Question

A vector A has a magnitude of 48.0 m and points in a direction 20° below the positive x axis. A second vector, B, has a magnitude of 75 m and points in a direction 60.0° above the positive x axis?

Using the component method of vector addition, how do you find the magnitude and direction of the vector C=A +B?

Answer 1

The magnitude of `C` is 90.808 m, nearly, and the direction of `C`

makes `31.234^o`, with the positive direction of the x-axis.

Explanation:

The vector

`A=(x, y)=r(cos theta, sin theta) =48(cos (-20^o), sin (-20^o))` and,

likewise, the vector

`B=75(cos 60^o, sin 60^o)`.

So, `A=(45.105, -16.417) and B=(37,5, 64.952)`, nearly.

Now, the components of `C=A+B` are

`(45.105, -16.417)+(37,5, 64.952)`

`=(45.105+37.5, -16.417+64.952)`

`=(82.605, 48.535)= r(cos theta, sin theta) =(x, y)`

`=96.808(cos 31.234^o, sin 31.234^o)`,

using `r = sqrt(x^2+y^2), cos theta = x/r and sin theta = y/r` .

The magnitude of `C` is 90.808 m, nearly, and the direction of `C`

makes `31.234^o`, with the positive direction of the x-axis.

Answer 2

Magnitude of resultant vector is `95.81m` and direction is `30.44^o` above `x`-axis

Explanation:

As vector `A` has magnitude of `48.0m` and points in a direction `20^o` below positive `x`-axis (i.e. `-20^o`) and vector `B` has magnitude of `75.0m` and points in a direction `60^o` above positive `x`-axis.

Hence the angle between two vectors is `60-(-20)=80^o` and hence

Magnitude of resultant vector is `sqrt(48^2+75^2+2xx48xx75xxcos80^o`

= `sqrt(2304+5625+7200xx0.17365)`

= `sqrt(2304+5625+1250.28)=sqrt(9179.28)=95.81m`

Direction wll be given by `alpha=tan^(-1)((75sin80^o)/(48+75cos80^o))`

= `tan^(-1)((75xx0.9848)/(48+75xx0.17365))`

= `tan^(-1)((73.86)/(48+13.02375))=tan^(-1)(73.86/61.02375)`

= `tan^(-1)1.21035=50.44^o`

But this is w.r.t. `A` and hence direction is `50.44^o-20^o=30.44^o` above `x`-axis