a body is at equlibrium under the action of three forces.one force is 10n acting due east and one is 5n in the direction 60degree north east.what is the magnitude and direction of the third force.
10 + 5[60o] + F = 0.
10 + 5*Cos60 + i5*sin60 + F = 0.
10 + 2.5 + 4.33i + F = 0.
12.5 + 4.33i + F = 0.
F = -12.5-4.33i = 13.2N[W19.1oS]
To find the magnitude and direction of the third force, we can use vector addition. The three forces can be represented as vectors:
Force 1 (F1) = 10 N, due east (0°)
Force 2 (F2) = 5 N, 60° north-east
To find the magnitude and direction of the third force (F3), we need to add these two forces (F1 and F2):
Step 1: Resolve the second force (F2) into its x and y components. Since it is at an angle of 60° north-east, we can split it into vertical (y-axis) and horizontal (x-axis) components.
The vertical component (F2y) can be found using trigonometry. Since the angle between F2 and the y-axis is 30° (complementary to 60° north-east), we can use the sine function:
F2y = F2 * sin(30°)
= 5 N * sin(30°)
= 2.5 N
The horizontal component (F2x) can be found using trigonometry as well. Since the angle between F2 and the x-axis is also 30°, we can use the cosine function:
F2x = F2 * cos(30°)
= 5 N * cos(30°)
= 4.33 N (rounded to two decimal places)
So now we have:
F2y = 2.5 N (in the positive y-axis direction)
F2x = 4.33 N (in the positive x-axis direction)
Step 2: Add the x and y components of the forces together.
Fx = F1 + F2x
= 10 N + 4.33 N
= 14.33 N (rounded to two decimal places)
Fy = F2y
= 2.5 N
Step 3: Find the resulting magnitude of the third force (F3) using the Pythagorean theorem:
|F3|² = Fx² + Fy²
|F3|² = (14.33 N)² + (2.5 N)²
|F3|² = 205.3489 N² + 6.25 N²
|F3|² = 211.5989 N²
|F3| ≈ √(211.5989 N²)
|F3| ≈ 14.54 N (rounded to two decimal places)
Therefore, the magnitude of the third force (F3) is approximately 14.54 N.
Step 4: Find the direction of the third force (F3) using trigonometry.
The direction of F3 (measured from the positive x-axis) can be found using the arctan function:
θ = arctan(Fy / Fx)
θ = arctan(2.5 N / 14.33 N)
θ = arctan(0.174)
θ ≈ 9.99° (rounded to two decimal places)
Therefore, the direction of the third force (F3) is approximately 10° (counterclockwise) from the positive x-axis.
In summary, the magnitude of the third force is approximately 14.54 N, and its direction is approximately 10° counterclockwise from the positive x-axis.