Given the vectors below, determine the resultant and the equilibrant using trigonometry. Show all your steps.
330.0 newtons @ 125 degrees
250.0 newtons @ 60.0 degrees
Fr = 330[125o] + 250[60o] = Resultant force.
X = 330*Cos125 + 250*Cos60 = -64.3 N.
Y = 330*sin125 + 250*sin60 = 487 N.
Q2.
Tan A = Y/X = 487/-64.3 = -7.57387.
A = -82.5o = 82.5o N. of W. = 97.5o CCW.
Fr=Y/sin A = 487/sin97.5 = 491 N.[97.5o}
Fr + Fe = 0, Fe = -Fr.
Fe = -(-64.3 + 487i) = 64.3 - 487i =
491[82.5o] S. of E. = 491 N.[277.5o]CCW.
= Equilibrant force.
Or Fe=-Fr=Fr[97.5o+180o] = 491N.[277.5]
To find the resultant and the equilibrant using trigonometry, we need to break down the given vectors into their horizontal and vertical components.
First, let's find the horizontal and vertical components of the first vector, 330.0 newtons @ 125 degrees.
Horizontal component: 330.0 newtons * cos(125 degrees) = -115.9 newtons
Vertical component: 330.0 newtons * sin(125 degrees) = 277.4 newtons
Now, let's find the horizontal and vertical components of the second vector, 250.0 newtons @ 60.0 degrees.
Horizontal component: 250.0 newtons * cos(60 degrees) = 125.0 newtons
Vertical component: 250.0 newtons * sin(60 degrees) = 216.5 newtons
Next, let's add up the horizontal components and vertical components separately to find the components of the resultant vector:
Horizontal component of the resultant = -115.9 newtons + 125.0 newtons = 9.1 newtons
Vertical component of the resultant = 277.4 newtons + 216.5 newtons = 493.9 newtons
Using the Pythagorean theorem, we can calculate the magnitude of the resultant vector:
Resultant = sqrt((9.1 newtons)^2 + (493.9 newtons)^2)
Resultant = sqrt(82.81 newtons^2 + 243942.21 newtons^2)
Resultant = sqrt(244025.02 newtons^2)
Resultant ≈ 493.97 newtons
To find the direction of the resultant vector, we can use the inverse tangent function:
Direction = atan(493.9 newtons / 9.1 newtons)
Direction ≈ 89.2 degrees
Therefore, the resultant vector is approximately 493.97 newtons @ 89.2 degrees.
To find the equilibrant vector, we need to find the negative of the resultant vector. The magnitude of the equilibrant vector will be the same as the magnitude of the resultant vector, but the direction will be opposite.
Equilibrant = 493.97 newtons @ (89.2 degrees + 180 degrees)
Equilibrant ≈ 493.97 newtons @ 269.2 degrees
To determine the resultant and equilibrant, we need to resolve the given vectors into their horizontal and vertical components.
First, let's find the horizontal and vertical components of the first vector, 330.0 newtons @ 125 degrees:
- The horizontal component is given by: H₁ = 330.0 * cos(125°)
- The vertical component is given by: V₁ = 330.0 * sin(125°)
Calculating H₁ and V₁:
H₁ = 330.0 * cos(125°) = -133.480 newtons
V₁ = 330.0 * sin(125°) = 289.590 newtons
Now, let's find the horizontal and vertical components of the second vector, 250.0 newtons @ 60.0 degrees:
- The horizontal component is given by: H₂ = 250.0 * cos(60°)
- The vertical component is given by: V₂ = 250.0 * sin(60°)
Calculating H₂ and V₂:
H₂ = 250.0 * cos(60°) = 125.000 newtons
V₂ = 250.0 * sin(60°) = 216.506 newtons
Next, we'll sum the horizontal and vertical components separately to find the resultant vector's components:
Horizontal component of resultant (Hᵣ) = H₁ + H₂
Vertical component of resultant (Vᵣ) = V₁ + V₂
Calculating Hᵣ and Vᵣ:
Hᵣ = -133.480 + 125.000 = -8.480 newtons
Vᵣ = 289.590 + 216.506 = 506.096 newtons
To find the magnitude and direction of the resultant vector, we'll use the Pythagorean theorem and trigonometry:
Magnitude of resultant vector (R) = sqrt(Hᵣ² + Vᵣ²)
Direction of resultant vector (θ) = atan(Vᵣ / Hᵣ)
Calculating R and θ:
R = sqrt((-8.480)² + (506.096)²) ≈ 506.54 newtons
θ ≈ atan(506.096 / -8.480) ≈ -88.794 degrees
Finally, to find the equilibrant vector, we just need to reverse the direction of the resultant vector:
Equilibrant vector magnitude = R
Equilibrant vector direction = θ + 180°
Equilibrant vector magnitude ≈ 506.54 newtons
Equilibrant vector direction ≈ -88.794 + 180° ≈ 91.206 degrees
Therefore, the resultant vector is approximately 506.54 newtons @ -88.794 degrees, and the equilibrant vector is approximately 506.54 newtons @ 91.206 degrees.