--- title: "Find T and θ for equilibrium" subject: "Applied Mechanics" chapter: 2 key: "semester_1/applied_mechanics:find-t-and-theta-equilibrium-10kg" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/find-t-and-theta-equilibrium-10kg site: BCT Engineering --- # Find T and θ for equilibrium Applied Mechanics · Chapter 2: Basic Concept in Statics and Static Equilibrium - Asked in: 2067 Asoj, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/find-t-and-theta-equilibrium-10kg - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-2 ## Question Find the value of T and θ in figure to be the system in equilibrium. [Figure: a 10 kg mass hangs from a junction; rope T acts up-left at angle θ and a 50 N force acts up-right at 30° to the horizontal.] ## Short answer Weight is 98.1 N. Equilibrium equations give T sin θ = 43.30 N and T cos θ = 73.10 N. Hence θ = 30.64° from the vertical and T = 84.96 N. ## Answer **Reading of the figure:** a 10 kg mass hangs from a junction. Rope T pulls up and to the left, making angle θ with the vertical. A 50 N force pulls up and to the right at 30° above the horizontal. Take g = 9.81 m/s², so W = 10 × 9.81 = 98.1 N. #### Equilibrium of the junction ΣFx = 0: T sin θ − 50 cos 30° = 0, so T sin θ = 43.30 N …(1) ΣFy = 0: T cos θ + 50 sin 30° − 98.1 = 0, so T cos θ = 73.10 N …(2) Dividing (1) by (2): tan θ = 43.30 / 73.10 = 0.5924, so **θ = 30.64°** (from the vertical). T = √(43.30² + 73.10²) = **84.96 N** **Answer: T = 84.96 N (≈ 85 N), θ = 30.64° with the vertical.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does SigmaFx = 0 give at the junction? A. T sin(theta) = 50 cos30° = 43.30 N B. T cos(theta) = 50 cos30° C. T sin(theta) = 98.1 N D. T sin(theta) = 50 sin30° Answer: A. T pulls left with horizontal component T sin(theta) (theta is from the vertical), balancing the 50 N force's horizontal component 50 cos30°. (Key point: Horizontal balance) 2. Why is T cos(theta) = 73.10 N and not 98.1 N? A. The 50 N force's vertical part (50 sin30° = 25 N) already supports part of the 98.1 N weight B. Because g is taken as 9.81 only for the rope C. Because the rope tension is reduced by friction D. Because only half of the weight acts on the junction Answer: A. SigmaFy = 0: T cos(theta) + 25 - 98.1 = 0, so T cos(theta) = 73.10 N. (Key point: Vertical balance) 3. How are theta and T obtained from the two equations? A. Divide (1) by (2) to get tan(theta), then T = sqrt(43.30^2 + 73.10^2) B. Add the two equations to get T, then subtract for theta C. Set theta = 30° because the other force is at 30° D. Use T = 98.1 N because it supports the mass Answer: A. tan(theta) = 43.30/73.10 gives theta = 30.64°, and the resultant of the components gives T = 84.96 N. (Key point: Finding theta and T) ## Related in Applied Mechanics - [Force-couple system about a point](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aforce-couple-system-beam-60kn-20knm-50kn) — Applied Mechanics, chapter 3 - [Find forces P and Q for equilibrium](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aunknown-forces-p-q-equilibrium) — Applied Mechanics, chapter 2 - [Find F1 and F2 for equilibrium](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aunknown-forces-f1-f2-equilibrium) — Applied Mechanics, chapter 2 - [Force P to give block acceleration of 10 m/s²](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aforce-p-block-2500n-acceleration-10) — Applied Mechanics, chapter 9 - [Force P to prevent block sliding down incline](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Ablock-50kg-force-p-prevent-sliding-down) — Applied Mechanics, chapter 5 - [Force-couple system at A; couple is free vector](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aforce-couple-system-point-a-box-couple-free-vector) — Applied Mechanics, chapter 3 ## Related in other subjects - [Two balls impact e = 0.9, 30 and 50 kg](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Atwo-balls-e09-30-50kg) — Applied Mechanics (Dynamics), chapter 2 - [Right circular cone, vertical angle 30°](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aright-circular-cone-vertical-angle) — Mathematics, chapter 3, asked 3× - [Horizontal length of 30 m span, 50 m tape](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atape-50m-span-30m-26c) — Surveying I, chapter 2 - [Derive resultant force, variable mass system](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Avariable-mass-resultant-force) — Applied Mechanics (Dynamics), chapter 3, asked 2×