Five forces act on an object: 1: 60 N at 90 degrees, 2: 40 N at 0 degrees, 3: 80 N at 270 degrees, 4: 40 N at 180 degrees, and 5: 50 N at 60 degrees. What is the magnitude and direction of the 6th froce that would produce equilibrium?

The vector sum of the six forces must be zero. Write that in the form of two equations (one each for the x and y components). Then solve for the unknown components of the sixth force.

At a point forces of 60N and 90N are acting and the angle between them is 60 degree. Determine the magnitude and the direction of their Resultant by graphical method

-30 N

Fr = 60[90o] + 40[0o] + 80[270o] + 40[180] + 50[60o].

X=60*Cos90+40*Cos0+80*Cos270+40*Cos180+50*Cos60 = 0 + 40 + 0 - 40 + 25 = 25 N.
Y=60*sin90+40*sin0+80*sin270+50*sin60 =
60 + 0 - 80 + 0 + 43.3 = 23.3 N.
Fr = X+Yi = 25 + 23.3i.

Fe = -25 - 23.3i = 34.2N[43.2o+180] =
34.2N[223.2o].

To find the magnitude and direction of the 6th force that would produce equilibrium, we need to consider the vector sum of the five forces acting on the object.

First, let's represent each force as a vector in a Cartesian coordinate system. We'll use the x-axis as the reference axis.

Force 1: 60 N at 90 degrees (counterclockwise from the positive x-axis)
This force can be represented as: F1 = 0 i + 60 j

Force 2: 40 N at 0 degrees (along the positive x-axis)
This force can be represented as: F2 = 40 i + 0 j

Force 3: 80 N at 270 degrees (clockwise from the positive x-axis)
This force can be represented as: F3 = 0 i - 80 j

Force 4: 40 N at 180 degrees (opposite direction to the positive x-axis)
This force can be represented as: F4 = -40 i + 0 j

Force 5: 50 N at 60 degrees (counterclockwise from the positive x-axis)
To represent this force, we need to find its x and y components. Using trigonometry:

Fx = magnitude * cos(angle) = 50 * cos(60) = 25
Fy = magnitude * sin(angle) = 50 * sin(60) = 43.30

Therefore, Force 5 can be represented as: F5 = 25 i + 43.30 j

Now, let's find the vector sum of all these forces:

F_total = F1 + F2 + F3 + F4 + F5

F_total = (0 i + 60 j) + (40 i + 0 j) + (0 i - 80 j) + (-40 i + 0 j) + (25 i + 43.30 j)

Simplifying the equation, we get:

F_total = 25 i + 23.30 j

Since the object is in equilibrium, the total force will be zero:

F_total = 0 i + 0 j = 0

Comparing the components of the total force equation, we can conclude that:

25 i + 23.30 j = 0 i + 0 j

Therefore, the magnitude of the 6th force required to produce equilibrium is 0 N, and its direction is irrelevant, as long as its x and y components cancel out the total force.