Question

why is it important to understand parallelogram method of vector addition and the head - to - tail method? it is important to understand the parallelogram method and the head - to - tail method of vector addition because they provide visual and mathematical ways to determine the resultant of two or more vectors. these methods are widely used in physics and engineering to analyze forces, motions, and other vector quantities in real - life situations. stamp it! write the gist statement that summarizes the main idea of the two methods of vector addition discussed in this lesson. explain - guided example review an example below of finding the resultant force of two vectors with an angle of 60° between them. what key observations stand out? after reviewing the example, try different angles of inclinations between any two vectors and find the resultant force. example: question two forces of 3 n and 4 n are acting at a point such that the angle between them is 60 degrees. find the resultant force answer magnitude r of the resultant force is r = √(3² + 4²+2×3×4 cos 60 deg)=√(9 + 16 + 12)=√(37 = 6.08 n direction of r is given by finding the angle q tan q=(3 sin 60 deg)/(4 + 3 cos 60 deg)=0.472 q = tan⁻¹ 0.472 = 25.3 deg

Explanation:

Step1: Recall the formula for magnitude of resultant vector

The formula for the magnitude $R$ of the resultant of two vectors $\vec{A}$ and $\vec{B}$ with an angle $\theta$ between them is $R = \sqrt{A^{2}+B^{2}+2AB\cos\theta}$. Here $A = 3N$, $B = 4N$ and $\theta=60^{\circ}$. So $R=\sqrt{3^{2}+4^{2}+2\times3\times4\times\cos60^{\circ}}$.

Step2: Calculate the value inside the square - root

First, calculate $3^{2}=9$, $4^{2}=16$, and $2\times3\times4\times\cos60^{\circ}=2\times3\times4\times\frac{1}{2}=12$. Then $9 + 16+12=37$. So $R=\sqrt{37}\approx6.08N$.

Step3: Recall the formula for the direction of the resultant vector

The formula for the direction $\alpha$ of the resultant vector with respect to the vector $\vec{B}$ is $\tan\alpha=\frac{A\sin\theta}{B + A\cos\theta}$. Here, $\tan\alpha=\frac{3\sin60^{\circ}}{4 + 3\cos60^{\circ}}$. Since $\sin60^{\circ}=\frac{\sqrt{3}}{2}$ and $\cos60^{\circ}=\frac{1}{2}$, we have $\tan\alpha=\frac{3\times\frac{\sqrt{3}}{2}}{4+3\times\frac{1}{2}}=\frac{\frac{3\sqrt{3}}{2}}{\frac{8 + 3}{2}}=\frac{3\sqrt{3}}{11}\approx0.472$.

Step4: Find the angle

We know that $\alpha=\tan^{- 1}(0.472)\approx25.3^{\circ}$.

Answer:

Magnitude of the resultant force is approximately $6.08N$ and its direction with respect to the $4N$ force is approximately $25.3^{\circ}$.