Vector Algebra
Moment of couple
Grade 12
Question:
<p>A couple is of moment \(\vec{G}\) and the force forming the couple is \(\vec{P}\). If \(\vec{P}\) is turned through a right angle, the moment of the couple thus formed is \(\vec{H}\). If instead, the forces \(\vec{P}\) are turned through an angle \(\alpha\), then the moment of couple becomes</p>
<p>\(\vec{G}\sin\alpha - \vec{H}\cos\alpha\)</p>
<p>\(\vec{H}\cos\alpha + \vec{G}\sin\alpha\)</p>
<p>\(\vec{G}\cos\alpha + \vec{H}\sin\alpha\)</p>
<p>\(\vec{H}\sin\alpha - \vec{G}\cos\alpha\)</p>
Step-by-Step Solution
Key Concept: The moment of a couple is perpendicular to the plane containing the two equal and opposite forces. When force direction changes, the moment vector rotates in 3D space, and its magnitude depends on the angle of rotation through the geometric relationship: |M(α)| = |M₀|√(2 - 2cos α) or using the constraint that G, H, and the new moment form a configuration in vector space.
Step 1: Let the initial force be P with moment G . The moment is perpendicular to P and lies in a specific direction. Step 2: When P is rotated by 90°, the new moment H is perpendicular to both the original moment G and the rotated force. Since the rotation is 90°, we have G ⊥ H (they are perpendicular), and |G| = |H| = | P | × d (where d is the perpendicular distance). Step 3: When force is rotated by angle α, the moment vector M (α) rotates in the plane containing G and H . The new moment is a linear combination: M (α) = G cos α + H sin α Step 4: Taking magnitude: | M (α)| = √(G^2cos^2α + H^2sin^2α + 2GH·cosα·sinα·cos 90°) = √(G^2cos^2α + H^2sin^2α) Step 5: Since |G| = |H|, this gives | M (α)| = |G| ∴ Answer: The moment of couple is √(G^2 + H^2) or the resultant of G and H at angle α, which equals G (constant magnitude for this symmetric case) or more generally √(G^2 + H^2 + 2GH cos(α - π/2))
Correct Answer: C