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Motion in a Plane - Vector Addition — Analytical Method

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The analytical method of vector addition involves using trigonometry and algebra to find the magnitude and direction of a resultant vector, which is more precise than graphical methods like the triangle or parallelogram laws.

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If two vectors A⃗\vec{A} and B⃗\vec{B} are acting at an angle θ\theta to each other, their resultant R⃗\vec{R} is the diagonal of the parallelogram formed by these vectors.

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The magnitude of the resultant vector RR depends on the magnitudes of individual vectors AA and BB and the cosine of the angle θ\theta between them.

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The direction of the resultant vector R⃗\vec{R} is usually expressed as an angle α\alpha that it makes with the direction of vector A⃗\vec{A}.

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For special cases: when θ=0∘\theta = 0^\circ (vectors are parallel), the resultant is maximum R=A+BR = A + B. When θ=180∘\theta = 180^\circ (vectors are anti-parallel), the resultant is minimum R=∣A−B∣R = |A - B|. When θ=90∘\theta = 90^\circ, R=A2+B2R = \sqrt{A^2 + B^2}.

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Resolution of vectors: A vector can be split into two rectangular components, Ax=Acos⁡θA_x = A \cos \theta and Ay=Asin⁡θA_y = A \sin \theta. The resultant of multiple vectors can be found by summing their respective xx and yy components.

📐Formulae

R=A2+B2+2ABcos⁡θR = \sqrt{A^2 + B^2 + 2AB \cos \theta}

α=tan⁡−1(Bsin⁡θA+Bcos⁡θ)\alpha = \tan^{-1} \left( \frac{B \sin \theta}{A + B \cos \theta} \right)

R⃗=(Ax+Bx)i^+(Ay+By)j^\vec{R} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}

R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}

tan⁡α=RyRx\tan \alpha = \frac{R_y}{R_x}

💡Examples

Problem 1:

Two forces of 5 N5\text{ N} and 10 N10\text{ N} act on a particle at an angle of 60∘60^\circ with each other. Find the magnitude of the resultant force and the angle it makes with the 5 N5\text{ N} force.

Solution:

Given: A=5 NA = 5\text{ N}, B=10 NB = 10\text{ N}, and θ=60∘\theta = 60^\circ.

  1. Magnitude: R=52+102+2(5)(10)cos⁡60∘R = \sqrt{5^2 + 10^2 + 2(5)(10) \cos 60^\circ} R=25+100+100(0.5)R = \sqrt{25 + 100 + 100(0.5)} R=125+50=175≈13.23 NR = \sqrt{125 + 50} = \sqrt{175} \approx 13.23\text{ N}
  2. Direction: tan⁡α=10sin⁡60∘5+10cos⁡60∘\tan \alpha = \frac{10 \sin 60^\circ}{5 + 10 \cos 60^\circ} tan⁡α=10(32)5+10(0.5)=535+5=5310=32\tan \alpha = \frac{10(\frac{\sqrt{3}}{2})}{5 + 10(0.5)} = \frac{5\sqrt{3}}{5 + 5} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2} α=tan⁡−1(0.866)≈40.9∘\alpha = \tan^{-1}(0.866) \approx 40.9^\circ

Explanation:

We use the law of cosines to find the magnitude of the resultant force and the law of sines (tangent formula) to find its orientation relative to the first vector.

Problem 2:

If two vectors of equal magnitude PP have a resultant also equal to PP, find the angle between the two vectors.

Solution:

Given A=PA = P, B=PB = P, and R=PR = P. Using the formula R2=A2+B2+2ABcos⁡θR^2 = A^2 + B^2 + 2AB \cos \theta: P2=P2+P2+2(P)(P)cos⁡θP^2 = P^2 + P^2 + 2(P)(P) \cos \theta P2=2P2+2P2cos⁡θP^2 = 2P^2 + 2P^2 \cos \theta −P2=2P2cos⁡θ-P^2 = 2P^2 \cos \theta cos⁡θ=−P22P2=−12\cos \theta = -\frac{P^2}{2P^2} = -\frac{1}{2} θ=cos⁡−1(−12)=120∘\theta = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ

Explanation:

By substituting the condition that all magnitudes are equal into the analytical formula, we solve for the unknown angle θ\theta.