A body is in equilibrium when three forces act on it. Two of the forces are F1 of magnitude 8N acting due east and F2 which is 5N acting N 60degree E. Find the magnitude and direction of the third force.
To find the magnitude and direction of the third force, we can use vector addition.
Step 1: Draw a vector diagram.
Begin by drawing a reference coordinate system with an x-axis (east) and y-axis (north). Draw the first force, F1, of magnitude 8N due east. Then draw the second force, F2, of magnitude 5N 60 degrees north of east.
Step 2: Resolve F2 into its x-component and y-component.
To calculate the x-component and y-component of F2, we can use trigonometry. The x-component, F2x, is the adjacent side of the angle (60 degrees) multiplied by the magnitude of F2 (5N) and cos(60). The y-component, F2y, is the opposite side multiplied by the magnitude of F2 and sin(60).
F2x = 5N * cos(60) = 2.5N
F2y = 5N * sin(60) = 4.33N
Step 3: Add the x-components and y-components of the forces.
To determine the magnitude and direction of the third force, we need to add the x-components and y-components of all the forces. Since the body is in equilibrium, the sum of the x-components and y-components should be zero.
∑Fx = F1x + F2x + F3x = 0
∑Fy = F1y + F2y + F3y = 0
Substituting the known values:
8N + 2.5N + F3x = 0
F3y + 4.33N = 0
Step 4: Solve for F3x and F3y.
From the above equations, we can solve for the x-component and y-component of the third force, F3.
F3x = -10.5N
F3y = -4.33N
Step 5: Calculate the magnitude and direction of the third force.
To find the magnitude and direction of the third force, we can use the Pythagorean theorem and trigonometry.
Magnitude of F3 = √(F3x^2 + F3y^2)
= √((-10.5N)^2 + (-4.33N)^2)
≈ 11.34N
Direction of F3 = tan^(-1)(F3y/F3x)
= tan^(-1)(-4.33N / -10.5N)
≈ 22.47 degrees north of west
Therefore, the magnitude of the third force is approximately 11.34N, and its direction is approximately 22.47 degrees north of west.
break up f1 and f2 into components, then f3 componnets are equal, but opposite directions.
F1+F2=8E +5cos60N+5sin60E
= E(8+5*.866)+5*1/2 N
F3=W(8+5*.866) + 2.5 S