The two vectors A and B in the figure have equal magnitudes of 13.5 m and the angles are θ1 = 33° and θ2 = 110°. How to find (a) the x component and (b) the y component of their vector sum R , (c) the magnitude of R, and (d) the angle R ?

1 Answer
Sep 22, 2015

Answer:

Here's what I got.

Explanation:

enter image source here

I don't wave a good way of drawing you a diagram, so I'll try to walk you through the steps as they come along.

So, the idea here is that you can find the #x#-component and the #y#-component of the vector sum, #R#, by adding the #x#-components and #y#-components, respectively, of the #vec(a)# and #vec(b)# vectors.

For vector #vec(a)#, things are pretty straighforward. The #x#-component will be the projection of the vector on the #x#-axis, which is equal to

#a_x = a * cos(theta_1)#

Likewise, the #y#-component will be the projection of the vector on the #y#-axis

#a_y = a * sin(theta_1)#

For vector #vec(b)#, things are a little more complicated. More specifically, finding the corresponding angles will be a little tricky.

The angle between #vec(a)# and #vec(b)# is

#theta_3 = 180^@ - theta_2 = 180^@ - 110^@ = 70^@#

Draw a parallel line to the #x#-axis that intersects the point where the tail of #vec(b)# and head of #vec(a)# meet.

http://www.regentsprep.org/regents/math/geometry/multiplechoicereviewg/quadrilaterals.htm

In your case, line #m# will be the #x#-axis and line #a# the parallel line you draw.

In this drawing, #angle6# is #theta_1#. You know that #angle6# is equal to #angle3#, #angle2#, and #angle7#.

The angle between #vec(b)# and the #x#-axis will be equal to

#180^@ - (theta_1 + theta_2) = 180^@ - 143^@ = 37^@#

This means that the #x#-component of vector #vec(b)# will be

#b_x = b * cos(37^@)#

Now, because the angle between the #x#-component and the #y#-component of a vector is equal to #90^@#, it follows that the angle for the #y#-component of #vec(b)# will be

#90^@ - 37^@ = 53^@#

The #y#-component will thus be

#b_y = b * sin(37^@)#

Now, keep in mind that the #x#-component of #vec(b)# is oriented in the opposite direction of the #x#-component of #vec(a)#. This means that the #x#-component of #vec(R)# will be

#R_x = a_x + b_x#

#R_x = 13.5 * cos(33^@) - 13.5 * cos(37^@)#

#R_x = 13.5 * 0.04 = color(green)("0.54 m")#

The #y#-components are oriented in the same direction, so you have

#R_y = a_y + b_y#

#R_y = 13.5 * [sin(110^@) + sin(37^@)]#

#R_y = 13.5 * 1.542 = color(green)("20.82 m")#

The magnitude of #vec(R)# will be

#R^2 = R_x^2 + R_y^2#

#R = sqrt(0.54""^2 + 20.82""^2)" m" = color(green)("20.83 m")#

To get the angle of #vec(R)#, simply use

#tan(theta_R) = R_y/R_x implies theta_R = arctan(R_y/R_x)#

#theta_R = arctan((20.82color(red)(cancel(color(black)("m"))))/(0.54color(red)(cancel(color(black)("m"))))) = color(green)(88.6""^@)#