Two forces of 20 N and 5 N are acting at an angle of 120 degree between them. Find the resultant force in magnitude and direction ?
To find the resultant force, we need to use vector addition. Given the magnitudes and angle between two forces, we can use the formula:
Resultant force (magnitude) = √(F₁² + F₂² + 2F₁F₂cosθ)
Where:
F₁ = Magnitude of the first force (20 N)
F₂ = Magnitude of the second force (5 N)
θ = Angle between the forces (120°)
Using the formula, we can calculate the resultant force:
Resultant force (magnitude) = √((20 N)² + (5 N)² + 2(20 N)(5 N)cos(120°))
Calculating this equation step-by-step:
1. Resultant force (magnitude) = √(400 N² + 25 N² + 2(20 N)(5 N)cos(120°))
2. Resultant force (magnitude) = √(425 N² + 200 N²cos(120°))
Now, let's calculate the cos(120°):
cos(120°) = -0.5
Substituting this value back into our equation:
Resultant force (magnitude) = √(425 N² + 200 N²(-0.5))
Resultant force (magnitude) = √(425 N² - 100 N²)
Resultant force (magnitude) = √(325 N²)
Resultant force (magnitude) ≈ 18.03 N
Therefore, the magnitude of the resultant force is approximately 18.03 N.
To find the direction of the resultant force, we can use the formula:
θ_r = tan^(-1)((F₂sinθ)/(F₁ + F₂cosθ))
Where:
θ_r = Angle of the resultant force
Using the formula, we can calculate the direction:
θ_r = tan^(-1)((5 Nsin(120°))/(20 N + 5 Ncos(120°)))
Calculating this equation step-by-step:
1. θ_r = tan^(-1)((5 Nsin(120°))/(25 Ncos(120°)))
2. θ_r = tan^(-1)((5 N(-0.866))/(25 N(-0.5)))
Now, let's calculate the values inside the tan^(-1):
sin(120°) = 0.866
cos(120°) = -0.5
3. θ_r = tan^(-1)((-4.33 N)/(25 N(-0.5)))
4. θ_r = tan^(-1)((8.66 N)/(12.5 N))
Simplifying further:
θ_r = tan^(-1)(0.693)
Therefore, the direction of the resultant force is approximately 35.26°.
Hence, the resultant force magnitude is approximately 18.03 N, and the direction is approximately 35.26°.
To find the resultant force, we can use the principle of vector addition. We need to find the sum of the two forces.
Step 1: Resolve the forces into their components:
To solve this problem, we need to resolve each force into its horizontal and vertical components.
Given:
Force 1 (F1) = 20 N
Force 2 (F2) = 5 N
Angle between the forces (θ) = 120 degrees
Let's resolve Force 1 (F1) into its horizontal and vertical components:
F1x = F1 * cos(θ)
F1y = F1 * sin(θ)
F1x = 20 * cos(120°)
F1y = 20 * sin(120°)
F1x = -10 N (negative value as this component is in the opposite direction of the positive x-axis)
F1y = 17.32 N (rounded to 2 decimal places)
Similarly, let's resolve Force 2 (F2):
F2x = F2 * cos(θ)
F2y = F2 * sin(θ)
F2x = 5 * cos(120°)
F2y = 5 * sin(120°)
F2x = -2.5 N
F2y = 4.33 N (rounded to 2 decimal places)
Step 2: Add the horizontal components and the vertical components separately:
Sum of x-components (Rx) = F1x + F2x
Sum of y-components (Ry) = F1y + F2y
Rx = -10 N + (-2.5 N)
Ry = 17.32 N + 4.33 N
Rx = -12.5 N
Ry = 21.65 N (rounded to 2 decimal places)
Step 3: Find the magnitude and direction of the resultant force:
Magnitude of the resultant force (R) can be found using the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
R = sqrt((-12.5 N)^2 + (21.65 N)^2)
R ≈ 25.16 N (rounded to 2 decimal places)
Direction of the resultant force (θr) can be found using trigonometry:
θr = arctan(Ry/Rx)
θr = arctan(21.65 N / -12.5 N)
θr ≈ -60.03° (rounded to 2 decimal places)
Note: The negative sign indicates that the resultant force is in the opposite direction of the positive x-axis.
Therefore, the resultant force in magnitude is approximately 25.16 N, and its direction is approximately -60.03° relative to the positive x-axis.
Use the 20N force as angle zero: <20,0>
Then the 5N force is <-4.33,2.5>
So, adding them up you get
<15.67,2.5>
Now just take the magnitude of that, and its direction angle θ has tanθ = 2.5/15.67