Vector Algebra
Equilibrium of Forces
Grade 12
Question:
<p>A weight of 13 kg is supported by two strings of lengths 5 and 12. Given that \(13^2 = 5^2 + 12^2\), the angle \(\angle AOB = \dfrac{\pi}{2}\). Using equilibrium conditions, \(T_1\) and \(T_2\) are:</p>
<p>\(T_1 = 12,\ T_2 = 5\)</p>
<p>\(T_1 = 5,\ T_2 = 12\)</p>
<p>\(T_1 = 13,\ T_2 = 5\)</p>
<p>\(T_1 = 5,\ T_2 = 13\)</p>
Step-by-Step Solution
Key Concept: In equilibrium, the vector sum of tensions must equal the weight force. Since 5² + 12² = 13² forms a right triangle, the tension vectors form a right angle, making this a classic 3-4-5 Pythagorean triple scaled by factor of 1 (or 5-12-13).
Step 1: Set up equilibrium condition. The weight acts downward with force W = 13g (in Newtons, where g ≈ 9.8 m/s^2). The two tension vectors T_1 and T_2 act along strings of lengths 5 and 12, meeting at angle ∠AOB = π/2. Step 2: Apply force balance. Since the strings are perpendicular (∠AOB = π/2), resolve tensions into components. The resultant of T_1 and T_2 must equal the weight: |T_1|^2 + |T_2|^2 = W^2. Step 3: Use the constraint from geometry. The tensions along the strings of lengths 5 and 12 satisfy: T_1^2 + T_2^2 = (13g)^2. By the Pythagorean relationship and force equilibrium: T_1 = 5g and T_2 = 12g (in appropriate units). Step 4: Verify: √[(5g)^2 + (12g)^2] = √[25g^2 + 144g^2] = √[169g^2] = 13g = W ✓ ∴ Answer: B (T_1 = 5g, T_2 = 12g, where g is gravitational acceleration)
Correct Answer: B