Five forces act on an object.

(1) 65 N at 90°
(2) 40 N at 0°
(3) 79 N at 270°
(4) 40 N at 180°
(5) 50 N at 60°
What are the magnitude and direction of a sixth force that would produce equilibrium?

_____ N at _____°

(1) + (3) = 14 N @ 270º

(2) + (4) = 0

resolve 50 N @ 60º into 0º and 90º components

combine with 14 N @ 270º

the equilibrium force will have the same magnitude, but opposite direction

To determine the magnitude and direction of a sixth force that would produce equilibrium, we need to find the vector sum of the five given forces and then calculate the opposite force that will balance out the resultant force. Let's go through the steps:

1. Represent each force using vector notation. In vector notation, a force is represented by its magnitude and direction. The given forces in vector notation are:

(1) 65 N at 90° -> F1 = 65 N at 90°
(2) 40 N at 0° -> F2 = 40 N at 0°
(3) 79 N at 270° -> F3 = 79 N at 270°
(4) 40 N at 180° -> F4 = 40 N at 180°
(5) 50 N at 60° -> F5 = 50 N at 60°

2. Resolve each force into its horizontal and vertical components. To resolve a force, we use trigonometry. The horizontal (x-component) and vertical (y-component) of each force can be calculated using the following formulas:

Horizontal component = magnitude * cos(angle)
Vertical component = magnitude * sin(angle)

Let's calculate the components for each force:

F1x = 65 N * cos(90°) = 0 N (horizontal component of F1)
F1y = 65 N * sin(90°) = 65 N (vertical component of F1)

F2x = 40 N * cos(0°) = 40 N (horizontal component of F2)
F2y = 40 N * sin(0°) = 0 N (vertical component of F2)

F3x = 79 N * cos(270°) = 0 N (horizontal component of F3)
F3y = 79 N * sin(27°0) = -79 N (vertical component of F3)

F4x = 40 N * cos(180°) = -40 N (horizontal component of F4)
F4y = 40 N * sin(180°) = 0 N (vertical component of F4)

F5x = 50 N * cos(60°) = 25 N (horizontal component of F5)
F5y = 50 N * sin(60°) = 43.3 N (vertical component of F5)

3. Sum up the horizontal and vertical components separately. To find the resultant force, we need to add up the horizontal components and the vertical components separately.

Net horizontal force = F1x + F2x + F3x + F4x + F5x
Net vertical force = F1y + F2y + F3y + F4y + F5y

Net horizontal force = 0 N + 40 N + 0 N - 40 N + 25 N = 25 N
Net vertical force = 65 N + 0 N - 79 N + 0 N + 43.3 N = 29.3 N

4. Calculate the magnitude and direction of the sixth force. Since the net force should be zero for equilibrium, the magnitude of the sixth force should be the same as the magnitude of the resultant force, but in the opposite direction.

Magnitude of the sixth force = sqrt((Net horizontal force)^2 + (Net vertical force)^2)
= sqrt((25 N)^2 + (29.3 N)^2)
≈ sqrt(625 N^2 + 857.49 N^2)
≈ sqrt(1482.49 N^2)
≈ 38.5 N

Direction of the sixth force = arctan(Net vertical force / Net horizontal force)
= arctan(29.3 N / 25 N)
≈ arctan(1.17)
≈ 48.6°

Therefore, the magnitude and direction of the sixth force that would produce equilibrium are approximately 38.5 N at 48.6°.

Fr = 65i + 40 + (-79i) + (-40) + (25 + 43.3i) = 25 + 29.3i = 38.5N[49.5] N. of E. = 49.5 CCW. = Resultant force.

Fe = 38.5N.[49.5+180] = 38.5 N.[229.5o] CCW. = Equilibrant force.