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Question #243415
F1 = 40N, 30º SE; F2 = 25N, 40º NW; and F3 = 30N, 50º NE, what is the resultant vector R?
Expert's answer
F
1
=
(
40
sin
(
180
°
−
30
°
)
,
40
cos
(
180
°
−
30
°
)
)
F_1=(40\sin(180\degree-30\degree), 40\cos(180\degree-30\degree))
F
1
=
(
40
sin
(
180°
−
30°
)
,
40
cos
(
180°
−
30°
))
F
2
=
(
25
sin
(
−
40
°
)
,
25
cos
(
−
40
°
)
)
F_2=(25\sin(-40\degree), 25\cos(-40\degree))
F
2
=
(
25
sin
(
−
40°
)
,
25
cos
(
−
40°
))
F
3
=
(
30
sin
(
50
°
)
,
30
cos
(
50
°
)
)
F_3=(30\sin(50\degree), 30\cos(50\degree))
F
3
=
(
30
sin
(
50°
)
,
30
cos
(
50°
))
R
=
F
1
+
F
2
+
F
3
R=F_1+F_2+F_3
R
=
F
1
+
F
2
+
F
3
=
(
40
sin
(
150
°
)
−
25
sin
(
40
°
)
+
30
sin
(
50
°
)
,
=(40\sin(150\degree)-25\sin(40\degree)+30\sin(50\degree),
=
(
40
sin
(
150°
)
−
25
sin
(
40°
)
+
30
sin
(
50°
)
,
40
cos
(
150
°
)
−
cos
(
40
°
)
+
30
cos
(
50
°
)
)
40\cos(150\degree)-\cos(40\degree)+30\cos(50\degree))
40
cos
(
150°
)
−
cos
(
40°
)
+
30
cos
(
50°
))
≈
(
26.911643
,
−
34.508499
)
\approx(26.911643, -34.508499)
≈
(
26.911643
,
−
34.508499
)
∣
R
∣
=
(
26.911643
)
2
+
(
−
34.508499
)
2
|R|=\sqrt{(26.911643)^2+(-34.508499)^2}
∣
R
∣
=
(
26.911643
)
2
+
(
−
34.508499
)
2
=
43.76
N
=43.76\ N
=
43.76
N
tan
θ
=
−
34.508499
26.911643
\tan \theta=\dfrac{-34.508499}{26.911643}
tan
θ
=
26.911643
−
34.508499
θ
=
−
52.05
°
\theta=-52.05\degree
θ
=
−
52.05°
R
=
43.76
N
,
37.95
º
S
E
R = 43.76\ N, 37.95º SE
R
=
43.76
N
,
37.95º
SE
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