A body is in equilibrium when three forces act on it. Two of the forces are F1 of magnitude 8N acting due east and F2 which is 5N acting N 60degree E. Find the magnitude and direction of the third force.
To find the magnitude and direction of the third force, we can use the concept of vector addition. Here's how we can approach it step-by-step:
Step 1: Resolve F2 into its horizontal and vertical components:
The magnitude of F2 is 5N, and it is inclined at an angle of 60 degrees to the east direction. Resolving it, we get:
F2_horizontal = 5N * cos(60°) = 5N * 0.5 = 2.5N (acting due east)
F2_vertical = 5N * sin(60°) = 5N * 0.866 = 4.33N (acting north)
Step 2: Draw a vector diagram to represent the given forces:
Start by drawing a line representing F1 of magnitude 8N due east. Then draw a line representing F2_horizontal of magnitude 2.5N due east, starting from the same point as F1. Next, draw a line representing F2_vertical of magnitude 4.33N in a direction perpendicular to F2_horizontal, pointing towards the north.
Step 3: Find the resultant vector:
To find the resultant vector (third force), we need to add the horizontal and vertical components of the forces. Add F1 and F2_horizontal first:
F1 + F2_horizontal = 8N + 2.5N = 10.5N
Step 4: Find the magnitude and direction of the resultant vector:
Using the Pythagorean theorem, we can find the magnitude of the resultant vector:
Magnitude = √((F1 + F2_horizontal)^2 + F2_vertical^2)
Magnitude = √((10.5N)^2 + (4.33N)^2)
Magnitude ≈ √(110.25N^2 + 18.76N^2)
Magnitude ≈ √129.01N^2
Magnitude ≈ 11.36N
To find the direction, we can use the inverse tangent function to calculate the angle:
Direction = atan(F2_vertical / (F1 + F2_horizontal))
Direction = atan(4.33N / 10.5N)
Direction ≈ atan(0.412)
Converting the angle from radians to degrees:
Direction ≈ 23.76 degrees
Therefore, the magnitude of the third force is approximately 11.36N, and its direction is approximately 23.76 degrees to the north of east.
To find the magnitude and direction of the third force, we can use vector addition.
Step 1: Draw a vector diagram.
Start by drawing the two given forces F1 and F2. Label F1 as 8N (east) and F2 as 5N (N 60° E).
Step 2: Resolve the vectors.
Resolve F2 into its vertical and horizontal components using trigonometry. Since F2 has a direction of N 60° E, the horizontal component is F2cos(60°) and the vertical component is F2sin(60°).
F2_horizontal = F2cos(60°)
F2_vertical = F2sin(60°)
F2_horizontal = 5N × cos(60°)
F2_horizontal = 5N × 0.5
F2_horizontal = 2.5N
F2_vertical = F2sin(60°)
F2_vertical = 5N × sin(60°)
F2_vertical = 5N × 0.866
F2_vertical = 4.33N
Now, draw the horizontal and vertical components for F2 in the vector diagram.
Step 3: Add the vectors.
Add the horizontal components of both forces separately and then add the vertical components separately to find the resultant force.
Horizontal component:
To find the horizontal component of the third force, add the horizontal components of F1 and F2:
F1_horizontal = 8N (east) = 8N
F2_horizontal = 2.5N (west) = -2.5N
F3_horizontal + 8N - 2.5N = 0N
F3_horizontal = 2.5N - 8N
F3_horizontal = -5.5N
Vertical component:
To find the vertical component of the third force, add the vertical components of F1 and F2:
F1_vertical = 0N (no vertical component)
F2_vertical = 4.33N (upward)
F3_vertical + 0N + 4.33N = 0N
F3_vertical = -4.33N
Step 4: Find the magnitude and direction of the third force.
To find the magnitude of the third force, use the Pythagorean theorem:
Magnitude of F3 = square root of (F3_horizontal^2 + F3_vertical^2)
Magnitude of F3 = square root of ((-5.5N)^2 + (-4.33N)^2)
Magnitude of F3 = square root of (30.25N^2 + 18.7489N^2)
Magnitude of F3 = square root of 49.9989N^2
Magnitude of F3 ≈ 7.071N
To find the direction of the third force, use trigonometry:
Direction of F3 = arctan(F3_vertical / F3_horizontal)
Direction of F3 = arctan(-4.33N / -5.5N)
Direction of F3 ≈ 39.07°
Therefore, the magnitude of the third force is approximately 7.071N, and its direction is approximately 39.07° (measured counterclockwise from the horizontal axis).