A body is in equilibrum when three forces act on it.Two of the forces are F1 of magnitude 8Newtons acting due east and F2 which is 5Newtons acting 060. find the magnitude and direction of the third force.

F1+F2+F3 = 0. F3 = -(F1+F2),

F3 = -(8N[0o]+5N.[60o]),
F3 = -((8+5*Cos60) + 5*sin60),
F3 = -10.5 - 4.33i = 11.4N[20.9o]
S. of W.

To find the magnitude and direction of the third force, we can use the concept of vector addition.

Step 1: Draw vectors F1 and F2.
- Draw a vector F1 of magnitude 8 Newtons in the east direction.
- Draw a vector F2 of magnitude 5 Newtons at an angle of 60 degrees with respect to the east direction.

Step 2: Determine the x and y components of F2.
- To find the x-component of F2, multiply the magnitude (5 Newtons) by the cosine of the angle (60 degrees). The formula is: F2x = F2 * cos(angle).
- To find the y-component of F2, multiply the magnitude (5 Newtons) by the sine of the angle (60 degrees). The formula is: F2y = F2 * sin(angle).

Step 3: Find the x-component of the third force.
- Since the body is in equilibrium, the net force acting on it should be zero.
- In the x-direction, the forces are F1 acting east and the x-component of the third force acting west.
- Therefore, the x-component of the third force should be equal to the x-component of F1. So, F3x = F1.

Step 4: Find the y-component of the third force.
- Since the body is in equilibrium, the net force acting on it should be zero.
- In the y-direction, the forces are the y-component of F2 acting south and the y-component of the third force acting north.
- Therefore, the y-component of the third force should be equal to the y-component of F2. So, F3y = F2y.

Step 5: Calculate the magnitude and direction of the third force.
- Use the x and y components of the third force (F3x and F3y) to find the magnitude of the third force using the Pythagorean theorem: magnitude of F3 = sqrt(F3x^2 + F3y^2).
- Use the equation tangent(theta) = F3y / F3x to find the direction of the third force (theta).

Following these steps:
- F3x = F1 = 8 Newtons (east).
- F2x = F2 * cos(60) = 5 * cos(60) = 5 * 0.5 = 2.5 Newtons (east).
- F2y = F2 * sin(60) = 5 * sin(60) = 5 * 0.866 = 4.33 Newtons (south).
- F3y = F2y = 4.33 Newtons (south).
- magnitude of F3 = sqrt(F3x^2 + F3y^2) = sqrt((8)^2 + (4.33)^2) = sqrt(64 + 18.76) = sqrt(82.76) ≈ 9.10 Newtons.
- theta = arctan(F3y / F3x) = arctan(4.33 / 8) ≈ 28.68 degrees (south of east).

Therefore, the magnitude of the third force is approximately 9.10 Newtons, and its direction is approximately 28.68 degrees south of east.