Question

Q1.At whats angle
do the two forces (P+Q) and (P-Q) act so
that resultant is
✓(3(P^2) + (Q^2))​

Answer 1

Step-by-step explanation:

Given that,

$\sf:\implies\:F_1=(P+Q)$

$\sf:\implies\:F_2=(P-Q)$

Resultant force;

$\sf:\implies\:F=\sqrt{3P^2+Q^2}$

We have to find angle between two force vectors.

❖ As per parallelogram law of vector addition, magnitude of resultant vector R of two vectors A and B inclined at an angle of θ is given by;

R² = A² + B² + 2AB cosθ

By substituting the given values;

$\sf\dashrightarrow\:(\sqrt{3P^2+Q^2})^2=(P+Q)^2+(P-Q)^2+2(P+Q)(P-Q)\cdot cos\theta</p><p>$

We know that,

(A + B)² = A² + 2AB + B²

(A - B)² = A² - 2AB + B²

(A² - B²) = (A + B) (A - B)

$\sf\dashrightarrow\:3P^2+Q^2=(P^2+2PQ+Q^2)+(P^2-2PQ+Q^2)+2(P^2-Q^2)\cdot cos\theta$

$\sf\dashrightarrow\:3P^2+Q^2=2P^2+2Q^2+2(P^2-Q^2)\cdot cos\theta$

$\sf\dashrightarrow\:(3P^2-2P^2)+(Q^2-2Q^2)=2(P^2-Q^2)\cdot cos\theta$

$\sf\dashrightarrow\:(P^2-Q^2)=2(P^2-Q^2)\cdot cos\theta$

$\sf\dashrightarrow\:2cos\theta=\dfrac{(P^2-Q^2)}{(P^2-Q^2)}$

$\sf\dashrightarrow\:cos\theta=\dfrac{1}{2}$

$\dashrightarrow\:\underline{\boxed{\bf{\orange{\theta=60^{\circ}}}}} </p><p>$

Answer 2

Answer:

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