Answer:
Magnitude of R = 16.7 lb
Angle = 55.7°
Step-by-step explanation:
Given:
a = 28 lb
b = 12 lb
a = 45 degree positive on x & y axis.
b = 30 degree below x-axis. negative on y & x.
Resolving the two forces to vector form, we have
a = (28cos45)i + (28sin45)j
b = (-12cos30)i + (-12sin30)j
The resultant force vector R = a +b
R = (28cos45 - 12cos30)i + (28sin45 - 12sin30)j
R = 9.41i + 13.80j
Magnitude of R = √(9.41^2 + 13.80^2) = 16.7 lb
Angle = taninverse (13.80/9.41) = 55.7°
To determine the magnitude of the resultant force, the two given forces have to be resolved as vectos. In the question, the magnitude of the resultant force is 39.7 N and makes an angle of [tex]49.5^{o}[/tex] with the x-axis.
The two given forces are vectors, thus they have to be resolved horizontally and vertically so as to determine their resultant.
So that:
Horizontal component = F Cos θ
Vertical component = F Sinθ
Thus, given that: a = 28 lb at [tex]45^{o}[/tex] on x & y axis, and b = 12 lb at [tex]30^{o}[/tex] below x-axis.
i. Resolving the vector horizontally, we have:
horizontal component = 28 Cos[tex]45^{o}[/tex] + 12 Cos [tex]30^{o}[/tex]
= 19.80 + 10.39
horizontal component, x = 30.19
ii. Resolving the vectors vertically, we have:
vertical component = 28 Sin [tex]45^{o}[/tex] + 12 Sin [tex]30^{o}[/tex]
= 19.80 + 6.0
vertical component, y = 25.80
a. Magnitude of the resultant force, |R| = [tex]\sqrt{x^{2} + y^{2} }[/tex]
= [tex]\sqrt{(30.19)^{2} + (25.80)^{2} }[/tex]
= 39.7124
|R| = 39.7 Newtons
b. The angle that the resultant makes with the positive x-axis can be determined by:
Tan θ = [tex]\frac{x}{y}[/tex]
θ = [tex]Tan^{-1}[/tex] [tex]\frac{30.19}{25.80}[/tex]
= 49.4832
θ = [tex]49.5^{o}[/tex]
Therefore, the magnitude of the resultant force is 39.7 N and makes an angle of [tex]49.5^{o}[/tex] with the x-axis.
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