Answer:
a) 202.7 N
b) 58.1 N
c) 74.1º N of E.
d) 210.9 N
Explanation:
a)
- The net force exerted in the y-direction, will be the sum of FH (which is directed northwards) and the y-component of FA.
- Since the magnitude of FA is 155 N and the angle of FA with the y-axis, is 22º (E of N), we can find the N-S component of FA, just applying the the definition of cosine, to the triangle defined by FA, the y- axis and a segment parallel to the x- axis between FA and the y-axis, as follows:
[tex]F_{Ay} = F_{A} * cos \theta = 155 N* cos 22 = 143.7 N (1)[/tex]
⇒ Fy = FH + FAy = 59 N + 143.7 N = 202.7 N (2)
b)
- We can proceed exactly in the same way for the x-axis.
- Since FH is directed due North, it has no component along the x-axis.
- So, Fx is directly the component of FA along the x-axis, which can be found applying the definition of sine to the same triangle than in a) as follows:
[tex]F_{x} = F_{A} * sin \theta = 155 N* sin 22 = 58.1 N (3)[/tex]
c)
- Taking the same triangle than in a) and b), we can apply the definition of tangent, in order to find the angle between F and the x-axis, as follows:
[tex]tg \theta = \frac{F_{y}}{F_{x}} = \frac{202.7N}{58.1N} = 3.5 (4)[/tex]
⇒ θ = tg⁻¹ (3.5) = 74.1º N of E. (5)
d)
- In order to be equal and opposite to the combined force FH+FA, it must have the same magnitude.
- This magnitude can be found applying the Pythagorean Theorem to the same triangle that we used in a), b) and c):
[tex]F_{DS} = \sqrt{(F_{x} ^{2} +F_{y} ^{2})} = \sqrt{(58.1N)^{2}) + (202.7N)^{2} } = 210.9 N (6)[/tex]
