Respuesta :
6m at 0.9 radian means 6m at [tex]51.57^0[/tex]
Since positive angles are counter clockwise
6m at [tex]51.57^0[/tex] can be written as 6*cos[tex]51.57^0[/tex]i+6*sin[tex]51.57^0[/tex]j = 3.73i+4.70j
5m at [tex]-75^0[/tex] can be written as 5*cos[tex](-75)^0[/tex]i+5*sin[tex](-75)^0[/tex]j = 1.294i-4.83j
4m at 1.2 radian means 4m at [tex]68.75^0[/tex]
4m at [tex]68.75^0[/tex] can be written as 4*cos[tex]68.75^0[/tex]i+4*sin[tex]68.75^0[/tex]j = 1.45i+3.73j
6m at [tex]-210^0[/tex] can be written as 6*cos[tex](-210)^0[/tex]i+6*sin[tex](-210)^0[/tex]j = -5.20i-3j
a) So sum of the vector = 1.274i+0.6j
b) Magnitude = [tex]\sqrt{1.274^2+0.6^2} = 1.98 m[/tex]
c) Angle, [tex]tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0[/tex]
The vector decomposition allows to find the results for the questions about the sum of the four vectors are:
a) The sum vector is R = (1.27 i ^ + 6.60 j ^) m
b) The modulus is: R = 6.72 m
c) The angle of the resultant is: θ = 79.1º
d) The angle in radians is: θ = 1,380 give
Given parameters
- Vectors
* A = 6.00 m with teat1 = 0.900 rad
* B = 5.00 m with tea2 = -75º
* C = 4.00 m with tea3 = 1.20 rad
* D = 6.00 with tea = -210º
To find
a) The sum of the four vectors
b) The magnitude
c) The angle in degrees
d) The angle in radians
Vectors are magnitudes that have direction and sense, therefore for their addition vector algebra must be used.
One of the simplest methods for adding vectors is to decompose the vectors into a coordinate system, perform the algebraic addition, and then compose the resulting vector.
Let's use a coordenate system with the x-axis is horizontal and the y-axis is vertical, with trigonometry we decompose the vectors.
Let's start by reducing the angles to degrees, to avoid problems of using different units in the calculations.
θ₁ = 0.9 rad ( [tex]\frac{180^{o} }{\pi \ rad}[/tex] ) = 51.57º
Let's measure the angle from the positive side of the x axis, counterclockwise.
θ₂ = 360 -75 = 285º
θ₃ = 1.20 rad ([tex]\frac{180^{o}}{\pi \ rad }[/tex] ) = 68.75º
θ₄ = 360-210º = 150º
Let's use trigonometry to decompose the vectors, see attached.
Vector A
sin θ₁ = [tex]\frac{A_y}{A}[/tex]
cos θ₁ = [tex]\frac{A_x}{A}[/tex]
[tex]A_y[/tex] = A sin θ₁
Aₓ = A cos θ₁
[tex]A_y[/tex]= 6 sin 51.57 = 4.700 m
Aₓ = 6 cos 51.57 = 3.729 m
Vector B
[tex]B_y[/tex] = B sin θ₂
Bₓ = B cos θ₂
[tex]B_y[/tex] = 5 sin 285 = -4.8296 m
Bₓ = 5 cos 285 = 1.294 m
Vector C
[tex]C_y[/tex] = C sin θ₃
Cₓ = C cos θ₃
[tex]C_y[/tex] = 4 sin 68.75 = 3.728 m
Cₓ = 4 cos 68.75 = 1.4498 m
Vector D
[tex]D_y[/tex] = D sin θ₄
Dₓ = D cos θ₄
[tex]D_y[/tex]= 6 sin 150 = 3.00 m
Dₓ = 6 cos 150 = -5.196 m
Let's do the sum of the components of the vectors.
x = Aₓ + Bₓ + Cₓ + Dₓ
x = 3,729 + 1,294 + 1,4498 - 5,196
x = 1.2768 m
y = A_y + B_y + C_y + D_y
y = 4.700 -4.8296 +3.728 + 3.00
y = 6.5984 m
a) The resulting vector in the form of unit vectors is: R = 1.27 i ^ + 6.60 j ^
b) Let's use the Pythagorean theorem to find the modulus
R =[tex]\sqrt{x^2 + y^2}[/tex]
R = [tex]\sqrt{1.2768^2 + 6.5984^2}[/tex]
R = 6.72 m
c) Let's use trigonometry to find the angle.
tan θ = [tex]\frac{y}{x}[/tex]
θ = tan⁻¹ [tex]\frac{y}{x}[/tex]
θ = tan⁻¹ [tex]\frac{6.5984}{1.3768}[/tex]
θ = 79.05º
d) Let's reduce to radians
θ = 79.05º ( [tex]\frac{\pi \ rad}{180^{o}}[/tex] )
θ = 1,380 rad
In conclusion, using the decomposition of vectors we can find the results for the questions about the sum of the four vectors are:
a) The sum vector is R = (1.27 i ^ + 6.60 j ^) m
b) The modulus is: R = 6.72 m
c) The angle of the resultant is: θ = 79.1º
d) The angle in radians is: θ = 1,38 give
Learn more here: brainly.com/question/24550128
