Find analytically the sum or resultant of the following displacements:
A= 10m Northwest , B= 20m 30 degree north of east and C = 35m due Souh
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All angles are measured CCW from +x-axis:
Fr = 10m[135o] + 20m[30o] + 35m[270o].
Fr = (-7.07+7.07i) + (17.32+10i) + (0-35i),
Combine like-terms and convert to polar form:
10.25 - 17.93i = 20.7m[-60o] = 20.7m[60o] S. of E.
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To find the sum or resultant of the displacements A, B, and C, we need to add these vectors together.
First, let's represent the given displacements as vectors:
A = 10m Northwest
B = 20m at 30 degrees north of east
C = 35m due South
Now, let's break down each displacement into its horizontal (x) and vertical (y) components.
For A:
A_x = -10m (movement to the left or West)
A_y = 10m (movement upward or North)
For B:
To find the x and y components of B, we can use trigonometry.
B_x = 20m * cos(30°) = 17.321m (movement towards the East)
B_y = 20m * sin(30°) = 10m (movement upward or North)
For C:
C_x = 0m (no horizontal movement)
C_y = -35m (movement downward or South)
Now, add up the horizontal and vertical components separately:
Resultant_x = A_x + B_x + C_x = -10m + 17.321m + 0m = 7.321m (movement towards the East)
Resultant_y = A_y + B_y + C_y = 10m + 10m - 35m = -15m (movement downward or South)
Now, we have the horizontal and vertical components of the resultant displacement. To find the magnitude and direction of the resultant displacement, we can use the Pythagorean theorem and trigonometry.
Magnitude of the resultant displacement (R):
R = sqrt((Resultant_x)^2 + (Resultant_y)^2)
R = sqrt((7.321m)^2 + (-15m)^2)
R ≈ 16.39m
Direction of the resultant displacement (θ):
θ = arctan(Resultant_y / Resultant_x)
θ = arctan(-15m / 7.321m)
θ ≈ -63.43°
Therefore, the sum or resultant of the given displacements A, B, and C is approximately 16.39m at an angle of -63.43° (measured counterclockwise from the positive x-axis).