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Understanding Motion through Experience - Vector Addition (Graphical Method)-advanced

Grade 9CBSE

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

🔑Concepts

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A scalar quantity has only magnitude (e.g., distance, speed), while a vector quantity has both magnitude and direction (e.g., displacement, velocity).

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Vectors are graphically represented by straight arrows. The length of the arrow is proportional to the magnitude using a chosen scale, and the arrowhead indicates the direction.

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The Triangle Law of Vector Addition states that if two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented by the third side of the triangle taken in the opposite order.

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The Parallelogram Law of Vector Addition states that if two vectors acting at a point are represented by the adjacent sides of a parallelogram, the diagonal passing through their point of intersection represents the resultant vector.

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The Head-to-Tail Method is used for adding multiple vectors graphically: place the tail of the second vector at the head of the first, the tail of the third at the head of the second, and so on. The resultant vector R⃗\vec{R} is drawn from the tail of the first to the head of the last vector.

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Scale Selection: In graphical addition, a suitable scale must be chosen (e.g., 1 cm=10 N1\text{ cm} = 10\text{ N} or 1 cm=5 m/s1\text{ cm} = 5\text{ m/s}) to accurately represent the vectors on paper.

📐Formulae

R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}

R=A2+B2 (When vectors are perpendicular, θ=90∘)R = \sqrt{A^2 + B^2} \text{ (When vectors are perpendicular, } \theta = 90^{\circ}\text{)}

tan⁡α=Bsin⁡θA+Bcos⁡θ (Direction of resultant)\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} \text{ (Direction of resultant)}

Rmax=A+B (When θ=0∘)R_{max} = A + B \text{ (When } \theta = 0^{\circ}\text{)}

Rmin=∣A−B∣ (When θ=180∘)R_{min} = |A - B| \text{ (When } \theta = 180^{\circ}\text{)}

💡Examples

Problem 1:

A hiker walks 4 km4\text{ km} due East and then 3 km3\text{ km} due North. Using the graphical method (or Pythagoras theorem for verification), find the magnitude and direction of the resultant displacement.

Solution:

  1. Let the scale be 1 cm=1 km1\text{ cm} = 1\text{ km}.
  2. Draw vector A⃗\vec{A} of length 4 cm4\text{ cm} towards the East.
  3. From the head of A⃗\vec{A}, draw vector B⃗\vec{B} of length 3 cm3\text{ cm} towards the North.
  4. Join the tail of A⃗\vec{A} to the head of B⃗\vec{B} to get the resultant R⃗\vec{R}.
  5. By Pythagoras theorem: R=42+32R = \sqrt{4^2 + 3^2} R=16+9R = \sqrt{16 + 9} R=25=5 kmR = \sqrt{25} = 5\text{ km}
  6. For direction θ\theta: tan⁡θ=34=0.75\tan \theta = \frac{3}{4} = 0.75 θ=tan⁡−1(0.75)≈36.87∘\theta = \tan^{-1}(0.75) \approx 36.87^{\circ}

Explanation:

The two displacements are perpendicular to each other. The triangle law is applied here. The resultant magnitude is the hypotenuse of the right-angled triangle formed by the Eastward and Northward paths.

Problem 2:

Two forces F1⃗=10 N\vec{F_1} = 10\text{ N} and F2⃗=10 N\vec{F_2} = 10\text{ N} act on a body at an angle of 120∘120^{\circ} to each other. Determine the resultant force graphically.

Solution:

  1. Choose a scale: 1 cm=2 N1\text{ cm} = 2\text{ N}. Thus, each force is represented by a 5 cm5\text{ cm} arrow.
  2. Draw F1⃗\vec{F_1} horizontally.
  3. Draw F2⃗\vec{F_2} at an angle of 120∘120^{\circ} from the tail of F1⃗\vec{F_1} (Parallelogram law).
  4. Complete the parallelogram.
  5. The diagonal R⃗\vec{R} represents the resultant.
  6. Measuring the diagonal, we find its length is 5 cm5\text{ cm}.
  7. Converting back from scale: 5 cm×2 N/cm=10 N5\text{ cm} \times 2\text{ N/cm} = 10\text{ N}. Using the formula for verification: R=102+102+2(10)(10)cos⁡(120∘)R = \sqrt{10^2 + 10^2 + 2(10)(10) \cos(120^{\circ})} Since cos⁡(120∘)=−0.5\cos(120^{\circ}) = -0.5: R=100+100+200(−0.5)R = \sqrt{100 + 100 + 200(-0.5)} R=200−100=100=10 NR = \sqrt{200 - 100} = \sqrt{100} = 10\text{ N}

Explanation:

When two equal vectors act at an angle of 120∘120^{\circ}, the magnitude of their resultant is equal to the magnitude of either vector. This is a unique property often tested in advanced conceptual problems.