Static Equilibrium: Exercise Notes (Video 2)

Problem 1: System in equilibrium with BC must remain horizontal

Context: A static system with two ropes (BC and AB) and a pulley. The BC rope stays horizontal. Mass and gravity given for one block; tensions in the ropes must balance to keep equilibrium.
Given:
- Mass m = 3 kg; gravity g = 10 m/s^2
- Weight of the 3 kg block: W_1 = m g = 3 \times 10 = 30\ \text{N}
- The weight acts downward (on the block connected to the rope at B) and the rope BC is horizontal.
- The rope AB forms an angle with the horizontal (used in the solution). In the derivation, the vertical component of AB is used with sin 30°.
Step 1: Identify the known horizontal/vertical components at the joint:
- The segment BC is horizontal, so its tension T_BC acts horizontally.
- The weight W1 = 30 N must be balanced by vertical components in the system via AB.
Step 2: Balance the vertical forces at the junction where AB connects (point B):
- The vertical component of the tension in AB must balance W1: T{AB} \sin(30^{\circ}) = W1 = 30\ \text{N}
- Solve for the tension in AB: T_{AB} = \frac{30}{\sin(30^{\circ})} = \frac{30}{0.5} = 60\ \text{N}
Step 3: Determine T_BC using the horizontal balance (as described in the video):
- The video notes that the tração in BC is linked to the balance of the forces; from the setup, its value is T_{BC} = 30\ \text{N}.
- (Note: The video text treats T_BC as equal to the weight involved in the horizontal balance for this arrangement.)
Result for Problem 1:
- T_AB = 60\ \text{N}
- T_BC = 30\ \text{N}
Key takeaway:
- Use weight W = m g to set the vertical balance: the appropriate component of the rope tension must equal the weight.
- The angle used here is the angle of AB relative to the horizontal, so the vertical component is T_AB sin(30°).
- When a rope segment is horizontal, its tension acts horizontally and contributes to horizontal balance with other horizontal components.
Important concept connections:
- Vector decomposition: split tensions into horizontal and vertical components.
- Equilibrium condition: sum of forces in each direction equals zero.
- The “macete” described in the video: when an angle is measured from the horizontal, horizontal component uses cos and vertical uses sin; when measured from the vertical, swap sin/cos accordingly.

Problem 2: Lantern (luminaire) of 80 N weight suspended by two cords at symmetric angles

Context: A luminária with weight 80 N is suspended by two cords CA and CB, making equal angles with the horizontal (30°).
Given:
- Weight of lamp: W = 80\ \text{N}
- Angles: each rope makes a 30° angle with the horizontal (the video uses the notation that the angle is measured from the horizontal, so vertical component uses sin).
Setup:
- Let the tensions be equal in the two cords: T{CA} = T{CB} = t (due to symmetry and equal angles).
Step 1: Horizontal balance:
- By symmetry, horizontal components cancel: T{CA} \cos(30^{\circ}) = T{CB} \cos(30^{\circ})\quad\Rightarrow\quad T{CA} = T{CB}
Step 2: Vertical balance:
- Vertical components must support the weight: 2 t \sin(30^{\circ}) = W = 80\ \text{N}
- Solve for t: t = \frac{W}{2 \sin(30^{\circ})} = \frac{80}{2 \times 0.5} = 80\ \text{N}
Result for Problem 2:
- TCA = TCB = 80\ \text{N}
Additional notes from the video:
- The speaker discusses that different trigonometric decompositions (using sin/cos depending on which angle is adjacent to vertical or horizontal) lead to the same result, provided the geometry is applied consistently.
- The video also mentions an alternative route using trigonometric identities or a triangle approach, but symmetry makes the two tensions equal in this configuration.
Key takeaway:
- For symmetric suspension, tensions are equal: T1 = T2 = W / (2 sin θ).
Notable angles and references:
- 30°, sin 30° = 1/2, cos 30° = √3/2.
- If different angles were present, you would have three possibly different tensions and you’d set up two independent equilibrium equations (sum of horizontal components and sum of vertical components).

