Respuesta :
Answer:
(a): 37.94 m.
(b): [tex]323.16^\circ.[/tex]
(c): 126.957 m.
(d): [tex]0.93^\circ.[/tex]
(e): 49.92 m.
(f): [tex]130.08^\circ.[/tex]
Explanation:
Given:
- Magnitude of [tex]\vec a[/tex] = 50 m.
- Direction of [tex]\vec a = 30^\circ.[/tex]
- Magnitude of [tex]\vec b[/tex] = 50 m.
- Direction of [tex]\vec b = 195^\circ.[/tex]
- Magnitude of [tex]\vec c[/tex] = 50 m.
- Direction of [tex]\vec c = 315^\circ.[/tex]
Any vector [tex]\vec A[/tex], making an angle [tex]\theta[/tex] with respect to the positive x-axis, can be written in terms of its x and y components as follows:
[tex]\vec A = A\cos\theta\ \hat i+A\sin\theta \ \hat j.[/tex]
where, [tex]\hat i,\ \hat j[/tex] are the unit vectors along the x and y axes respectively.
Therefore, the given vectors can be written as
[tex]\vec a = 50\cos30^\circ \ \hat i+50\sin 30^\circ\ \hat j = 43.30\ \hat i +25\ \hat j\\\vec b = 50\cos195^\circ \ \hat i+50\sin 195^\circ\ \hat j = -48.29\ \hat i +-12.41\ \hat j\\\vec c = 50\cos 315^\circ \ \hat i+50\sin 315^\circ\ \hat j = 35.35\ \hat i +-35.35\ \hat j\\[/tex]
(a):
[tex]\vec a +\vec b + \vec c= (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29+35.35)\hat i+(25-12.41-35.35)\hat j\\=30.36\hat i-22.75\hat j.\\\\\text{Magnitude }=\sqrt{30.36^2+(-22.75)^2}=37.94\ m.[/tex]
(b):
Direction [tex]\theta[/tex] can be found as follows:
[tex]\tan\theta = \dfrac{\text{x component of }(\vec a + \vec b +\vec c)}{\text{y component of }(\vec a + \vec b +\vec c)}=\dfrac{-22.75}{30.36}=-0.749\\\Rightarrow \theta = \tan^{-1}(-0.749)=-36.84^\circ.[/tex]
The negative sign indicates that the sum of the vectors is [tex]36.84^\circ.[/tex] below the positive x axis.
Therefore, direction of this vector sum counterclockwise with respect to positive x-axis = [tex]360^\circ-36.84^\circ=323.16^\circ.[/tex]
(c):
[tex]\vec a -\vec b + \vec c= (43.30\ \hat i +25\ \hat j)-(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30+48.29+35.35)\hat i+(25+12.41-35.35)\hat j\\=126.94\hat i+2.06\hat j.\\\\\text{Magnitude }=\sqrt{126.94^2+2.06^2}=126.957\ m.[/tex]
(d):
Direction [tex]\theta[/tex] can be found as follows:
[tex]\tan\theta = \dfrac{\text{x component of }(\vec a - \vec b +\vec c)}{\text{y component of }(\vec a - \vec b +\vec c)}=\dfrac{2.06}{126.94}=0.01623\\\Rightarrow \theta = \tan^{-1}(0.01623)=0.93^\circ.[/tex]
(e):
[tex](\vec a + \vec b)-(\vec c + \vec d)=0\\(\vec a + \vec b)=(\vec c + \vec d)\\\vec d = \vec a + \vec b -\vec c.[/tex]
[tex]\vec d = \vec a +\vec b - \vec c= (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)-(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29-35.35)\hat i+(25-12.41+35.35)\hat j\\=-40.34\hat i+47.94\hat j.\\\\\text{Magnitude }=\sqrt{(-40.34)^2+47.94^2}=62.65\ m.[/tex]
(f):
Direction [tex]\theta[/tex] can be found as follows:
[tex]\tan\theta = \dfrac{\text{x component of }\vec d}{\text{y component of }\vec d}=\dfrac{47.94}{-40.34}=-1.188\\\Rightarrow \theta = \tan^{-1}(-1.188)=-49.92^\circ.[/tex]
The x component of this vector is negative and y component is positive therefore the vector lie in second quadrant, which means, the direction of this vector, counterclockwise with respect to positive x axis = [tex]180^\cir.
c-49.92^\circ=130.08^\circ.[/tex]
