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?

2 Answers
Aug 5, 2016

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.

Aug 5, 2016

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