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Motion in a Plane - Addition and Subtraction of Vectors — Graphical Method

Grade 11CBSEPhysics

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

🔑Concepts

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Vectors are physical quantities that possess both magnitude and direction. Graphically, they are represented by an arrow where the length is proportional to the magnitude and the tip indicates the direction.

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Triangle Law of Vector Addition: If two vectors A⃗\vec{A} and B⃗\vec{B} are represented in magnitude and direction by two sides of a triangle taken in the same order (head of A⃗\vec{A} to tail of B⃗\vec{B}), then their resultant R⃗\vec{R} is represented by the third side of the triangle taken in the opposite order (tail of A⃗\vec{A} to head of B⃗\vec{B}).

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Parallelogram Law of Vector Addition: If two vectors A⃗\vec{A} and B⃗\vec{B} are represented by two adjacent sides of a parallelogram directed away from a common point, then the diagonal of the parallelogram passing through that same point represents the resultant vector R⃗\vec{R}.

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Vector Subtraction: Subtraction of vector B⃗\vec{B} from vector A⃗\vec{A} is defined as the addition of the negative of B⃗\vec{B} to A⃗\vec{A}. Mathematically: A⃗−B⃗=A⃗+(−B⃗)\vec{A} - \vec{B} = \vec{A} + (-\vec{B}). Graphically, you flip the direction of B⃗\vec{B} by 180∘180^\circ and add it to A⃗\vec{A} using the triangle or parallelogram law.

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Polygon Law: Used for adding more than two vectors. The vectors are arranged such that the head of the first is connected to the tail of the second, the head of the second to the tail of the third, and so on. The resultant is the vector closing the polygon from the tail of the first to the head of the last.

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Commutative and Associative Property: Vector addition is commutative: A⃗+B⃗=B⃗+A⃗\vec{A} + \vec{B} = \vec{B} + \vec{A}. It is also associative: (A⃗+B⃗)+C⃗=A⃗+(B⃗+C⃗)(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C}).

📐Formulae

R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}

S⃗=A⃗−B⃗=A⃗+(−B⃗)\vec{S} = \vec{A} - \vec{B} = \vec{A} + (-\vec{B})

∣R⃗∣=A2+B2+2ABcos⁡θ|\vec{R}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}

tan⁡α=Bsin⁡θA+Bcos⁡θ\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}

A⃗+0⃗=A⃗\vec{A} + \vec{0} = \vec{A}

💡Examples

Problem 1:

A car travels 40 km40\text{ km} North and then 30 km30\text{ km} East. Find the magnitude of the resultant displacement using the graphical concept of a right-angled triangle.

Solution:

Let A⃗\vec{A} be the northward displacement and B⃗\vec{B} be the eastward displacement. Since North and East are perpendicular, θ=90∘\theta = 90^\circ. Using the Triangle Law: ∣R⃗∣=402+302|\vec{R}| = \sqrt{40^2 + 30^2} ∣R⃗∣=1600+900=2500=50 km|\vec{R}| = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{ km}

Explanation:

The two displacements form the legs of a right triangle. The resultant is the hypotenuse connecting the starting point to the final position.

Problem 2:

Two forces F⃗1\vec{F}_1 and F⃗2\vec{F}_2 are acting in the same direction. If ∣F⃗1∣=15 N|\vec{F}_1| = 15\text{ N} and ∣F⃗2∣=10 N|\vec{F}_2| = 10\text{ N}, find the total force magnitude.

Solution:

Since the forces are in the same direction, the graphical addition involves placing them head-to-tail in a straight line: 15+1025\begin{array}{r} 15 \\ + 10 \\ \hline 25 \end{array} The resultant magnitude is 25 N25\text{ N}.

Explanation:

When vectors are collinear and in the same direction, their resultant magnitude is the simple algebraic sum of their individual magnitudes.

Problem 3:

If A⃗\vec{A} is a vector pointing East with magnitude 1010 units, describe the vector −A⃗-\vec{A} used in subtraction.

Solution:

The vector −A⃗-\vec{A} has the same magnitude as A⃗\vec{A} but points in the exactly opposite direction. ∣−A⃗∣=10 units|-\vec{A}| = 10 \text{ units} Direction: West.

Explanation:

The negative of a vector is used to perform subtraction graphically by reversing the direction of the vector being subtracted.