Cable Systems - Statics and Dynamics

1. If the sag of the main cable at the lowest point is 4 m below the chord, what is the load $W$ suspended from the lowest point?
   Answer: $W = H \times (\tan\theta_L + \tan\theta_R)$ where $\theta_L$ and $\theta_R$ are the slopes of the left and right cable segments at the low point; numerically $W = 50(4/10 + 4/20) = 50(0.4 + 0.2) = 30 \text{ kN}$. When analyzing a cable with a concentrated load at its lowest point, the key is vertical equilibrium at that point. The horizontal tension component $H$ remains constant throughout the cable, and the vertical force components from each cable segment must together balance the applied load $W$. Each segment contributes a vertical force equal to $H \times \tan\theta$, where $\theta$ is that segment's slope angle. Here, the lowest point sits 10 m from the left support and 20 m from the right, with a 4 m sag. The left segment rises 4 m over 10 m horizontally, giving $\tan\theta_L = 4/10 = 0.4$. The right segment rises 4 m over 20 m, giving $\tan\theta_R = 4/20 = 0.2$. Vertical equilibrium then yields $W = H(\tan\theta_L + \tan\theta_R) = 50(0.4 + 0.2) = 30 \text{ kN}$, confirming answer A. Answer B incorrectly assumes the lowest point must be at midspan, forcing a symmetric geometry that doesn't exist here. Midspan placement only occurs when the load is centered between symmetric supports — a special case, not a general rule. Answer C makes the error of assuming the right cable segment is horizontal at the low point, contributing zero vertical force. In reality, both segments slope upward away from the lowest point, so both carry vertical components. Answer D uses $2H\sin\theta$ with a single averaged angle, incorrectly assuming symmetry. It also mixes up when to use sine versus tangent — since $H$ is the horizontal component, you multiply by $\tan\theta$, not $\sin\theta$. As a study tip: whenever a cable's low point is off-center, always compute separate slope angles for each segment and sum both vertical contributions.
2. Assuming the system is in equilibrium and all cables are taut, which of the following correctly characterizes the static determinacy of this system?
   Answer: The system is statically determinate. Symmetry forces $T_2 = T_3$, reducing the problem to two unknowns. The vertical force equation ΣFy = 0 and the moment equation ΣM = 0 about the left edge then provide two independent equations sufficient to uniquely solve for $T_1$ and $T_2$. When analyzing a suspended platform problem, your first move should always be to count unknowns versus independent equations — but don't stop there. Look for geometric conditions like symmetry that reduce the unknown count before declaring indeterminacy. Here, three cables create three unknown tensions: $T_1$, $T_2$, and $T_3$. In 2D, rigid-body equilibrium gives you three scalar equations: $\Sigma F_x = 0$, $\Sigma F_y = 0$, and $\Sigma M = 0$. Naively, three unknowns and three equations sounds determinate — but notice that $\Sigma F_x = 0$ yields $T_2 \sin 20° = T_3 \sin 20°$, which simply confirms $T_2 = T_3$. This isn't a new independent equation; it's just the mathematical expression of symmetry. So you effectively have two unknowns ($T_1$ and $T_2$) and two genuinely independent equations — $\Sigma F_y = 0$ and $\Sigma M = 0$ about any convenient point. That's exactly determined, making C correct. Choice A overcounts the degrees of indeterminacy by ignoring how symmetry eliminates one unknown, leaving the system over-constrained in its analysis. Choice B correctly identifies that symmetry reduces unknowns but wrongly claims the moment equation is automatically satisfied regardless of tension values — it's not; taking moments about, say, the left edge still gives a real constraint that uniquely determines $T_1$. Choice D makes the subtle but critical error of treating $\Sigma F_x = 0$ as an independent third equation, when in fact it merely restates the symmetry condition already used to set $T_2 = T_3$. Your study takeaway: whenever symmetry is present, use it to reduce unknowns first, then count truly independent equations. Don't double-count symmetry as both a simplification and an equation.
3. Which of the following correctly expresses both $T_1$ and $T_2$ obtained by applying equilibrium at the joint?
   Answer: $T_1 = 10 \text{ kN}$ and $T_2 = 10\sqrt{3} \approx 17.3 \text{ kN}$, found by resolving horizontal and vertical equilibrium with the sine rule applied to the triangle of forces. When three cables meet at a frictionless joint in static equilibrium, the net force in every direction must be zero. Your tool is two scalar equations: $\sum F_x = 0$ and $\sum F_y = 0$. Set up a coordinate system, decompose each tension into horizontal and vertical components using the cable's angle from horizontal, then solve the resulting system. Cable 1 pulls left at 30° above horizontal, so its components are $(-T_1\cos30°,\; +T_1\sin30°)$. Cable 2 pulls right at 60° above horizontal: $(+T_2\cos60°,\; +T_2\sin60°)$. Cable 3 pulls straight down: $(0,\; -20)$. Horizontal equilibrium ($\sum F_x = 0$): $T_2\cos60° = T_1\cos30° \implies \frac{T_2}{2} = \frac{T_1\sqrt{3}}{2} \implies T_2 = T_1\sqrt{3}$ Vertical equilibrium ($\sum F_y = 0$): $T_1\sin30° + T_2\sin60° = 20 \implies \frac{T_1}{2} + \frac{T_1\sqrt{3}\cdot\sqrt{3}}{2} = 20 \implies 2T_1 = 20 \implies T_1 = 10\text{ kN},\; T_2 = 10\sqrt{3} \approx 17.3\text{ kN}$ This confirms B is correct. A swaps the two values. Because Cable 2 makes the steeper angle (60° vs 30°), it carries more vertical load, so it must have the larger tension — that's $T_2$, not $T_1$. C treats each cable as if it independently supports the entire weight $W$, ignoring the coupling between them through horizontal equilibrium. You cannot set each vertical component equal to $W$ separately. D invents a decomposition of $W$ along cable directions using cosines, which has no physical basis in Cartesian equilibrium. A useful habit: before solving, ask which cable is steeper. The steeper cable always carries the greater tension — a quick sanity check to catch swapped answers like A.
