symmetry. The horizontal distance to the lowest point from each support is different (a ≠ b), and the two height differences give two equations in two unknowns (H and a or b).
Wrong Answer
Yes: H = 3(24²)/(8×2) = 108 kN.
Correct Answer
No. With unequal support heights, the lowest point is not at midspan and the symmetric formula does not apply. Set up y = (w/2H)x² with origin at the lowest point and two boundary conditions for H.
Misconception Id
M9
Correct Vs Incorrect
Correct Approach
Set origin at lowest point. y = (w/2H)x². At left support: y₁ = h₁, x = a (horizontal distance to left support). At right support: y₂ = h₂, x = b (horizontal distance to right support), with a + b = L = 30 m. Solve for H and a, b simultaneously.
Incorrect Approach
Left support is at elevation 0, right support at elevation 6 m, span 30 m, w = 2 kN/m, sag (from left support to lowest point) = 3 m. Apply H = wL²/8d = 2(30²)/(8×3) = 75 kN.
Why Students Believe It
The formula is memorized without understanding its derivation. Students apply it universally without checking the assumed geometry (both supports at the same elevation, load symmetric).
In a machine analysis, you can apply equilibrium to the machine as a whole to find the input-output force relationship directly.
Tags
- procedural_error
- moment_equation
- mechanical_advantage
Topic
Machines — Analysis Procedure
Severity
minor
Exam Impact
Students trying to use the whole-machine FBD for a compound machine end up with too many unknowns and cannot isolate the relationship between effort and load.
The Reality
For machines, the whole-machine FBD gives only external reactions from the machine's support. The input-output (effort-load) relationship is found by dismembering and applying equilibrium to individual moving parts. The reason: the effort and load are both external forces on the machine, but they act on DIFFERENT parts — you must isolate the part where both effort and load appear simultaneously, then take moments about the pivot of that part.
Trap Question
Question
A pair of bolt cutters has a pivot at P. The hand squeeze force is 150 N applied 300 mm from P. The cutting jaw contacts the bolt 20 mm from P. What is the cutting force on the bolt?
Explanation
Taking moments about pin P on ONE handle eliminates the unknown pin reaction. The ratio 300/20 = 15 is the mechanical advantage. F_cut = 150 × 15 = 2250 N. The whole-machine FBD is unnecessary and unhelpful for finding this relationship.
Wrong Answer
Cannot determine without the pin reaction — I need to analyze the whole bolt cutter.
Correct Answer
Isolate one handle-jaw assembly. ΣM_P = 0: 150(0.300) = F_cut(0.020) → F_cut = 2250 N.
Misconception Id
M10
Correct Vs Incorrect
Correct Approach
Dismember the pliers at the pivot pin. Analyze ONE jaw/handle as a free body — it has the hand force (effort), the jaw contact force (load), and the pivot pin reaction. Apply ΣM about the pivot pin to eliminate the pin reaction and find Load/Effort directly.
Incorrect Approach
I'll apply ΣM = 0 to the whole pliers (both handles + jaw) to get the grip force from the hand force.
Why Students Believe It
Students apply the 'whole-structure FBD' shortcut (which works for finding external reactions) to machines, thinking they can get internal forces the same way.
Cable tension equals zero at the lowest point of the cable.
Tags
- conceptual_gap
- common_error
- minimum_tension
Topic
Cables — Tension at Lowest Point
Severity
minor
Exam Impact
A board question may ask for the 'tension at the lowest point.' Answering zero loses marks. The correct answer is T = H, which should have been computed earlier in the problem.
The Reality
At the lowest point, the vertical component V = 0, but the horizontal component H ≠ 0. Since T = √(H² + V²) = √(H² + 0²) = H, the tension at the lowest point equals H — the constant horizontal component. This is actually the MINIMUM tension in the cable, not zero. A cable always has tension everywhere (otherwise it would be slack, not in equilibrium).
Trap Question
Question
A parabolic suspension cable has H = 400 kN. What is the cable tension at the lowest point of the cable?
Explanation
At the lowest point, the cable direction is horizontal, meaning the entire tension is in the horizontal direction. T = H = 400 kN. This is the minimum tension in the cable. The vertical component is zero, but the tension itself is not zero.
Wrong Answer
Zero, because the cable is horizontal at its lowest point.
Correct Answer
T = H = 400 kN
Misconception Id
M11
Correct Vs Incorrect
Correct Approach
At the lowest point, V = 0 (no vertical component). T = √(H² + V²) = √(H² + 0) = H. Tension is minimum (= H) at the lowest point, not zero.
Incorrect Approach
At the lowest point, the cable is horizontal, so the tension component along the cable direction in the vertical direction is zero — therefore tension T = 0 at the lowest point.
Why Students Believe It
At the lowest point of a parabola or catenary, the cable slope is horizontal (dy/dx = 0), meaning the vertical component of tension V = 0. Students incorrectly conclude that T = 0 as well.
Quick Self Check
A two-force member must be loaded at ONLY its two pin endpoints with no intermediate forces or moments. A member pinned at both ends but carrying a transverse load at its midpoint is a multi-force member.
Statement
Any member that is pinned at both ends is automatically a two-force member.
With no horizontal external loads between any two points, horizontal equilibrium (ΣFx = 0) demands that H is the same in every segment. This is the fundamental property of cable analysis.
Statement
The horizontal component of tension H is constant throughout a cable loaded by vertical forces only.
Maximum tension occurs at the supports where the cable is steepest. At midspan, T = H (minimum). At supports, T_max = √(H² + (wL/2)²), which is always greater than H.
Statement
The maximum tension in a symmetrically loaded parabolic cable occurs at midspan.
Self-weight distributed per unit cable length produces a CATENARY (hyperbolic cosine shape). A parabola results from load distributed per unit horizontal length (e.g., bridge deck load). For shallow sags, the catenary is well-approximated by a parabola.
Statement
A cable hanging under its own self-weight forms a parabolic shape.
Newton's Third Law: if member A pushes member B upward through pin C, then member B pushes member A downward through the same pin C. FBDs of dismembered parts must show equal-and-opposite pin forces.
Statement
When dismembering a frame, pin forces on the two members sharing a pin must be drawn in opposite directions.
H = wL²/8d is valid only for symmetric cables with both supports at the same elevation. For supports at different heights, the general parabolic equation with two boundary conditions must be used.
Statement
The formula H = wL²/8d applies to any cable with a uniform distributed load regardless of support elevation difference.
At the lowest point, the vertical component V = 0, so T = √(H² + 0²) = H. This is the MINIMUM tension in the cable, equal to H, not zero.
Statement
At the lowest point of a parabolic cable, the tension is equal to the constant horizontal component H.
MA = effort arm/load arm. If the effort arm is shorter than the load arm, MA < 1. Such machines trade force for speed or range of motion (e.g., tweezers, fishing rod, broom). MA can be any positive value.
Statement
Mechanical advantage (MA) of a machine is always greater than or equal to 1.
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