Chapter 8.1, Problem 26E

To determine

Find the magnitude and direction of resultant.

Answer to Problem 26E

  $3.6cmat132°$

Explanation of Solution

Given information:

  $\stackrel{\rightharpoonup }{r}+\stackrel{\rightharpoonup }{t}+\stackrel{\rightharpoonup }{u}$

Calculation:

Consider vector $\stackrel{\rightharpoonup }{r},\stackrel{\rightharpoonup }{t},\stackrel{\rightharpoonup }{u}$

The magnitude of vector $\stackrel{\rightharpoonup }{r}$ is $2.6cmat128°$ .

The magnitude of vector $\stackrel{\rightharpoonup }{t}$ is $2.1cmat14°$ .

The magnitude of vector $\stackrel{\rightharpoonup }{u}$ is $3cmat340°$ .

Now calculate $\stackrel{\rightharpoonup }{r}+\stackrel{\rightharpoonup }{t}+\stackrel{\rightharpoonup }{u}$ .

Apply triangle and Head −to −tail method to find the resultant of both vectors $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}$ . The vector $m\stackrel{\rightharpoonup }{a}$ represents $m$ times the magnitude of vector $\stackrel{\rightharpoonup }{a}$ when $m$ is negative the direction of $m\stackrel{\rightharpoonup }{a}$ is opposite to that of $\stackrel{\rightharpoonup }{a}$ .

We have three vectors, with the help of rural and protractor draw first vector in standard position, now draw second vector with initial point as terminating point of first and so on.

Draw the vectors with resultant,

The length and angle of resultant vector, $\stackrel{\rightharpoonup }{r}+\stackrel{\rightharpoonup }{t}+\stackrel{\rightharpoonup }{u}$ is $3.6cmat132°$

Hence, the magnitude and direction of resultant vector $\stackrel{\rightharpoonup }{r}+\stackrel{\rightharpoonup }{t}+\stackrel{\rightharpoonup }{u}$ is $3.6cmat132°$