Question

two vectors have magnitude 4 and 5 units . the angle between 30 degree . taking the first vector along x_axies , calculate the magnitude and the direction of the resultant?

Answer 1

pleasee find the attachment for solution.

Answer 2

Answer:

The resultant vector's magnitude is found to be 8.7, with a direction making an angle of 16.7° from the x-axis.

Explanation:

The resultant (result vector) is a vector sum of two or more vectors. This is the result of adding two or more vectors together. Adding the displacement vectors A, B, C and so on results in the vector R. The vector R can be determined using an accurately drawn and scaled vector addition diagram.

We know that the resultant vector of any two vectors can be calculated by the following expression:

$\vec{R}=\vec{A}+\vec{B}$

Simplifying which, we get:

$|R|^2=|A|^2+|B|^2+2|A||B|cos\theta$

Where $|A|$ and $|B|$ are the magnitudes of the two vectors in consideration and $\theta$ is the angle between them.

Now, we are given that:

|A| = 4, |B| = 5

and the angle between them is: $\theta=30^\circ$

Substituting these in the expression, we get:

$|R|^2=|4|^2+|5|^2+2|4||5|cos30^\circ$

$|R|^2=16+25+(40\times\frac{\sqrt3}{2})$

or we can say:

$|R|=\sqrt{(16+25+(20\sqrt3))}$

using $\sqrt3=1.732$, we get:

$|R|=\sqrt{75.64}$

which gives us:

|R| = 8.697 or 8.7

Now, we know that the first vector is along x-axis, thus the expression for finding the angle between the x-axis and the resultant will be:

$tan\alpha=\frac{|B|sin \theta}{|A|+|B|cos\theta}$

Now, substituting the values, we get:

$tan\alpha=\frac{5sin 30^\circ}{4+5cos 30^\circ}=\frac{5\times\frac{1}{2}}{4+5(\frac{\sqrt3}{2})}$

$tan\alpha =\frac{5}{8+5\sqrt3}$

$tan\alpha =\frac{5}{16.660}=0.300$

Which gives us the angle as:

$\alpha =tan^{-1}(0.3)$

α = 16.7°

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