Find the magnitude and direction of the equilibrant force for these two concurrent forces: 13.5 N, E and 32.0 N, N. Pls HELP me answer this. Thank u very much.
X = 13.5 N.
Y = 32 N.
tan A = Y/X = 32/13.5 = 2.37037
A = 67.1o
Fr = 32/sin 67.1 = 34.7 N[67.1o] = Resultant force.
Equilibrant=34.7N[67.1+180]=34.7N[247.1]
To find the magnitude and direction of the equilibrant force for these two concurrent forces, we can start by drawing a vector diagram.
Step 1: Draw a scale diagram using a ruler to represent the two concurrent forces. The force of 13.5 N, E can be represented by a vector pointing east (→) with a length proportional to 13.5 units. The force of 32.0 N, N can be represented by a vector pointing north (↑) with a length proportional to 32.0 units.
Step 2: Using a protractor, draw a line from the tail of the force vector representing the 13.5 N, E force to the head of the force vector representing the 32.0 N, N force. This line represents the resultant force, which is the vector sum of the two concurrent forces.
Step 3: Draw a line from the head of the resultant force vector back to the tail of the original 13.5 N, E force vector. This line represents the equilibrant force.
Step 4: Use a ruler and protractor to measure the length of the equilibrant force vector. Let's say it measures 28 units.
Step 5: The magnitude of the equilibrant force is 28 units, which, in this case, represents 28 N.
Step 6: To determine the direction of the equilibrant force, we need to measure the angle between the equilibrant force vector and the opposite direction of the original force vector. Let's say we measure this angle as 45 degrees (counterclockwise from the opposite direction of the original force vector).
Step 7: Therefore, the direction of the equilibrant force is 45 degrees clockwise from the opposite direction of the original force vector. In this case, we can say it is 45 degrees west of south (WSW).
So, the magnitude of the equilibrant force is 28 N, and its direction is 45 degrees west of south (WSW).
To find the magnitude and direction of the equilibrant force for two concurrent forces, we need to determine the resultant force first. The resultant force is the net force when multiple forces act at the same point.
Given that the two forces are 13.5 N, east (in the positive x-axis direction) and 32.0 N, north (in the positive y-axis direction), we can consider these forces as vectors.
Step 1: Resolve the forces into their x and y components:
The force of 13.5 N, E can be broken down into its x and y components as:
Fx = 13.5 N * cos(90°) = 0 N (since it is purely in the y-axis direction)
Fy = 13.5 N * sin(90°) = 13.5 N
The force of 32.0 N, N can be broken down into its x and y components as:
Fx = 32.0 N * cos(0°) = 32.0 N
Fy = 32.0 N * sin(0°) = 0 N (since it is purely in the x-axis direction)
Step 2: Add the x and y components of the forces:
Resolving the x-components gives us: Fx = 32.0 N + 0 N = 32.0 N
Resolving the y-components gives us: Fy = 0 N + 13.5 N = 13.5 N
Step 3: Find the resultant force:
To find the magnitude of the resultant force, we can use the Pythagorean theorem:
Resultant force = √(Fx^2 + Fy^2)
= √((32.0 N)^2 + (13.5 N)^2)
= √(1024 N^2 + 182.25 N^2)
= √1206.25 N^2
= 34.7 N (rounded to one decimal place)
Step 4: Find the direction of the resultant force:
To find the direction, we can use trigonometry. The angle θ can be found using the formula:
θ = tan^(-1)(Fy/Fx)
= tan^(-1)(13.5 N/32.0 N)
= tan^(-1)(0.42)
Using a calculator, θ ≈ 22.6° (rounded to one decimal place)
Since the equilibrant force is equal in magnitude and opposite in direction to the resultant force, the magnitude of the equilibrant force is 34.7 N, and its direction is 180° + θ. In this case, it would be 180° + 22.6° = 202.6°.
Therefore, the magnitude of the equilibrant force is 34.7 N, and its direction is 202.6°.