Problem 3: Luminária between two walls with two equal ropes at equal angles to horizontal

Context: A 60 N weight (mass 6 kg, g = 10) lamp is suspended by two cords that connect to opposite walls. The two ropes CA and CB make equal angles with the horizontal, meeting at the lamp, forming a symmetric setup. Distances in the supporting geometry give a 3-4-5 right triangle in the projection.
Given:
- Mass m = 6 kg; gravity g = 10 m/s^2
- Weight: W = m g = 60\ \text{N}
- Geometry (from the diagram and the video): the half-triangle has legs 30 cm (vertical) and 40 cm (horizontal), so the rope length (hypotenuse) is 50 cm. Thus the angle θ between each rope and the horizontal satisfies:
- \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{30}{50} = \frac{3}{5}
- \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{40}{50} = \frac{4}{5}
Step 1: Tensions in each rope are equal due to symmetry: denote the common tension by t (i.e., TCA = TCB = t).
Step 2: Vertical balance at the lamp:
- The vertical components of the two tensions must sum to the weight: 2 t \sin\theta = W = 60\ \text{N}
- Using sin θ = 3/5: 2 t \left(\frac{3}{5}\right) = 60 \Rightarrow t \left(\frac{3}{5}\right) = 30
- Solve for t: t = \frac{30}{(3/5)} = 50\ \text{N}
Result for Problem 3:
- TCA = TCB = 50\ \text{N}
Alternative verification via Law of Cosines (conceptual):
- The two tensions form an angle φ between them. Using the vector sum T1 + T2 = -W and |T1| = |T2| = t, one can write
- |W|^2 = |T1 + T2|^2 = t^2 + t^2 + 2 t^2 \cos\phi = 2 t^2 (1 + \cos\phi)
- For this symmetric configuration with equal angle θ relative to horizontal, φ = 180° - 2θ, so cos φ = cos(180° - 2θ) = -cos(2θ).
- Using θ with sin θ = 3/5 and cos θ = 4/5, cos(2θ) = cos^2 θ - sin^2 θ = (16/25) - (9/25) = 7/25, so cos φ = -7/25.
- Then 60^2 = 2 t^2 [1 + (-7/25)] = 2 t^2 (18/25) = (36/25) t^2, giving t^2 = 3600 / (36/25) = 2500, so t = 50 N, consistent with the vertical balance.
Key takeaways and cautions:
- Equal angles lead to equal tensions (T1 = T2 = t).
- If the angles were not equal, you’d have three different tensions and you’d set up separate equilibrium equations for horizontal and vertical directions.
- In this problem, the 3-4-5 triangle gives a convenient sin θ = 3/5 and cos θ = 4/5, which makes the numeric solution straightforward.
Quick recap of core formulas used in these problems:
- Weight: W = m g
- Decomposition of a tension into horizontal/vertical components when the rope makes an angle θ with the horizontal: Tx = T \cos\theta, \quad Ty = T \sin\theta
- Equilibrium conditions (sum of forces in each direction equals zero):
- Horizontal: sum F_x = 0
- Vertical: sum F_y = 0
- For symmetric suspensions with angle θ to the horizontal and weight W, vertical balance gives: 2 T \sin\theta = W leading to T = \frac{W}{2 \sin\theta}
- If necessary, Law of Cosines check for the resultant of two equal tensions at angle φ: |W|^2 = 2t^2(1 + \cos\phi) with φ the angle between the two tension vectors.

Summary and practical notes

Always compute weights first: W = m g
Identify which rope tensions balance vertical vs horizontal components at each joint.
For symmetric configurations (equal angles), tensions tend to be equal; use this to reduce unknowns to a single variable.
Use the known special triangles (30-60-90 and 3-4-5) to get sin and cos values quickly:
- 30-60-90: \sin 30° = \tfrac{1}{2}, \cos 30° = \tfrac{\sqrt{3}}{2}, etc.
- 3-4-5 triangle: \sin\theta = \tfrac{3}{5}, \cos\theta = \tfrac{4}{5} when the opposite side is 3 and the hypotenuse is 5.
When there’s a triangular geometry (like a rope length forming a right triangle with known legs), you can extract sin/cos directly from the ratio of legs to the hypotenuse.
The video also demonstrates flexibility in approaches: decomposing vectors directly, using sine/cosine projections, or applying the law of cosines as a cross-check.
Real-world relevance: these static-equilibrium problems model supports for ceilings, bridges, electrical pylons, and any system where multiple cables/ropes hold a load in place.

Quick reference: key equations used in these problems

Weight: W = m g
Vertical component of a tension at angle θ: T_y = T \sin\theta
Horizontal component of a tension at angle θ: T_x = T \cos\theta
Horizontal equilibrium (general): \sum F_x = 0
Vertical equilibrium (general): \sum F_y = 0
For symmetric two-cable support with angle θ to the horizontal and load W: 2T\sin\theta = W \Rightarrow T = \frac{W}{2\sin\theta}
For a 3-4-5 triangle: if opposite side is 3 and hypotenuse is 5, then \sin\theta = \tfrac{3}{5}, \; \cos\theta = \tfrac{4}{5}
Law of Cosines check for two equal tensions at angle φ: |W|^2 = 2t^2(1 + \cos\phi) with φ the angle between the two tension vectors.