An 11N force acting due north and a 22-N force acting 30° North of east act concurrently on a point. What is the equilibrant force at that point? I'm having such a hard time.=(
find resultant and do the opposite.
north
11 + 22 sin 30
east
22 cos 30
magnitude = sqrt(north^2 + east^2)
tan angle north of east
= north/east
you want that magnitude
but opposite direction, same angle south of west
To find the equilibrant force, we need to determine the magnitude and direction of the force that will balance out the given forces.
First, let's break down the 22-N force into its horizontal and vertical components. We can use trigonometry to calculate these components.
Horizontal component:
Fhorizontal = F * cos(theta)
Fhorizontal = 22 N * cos(30°)
Fhorizontal ≈ 22 N * 0.866
Fhorizontal ≈ 19.05 N
Vertical component:
Fvertical = F * sin(theta)
Fvertical = 22 N * sin(30°)
Fvertical ≈ 22 N * 0.5
Fvertical ≈ 11 N
Now, let's draw a vector diagram to visualize the forces. Start with a point representing the origin and draw a line representing the 11-N force due north and another line representing the 19.05-N force due east. Place the arrowheads at the end of each line to indicate the direction of the forces.
To find the equilibrant force, we need to find a force that cancels out both the horizontal and vertical components. Therefore, we need to find a force that is equal in magnitude but opposite in direction to the sum of those components.
Horizontal equilibrant component:
Fhorizontal_eq = - 19.05 N
Vertical equilibrant component:
Fvertical_eq = - 11 N
Now, we can use the Pythagorean theorem to find the magnitude of the equilibrant force.
Magnitude of the equilibrant force:
|F_eq| = sqrt(Fhorizontal_eq^2 + Fvertical_eq^2)
|F_eq| = sqrt((-19.05 N)^2 + (-11 N)^2)
|F_eq| ≈ sqrt(363.20 N^2 + 121 N^2)
|F_eq| ≈ sqrt(484.20 N^2)
|F_eq| ≈ 22 N
Therefore, the magnitude of the equilibrant force is approximately 22 N. Now, let's determine its direction.
Direction of the equilibrant force can be found by finding the angle it makes with the positive x-axis (east direction).
Angle of the equilibrant force (theta_eq):
theta_eq = atan(Fvertical_eq / Fhorizontal_eq)
theta_eq = atan(-11 N / -19.05 N)
theta_eq = atan(0.577)
Using a calculator, we find that atan(0.577) ≈ 29.35°
Therefore, the direction of the equilibrant force is approximately 29.35° south of west.
To summarize, the equilibrant force at the given point has a magnitude of approximately 22 N and is directed 29.35° south of west.