Question

The two vectors a and b in the figure have equal magnitude of 28 m and the angles are theta1 = 23degree and theta2 = 104degree. Find the x-component of their vector sum r. Find the y-component of their vector sum r. What is the magnitude of their vector sum r? Find the angle that their vector sum r makes with the positive direction of the x-axis.

Answer 1

Resolve the each vector into rectangular components as shown below,

For $\vec{a}$ , as given in question, the magnitude is 28 m and the angle with the x-axis is $23^{\circ}$ ,

$\vec{a}_{x}=28 cos23^{\circ}$

   $\vec{a}_{y}=28 sin23^{\circ}$

Similarily for $\vec{b}$ the magnitude of the vector is 28 m and the angle with x-axis is $104^{\circ}+23^{\circ}=127^{\circ}$ . we can write the angle in -x direction as $180^{\circ}-127^{\circ}=53^{\circ}$ . Therefore,

   $\vec{b}_{x}=28cos53^{\circ}$

   $\vec{b}_{y}=28sin53^{\circ}$

a) The x component of their vector sum r is,

$r_{x}=28cos23^{\circ}-28cos53^{\circ}$

$r_{x}=25.77-16.85=8.92m$

b) The y component of their vector sum r is,

$r_{y}=28sin23^{\circ}+28sin53^{\circ}$

c) The magnitude of vector sum r is,

   $r=\sqrt{(r{_{x}}^{2})+(r{_{y}}^{2})}$

$r=\sqrt{(8.92)^{2}+(33.3)^{2}}=34.47m$

d) The angle that their vector sum r makes with the positive direction of the x-axis is,

$tan\theta =r_{y}/r_{x}=33.3/8.92=3.73$

   $\theta =tan^{-1}(3.73)=75^{\circ}$