three forces, each of magnitude 1450 LB act in the same point. The angle between adjacent forces is 45.0 degree. Find the resultant force?
say one is along x axis, second is 45 deg toward y axis (East and Northeast), third is 45 degrees down toward - y axis (Southeast)
x forces:
1450 + 1450 cos 45 + 1450 cos(-45)
= 1450 (1 + .707 + .707)
= 1450(2.414)
= 3501 Lb
y forces
0 + 1450 sin 45 + 1450 sin (-45)
= 0
so
1450 Lb in direction of middle force
To find the resultant force, we need to use the concept of vector addition. We can find the horizontal and vertical components of each force and then add them together.
Given:
Magnitude of each force (F) = 1450 lb
Angle between adjacent forces (θ) = 45 degrees
First, let's consider the horizontal components:
Since the angle between adjacent forces is 45 degrees, each horizontal component is equal to the force multiplied by the cosine of the angle.
Horizontal component (F_horizontal) = F * cos(θ)
Next, let's consider the vertical components:
Similarly, the vertical component for each force is equal to the force multiplied by the sine of the angle.
Vertical component (F_vertical) = F * sin(θ)
Now, let's calculate these components for each force:
For the first force:
F1_horizontal = 1450 lb * cos(45°) = 1024.9 lb (rounded to one decimal place)
F1_vertical = 1450 lb * sin(45°) = 1024.9 lb (rounded to one decimal place)
For the second force:
F2_horizontal = 1450 lb * cos(45°) = 1024.9 lb (rounded to one decimal place)
F2_vertical = 1450 lb * sin(45°) = 1024.9 lb (rounded to one decimal place)
For the third force:
F3_horizontal = 1450 lb * cos(45°) = 1024.9 lb (rounded to one decimal place)
F3_vertical = 1450 lb * sin(45°) = 1024.9 lb (rounded to one decimal place)
Now, let's add up the horizontal and vertical components separately:
Total horizontal component (F_total_horizontal) = F1_horizontal + F2_horizontal + F3_horizontal
= 1024.9 lb + 1024.9 lb + 1024.9 lb
= 3074.7 lb (rounded to one decimal place)
Total vertical component (F_total_vertical) = F1_vertical + F2_vertical + F3_vertical
= 1024.9 lb + 1024.9 lb + 1024.9 lb
= 3074.7 lb (rounded to one decimal place)
Finally, the magnitude of the resultant force (F_resultant) can be calculated using the Pythagorean theorem:
F_resultant = √(F_total_horizontal^2 + F_total_vertical^2)
= √((3074.7 lb)^2 + (3074.7 lb)^2)
= √(9455352.09 lb^2 + 9455352.09 lb^2)
= √18910704.18 lb^2
= 4349.9 lb (rounded to one decimal place)
Therefore, the magnitude of the resultant force is approximately 4349.9 lb.
To find the resultant force, we can use the concept of vector addition.
First, let's assume the three forces are labeled as F1, F2, and F3. Each force magnitude is 1450 lb.
Since the forces are acting at the same point, we can represent them as vectors starting from the same point.
Using trigonometry, we can split each force into its horizontal and vertical components. Since the angle between adjacent forces is 45 degrees, the horizontal and vertical components will be equal in magnitude.
Let's denote the horizontal and vertical components of each force as Fx and Fy, respectively.
Since the angle between adjacent forces is 45 degrees, the horizontal and vertical components will be equal in magnitude.
So, F1x = F1y = 1450 lb, F2x = F2y = 1450 lb, and F3x = F3y = 1450 lb.
To find the resultant force, we need to add up the horizontal and vertical components separately.
Horizontal component (Rx):
Rx = F1x + F2x + F3x = 3 * 1450 lb = 4350 lb
Vertical component (Ry):
Ry = F1y + F2y + F3y = 3 * 1450 lb = 4350 lb
Now, using the Pythagorean theorem, we can calculate the magnitude of the resultant force (R):
R = sqrt(Rx^2 + Ry^2)
= sqrt(4350 lb^2 + 4350 lb^2)
= sqrt(2 * (4350 lb)^2)
= sqrt(2) * 4350 lb
Therefore, the magnitude of the resultant force is sqrt(2) * 4350 lb.