Question

Q. State triangle law of vector addition. Find magnitude of resultant of two vectors

inclined at angle Ѳ. Also find the direction of resultant.​

Answer 1

Explanation:

If two vectors are represented by the sides of a triangle both in magnitude and direction taken in order, the resultant sum of the vectors is given by the closing third side of the triangle taken in the reverse order both in the magnitude and direction.

The resultant is found by adding vectors together. You can use the trig functions to find the x and y components of each vector, add them, and find the magnitude of the new vector...or, you can notice that 110° - 20° = 90°. So what you have is a 3, 4, 5 right triangle. The magnitude is 5.

Derivation of the law

Let θ be the angle between P and Q and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q. Now, expand A to C and draw BC perpendicular to OC. which is the magnitude of resultant.

thank you

Answer 2

Answer:

Triangle law of vector addition states that if two vectors can be represent the magnitude and direction by the two sides of a triangle take in the same order then the resultant is represented completely by the third side of the triangle taken in the in the opposite direction.

Resultant vector:  $\vec R= \vec P+ \vec Q$

Consider that two vectors P and Q represented by the sides AD and DC. Consider that R is the resultant vector of P and Q, represented by AC.

Now, from the triangle ABC (from the attached figure below), use Pythagoras theorem:

AC² = AB² + BC²                                                          ..................(1)

Now as shown in the figure AB = AD + BD

Put in equation (1) we get;

AC² = (AD + BD)² + BC²                                               ..................(2)

Now from the triangle DCB,

$cos\theta = \frac{BD}{DC}$

$BD = DC cos \theta$

$BD = Qcos \theta$

and, $sin\theta =\frac{BC}{DC}$

$BC = DC sin\theta$

$BC = Qsin\theta$

Substitute the value of BC and DB in equation (2);

R² = (P + Qcosθ)² + (Qsinθ)²

R² = P² + Q²cos²θ+ 2PQcosθ + Q²sin²θ

R² = P² + 2PQcosθ + (Q²cos²θ + Q²sin²θ)

R² = P² + 2PQcosθ + Q²(cos²θ + sin²θ)

R² = P² + 2PQcosθ + Q²

$R=\sqrt{P^2 + Q^2 + 2PQcos\theta}$

The above expression gives magnitude of resultant of two vectors inclined at angle θ.

Consider that Ф is the angle between vectors P and R.

From the triangle, ABC

$tan\phi =\frac{BC}{AB}$

$tan\phi =\frac{BC}{AD+ DB}$

Put the value of AD and DB in above equation:

$tan\phi=\frac{Qsin\theta}{P+ Q cos\theta}$

$\phi= tan^{-1}\large (\frac{Qsin\theta}{P+Qcos\theta}\large )$

The above equation will give the  the direction of resultant vector.