Name: Motion in a Plane With Constant Acceleration
Uploaded: 2019-07-19
Duration: 6 min 42 s
Description: In this video, we are going to look at the simple concept of motion in a plane with constant acceleration. You'll be able to master the concept in just a few...

Question

State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.

Accepted Answer

The parallelogram law of vectors states that if two vectors are represented by the two adjacent sides of a parallelogram then the diagonal passing through the intersection of given vectors represent their resultant both in direction and magnitude.

Consider two vectors $\vec{P} a n d \vec{Q}$ being represented by two consecutive sides of a parallelogram the angle between the two vectors is $θ$ and the resultant is $R$ .
Applying the Pythagorean theorem in the right-angled triangle. Then we have
$H^{2} = P^{2} + B^{2} {(A C)}^{2} = {(A E)}^{2} + {(C E)}^{2} {(A C)}^{2} = {(A B + B E)}^{2} + {(C E)}^{2} {(A C)}^{2} = {(A B)}^{2} + {(B E)}^{2} + (2 \times A B \times B E) + {(C E)}^{2} . . . . . (1)$
Applying trigonometry in the right angle triangle $B C E$ , we have:
$\frac{B E}{B C} = \cos θ B E = B C \cos θ . . . . . (2) A n d \frac{C E}{B C} = \sin θ C E = B C \sin θ . . . . . (3)$
Put equation (2) and (3) in equation (1)
${(A C)}^{2} = {(A B)}^{2} + (B C \cos θ)^{2} + (2 \times A B \times B C \cos θ) + {(B C \sin θ)}^{2} {(A C)}^{2} = {(A B)}^{2} + {(B C)}^{2} + (2 \times A B \times B C \cos θ) . . . . . (4)$
But for a parallelogram
$B C = A D . . . . . (5)$
Put equation (5) in equation (4)
${(A C)}^{2} = {(A B)}^{2} + {(A D)}^{2} + (2 \times A B \times A D \cos θ) (R)^{2} = (P)^{2} + {(Q)}^{2} + (2 \times P \times Q \cos θ) R = \sqrt{(P)^{2} + {(Q)}^{2} + (2 \times P \times Q c o s θ)}$
We can also find out the angle made by the resultant with the $\vec{P}$ vector. Applying trigonometry in the right angle triangle $A C E$ we get:
$\tan α = \frac{C E}{A C} \tan α = \frac{C E}{A B + B E} \tan α = \frac{B C \sin θ}{A B + B E} \tan α = \frac{A D \sin θ}{A B + A D \cos θ} \tan α = \frac{Q \sin θ}{P + Q \cos θ}$
Resultant acts in the direction making an angle $α = \tan^{- 1} (\frac{Q \sin θ}{P + Q \cos θ})$ with direction of vector $\vec{P}$ .

State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.

Important Questions on Motion in a Plane

Learn and Prepare for any exam you want !