Question

key move: to solve a parallelogram law of vectors question, first draw the two vectors from a common point, complete the parallelogram, then draw the diagonal from the common point, and finally use the cosine rule or sine rule to calculate the magnitude and direction of the resultant. elaborate - practice complete the following practice tasks. show all work and thinking for each task 1. two forces of magnitudes 5 n and 8 n act on a body at an angle of 60°. use the parallelogram law of vectors to find the resultant force. 2. two vectors, each of magnitude 10 units, are inclined at an angle of 120°. find the magnitude of their resultant using the parallelogram law 3. a boy pulls a cart with a force of 6 n east, while another boy pulls with a force of 8 n making an angle of 90° with the first force. use the parallelogram law to determine the resultant force. 4. a particle is acted upon by two forces of 12 n and 9 n, with an angle of 135° between them. find the resultant force using the parallelogram law of vectors

Explanation:

Step1: Recall the formula for the magnitude of the resultant of two - vectors using the parallelogram law

The magnitude of the resultant $\vec{R}$ of two vectors $\vec{A}$ and $\vec{B}$ with an angle $\theta$ between them is given by $R=\sqrt{A^{2}+B^{2}+2AB\cos\theta}$.

Step2: Solve problem 1

Let $A = 5$ N, $B = 8$ N and $\theta=60^{\circ}$, $\cos\theta=\frac{1}{2}$.
\[

$$\begin{align*} R&=\sqrt{5^{2}+8^{2}+2\times5\times8\times\frac{1}{2}}\\ &=\sqrt{25 + 64+40}\\ &=\sqrt{129}\\ &\approx11.36\text{ N} \end{align*}$$

\]

Step3: Solve problem 2

Let $A = 10$ units, $B = 10$ units and $\theta = 120^{\circ}$, $\cos\theta=-\frac{1}{2}$.
\[

$$\begin{align*} R&=\sqrt{10^{2}+10^{2}+2\times10\times10\times(-\frac{1}{2})}\\ &=\sqrt{100 + 100-100}\\ &=\sqrt{100}\\ &=10\text{ units} \end{align*}$$

\]

Step4: Solve problem 3

Let $A = 6$ N, $B = 8$ N and $\theta = 90^{\circ}$, $\cos\theta = 0$.
\[

$$\begin{align*} R&=\sqrt{6^{2}+8^{2}+2\times6\times8\times0}\\ &=\sqrt{36 + 64}\\ &=\sqrt{100}\\ &=10\text{ N} \end{align*}$$

\]

Step5: Solve problem 4

Let $A = 12$ N, $B = 9$ N and $\theta=135^{\circ}$, $\cos\theta=-\frac{\sqrt{2}}{2}$.
\[

$$\begin{align*} R&=\sqrt{12^{2}+9^{2}+2\times12\times9\times(-\frac{\sqrt{2}}{2})}\\ &=\sqrt{144 + 81-108\sqrt{2}}\\ &=\sqrt{225-108\sqrt{2}}\\ &\approx\sqrt{225 - 108\times1.414}\\ &=\sqrt{225-152.712}\\ &=\sqrt{72.288}\\ &\approx8.5\text{ N} \end{align*}$$

\]

Answer:

1. Approximately $11.36$ N
2. $10$ units
3. $10$ N
4. Approximately $8.5$ N