Vector Algebra
Cross Product / Moments
Grade 12
Question:
<p>In a right angle \(\triangle ABC\), \(\angle A = 90°\) and sides \(a, b, c\) are, respectively, 5 cm, 4 cm and 3 cm. If a force \(\vec{F}\) has moments 0, 9 and 16 in N cm units, respectively, about vertices \(A\), \(B\) and \(C\), then magnitude of \(\vec{F}\) is</p>
<p>3</p>
<p>4</p>
<p>5</p>
<p>9</p>
Step-by-Step Solution
Key Concept: The moment of a force about a point equals |F| × d, where d is the perpendicular distance from the point to the line of action. Use the three moment equations simultaneously to find both the magnitude and position of the force's line of action.
Step 1: Set up the problem. In right triangle ABC with ∠A = 90°, place A at origin, B at (3,0), and C at (0,4). Note: a=BC=5, b=AC=4, c=AB=3. Step 2: Let force F have magnitude F and act along a line. The moment about point P is M_P = F × d_P, where d_P is perpendicular distance from P to the line of action. Step 3: Given: M_A = 0, M_B = 9, M_C = 16 (in N cm). From M_A = 0: The force's line of action passes through A (distance = 0). Step 4: Since the line passes through A(0,0), the line equation is: αx + βy = 0 (where α^2 + β^2 = 1 for unit normal). Step 5: Distance from B(3,0) to line: d_B = |3α| = 3|α|
Distance from C(0,4) to line: d_C = |4β| = 4|β| Step 6: Using moment equations:
M_B: F × 3|α| = 9 → F|α| = 3
M_C: F × 4|β| = 16 → F|β| = 4 Step 7: Since α^2 + β^2 = 1:
(F|α|)^2 + (F|β|)^2 = F^2(α^2 + β^2) = F^2
9 + 16 = F^2
F^2 = 25 Step 8: Therefore, F = 5 N ∴ Answer: C (Magnitude of force = 5 N)
Correct Answer: C