4. A rigid horizontal beam AB (length 6 m, weight negligible) is supported at A by a pin and at point C (4 m from A) by a cable that makes 35° with the horizontal. A vertical load of 24 kN is applied at B. A student attempts to find the cable tension $T$ by resolving the vertical component of $T$ directly: $T\sin35° = 24 \text{ kN}$, giving $T = 41.8 \text{ kN}$. What error has the student made, and what is the correct cable tension?
   Answer: The student ignored the pin reaction at A. Taking moments about A: $T\sin35° \times 4 = 24 \times 6$, giving $T = 62.7 \text{ kN}$. The pin at A carries the remaining vertical and horizontal forces. When analyzing a beam with multiple supports, your first instinct should be to use moment equilibrium, not force equilibrium alone. A pin support at A can provide both vertical and horizontal reactions, meaning vertical force balance alone gives you one equation with two unknowns ($A_y$ and $T$) — you need more information. The smart move is to take moments about A, which eliminates the pin reactions entirely (they act at A, so their moment arm is zero). The cable at C applies an upward vertical force of $T\sin35°$ at 4 m from A, and the 24 kN load acts downward at 6 m from A. Setting the sum of moments about A equal to zero: $T\sin35° \times 4 = 24 \times 6$ $T = \frac{144}{4\sin35°} = \frac{144}{2.294} \approx 62.7 \text{ kN}$ This confirms A is correct. Once you have $T$, you can return to force equilibrium to find the pin reactions at A. The student's error was treating the beam as if the cable were the only support — essentially ignoring the pin at A. The vertical forces don't have to balance between just the cable and the load; the pin contributes too. B is wrong because horizontal force components never appear in vertical equilibrium — mixing them creates a physically meaningless equation. C introduces an incorrect angle substitution; the moment arm for a vertical cable component on a horizontal beam is simply the horizontal distance, not adjusted by $\cos35°$. D falsely claims $A_y = 0$; a pin on a beam with an offset load absolutely carries vertical reaction. Study tip: Whenever a beam has a pin support, take moments about that pin first — it's almost always the fastest path to finding unknown forces.
5. A pin-jointed frame has a joint P where four members meet. Three members are rigid bars (can carry tension or compression), and one member is a cable (tension only). The cable runs at 50° above horizontal toward the upper-right. The three rigid bars run: (1) horizontally to the left, (2) vertically downward, and (3) at 30° below horizontal to the lower-left. An unknown external force $\mathbf{F}$ is applied at P. A student argues that because the cable is a tension-only member, if analysis of joint P — ignoring the cable — results in a net force pointing away from the upper-right at joint P, the cable must be slack (zero tension). Under what condition is this reasoning correct?
   Answer: The reasoning is correct only if the three rigid bars are statically determinate on their own — that is, the three bar forces can be uniquely found from the two equilibrium equations at P without the cable — and the resulting net residual force at P points in the direction the cable would need to push (compress) rather than pull. Whenever you encounter a tension-only member at a joint, your first instinct should be to ask: can I solve for all other member forces independently? That question is the heart of this problem. At joint P, you have two equilibrium equations ($\sum F_x = 0$ and $\sum F_y = 0$) and four unknowns. If the cable is removed temporarily, you're left with three bar forces and one external force F — still potentially more unknowns than equations. The student's reasoning only works if those three bar forces are uniquely determinable from the two equations without the cable. This happens when the geometry of the three bars and the known external load make the system statically determinate by themselves — for example, if one bar's orientation directly isolates a force, allowing back-substitution. Once you've uniquely solved for all three bar forces, you can compute the net residual force that the cable would need to supply. If that residual points in the cable's compression direction (pushing rather than pulling), the cable cannot provide it and must be slack. That's exactly what D states, making it correct. A is wrong because the cable's direction relative to F is irrelevant — what matters is whether the three bars alone are statically determinate, not whether F has a component along the cable axis. B is wrong because the reasoning is not always valid for any joint; if the bars themselves are indeterminate, you can't isolate the residual force without additional information. C is wrong because four-member joints are not automatically indeterminate — the specific geometry can still allow a determinate sub-system when one member is removed. Study tip: When analyzing tension-only members, always verify determinacy of the remaining structure first. If removing the cable leaves a determinate system, check the residual's direction — compression demand means the cable is slack.

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