Get the resultant and its direction along the horizontal for the group of forces
F1 = 12,000 N at 42° NE F3 = 13,500 N at 37° SE
F2 = 15,000 N at 40° SW F4 = 3,500 N at 30° SW
B.2. The force F2 is missing, however resultant R having 5,432.24 N was acting 39° SE , F1 is 4,000 at 49° NW , F3 is 3,000 N at 27° NE and F4 is 5,000 N at 20° SW . Calculate F2 and its direction
To find the resultant and its direction along the horizontal for the first group of forces:
- Convert each force vector into its horizontal and vertical components:
- F1: horizontal = 12,000 cos(42°) = 8,987.34 N, vertical = 12,000 sin(42°) = 8,146.29 N
- F2: horizontal = 15,000 cos(40°) = 11,481.97 N, vertical = -15,000 sin(40°) = -9,700.66 N
- F3: horizontal = 13,500 cos(37°) = 10,822.37 N, vertical = -13,500 sin(37°) = -8,150.50 N
- F4: horizontal = 3,500 cos(30°) = 3,032.12 N, vertical = -3,500 sin(30°) = -1,750.00 N
- Add up the horizontal and vertical components separately:
- Horizontal: 8,987.34 N + 11,481.97 N + 10,822.37 N + 3,032.12 N = 34,323.80 N
- Vertical: 8,146.29 N - 9,700.66 N - 8,150.50 N - 1,750.00 N = -11,454.87 N
- Use the Pythagorean theorem to find the magnitude of the resultant:
- Magnitude: sqrt((34,323.80 N)^2 + (-11,454.87 N)^2) = 35,973.26 N
- Use inverse tangent to find the direction of the resultant along the horizontal:
- Direction: tan^-1(-11,454.87 N / 34,323.80 N) = -18.94° (measured clockwise from east)
Therefore, the resultant for the first group of forces is 35,973.26 N at 18.94° W of S.
To find F2 and its direction for the second group of forces:
- Convert each force vector into its horizontal and vertical components:
- R: horizontal = 5,432.24 cos(39°) = 4,169.21 N, vertical = -5,432.24 sin(39°) = -3,398.39 N
- F1: horizontal = 4,000 cos(49°) = 2,572.98 N, vertical = 4,000 sin(49°) = 2,996.07 N
- F3: horizontal = 3,000 cos(27°) = 2,588.19 N, vertical = 3,000 sin(27°) = 1,336.68 N
- F4: horizontal = 5,000 cos(20°) = 4,723.17 N, vertical = -5,000 sin(20°) = -1,710.12 N
- Add up the horizontal and vertical components separately and negate both:
- Horizontal: -(4,169.21 N + 2,572.98 N + 2,588.19 N + 4,723.17 N) = -14,053.55 N
- Vertical: -(-3,398.39 N + 2,996.07 N + 1,336.68 N - 1,710.12 N) = 6,441.06 N
- Use the Pythagorean theorem to find the magnitude of F2:
- Magnitude: sqrt((-14,053.55 N)^2 + (6,441.06 N)^2) = 15,415.69 N
- Use inverse tangent to find the direction of F2:
- Direction: tan^-1(6,441.06 N / 14,053.55 N) = 24.93° (measured counterclockwise from south)
Therefore, F2 for the second group of forces is 15,415.69 N at 24.93° N of W.
For the first part, we will find the horizontal components of each force:
F1 horizontal component: 12000 * cos(42) = 12000 * 0.7431 = 8917.2 N
F2 horizontal component: -15000 * cos(40) = -15000 * 0.7660 = -11490 N (negative sign because it's SW)
F3 horizontal component: 13500 * cos(37) = 13500 * 0.7986 = 10781.1 N
F4 horizontal component: -3500 * cos(30) = -3500 * 0.8660 = -3031 N (negative sign because it's SW)
Sum of horizontal components:
8917.2 - 11490 + 10781.1 - 3031 = 177.3 N
Now we will find the resultant and its angle along the horizontal axis:
Resultant horizontal force (Rh) = 177.3 N
For the second part:
Given that the resultant R is 5432.24 N and acts at an angle of 39° SE, we can find its horizontal component:
Horizontal component of R: 5432.24 * cos(39) = 5432.24 * 0.7771 = 4222.07 N
Now we will calculate the horizontal components of F1, F3, and F4:
F1 horizontal component: 4000 * cos(49) = 4000 * 0.6561 = 2624.4 N
F3 horizontal component: 3000 * cos(27) = 3000 * 0.8910 = 2673 N
F4 horizontal component: -5000 * cos(20) = -5000 * 0.9397 = -4698.5 N (negative sign because it's SW)
We can find the horizontal component of F2 by subtracting the horizontal components of F1, F3, and F4 from the horizontal component of R:
Horizontal component of F2 = Horizontal component of R - (Horizontal components of F1, F3, and F4)
= 4222.07 - 2624.4 - 2673 + 4698.5
= 3623.17 N
As the horizontal component of F2 is positive and the overall resultant force is acting in the southeast direction, the horizontal component of F2 is also in the Southeast direction. Thus, the force F2 will have the direction acting to the Southeast.
Now, to find the magnitude of F2, we can use the tangent of the resultant angle and the sum of the vertical components to find the vertical component of F2:
tan(39) = (Sum of vertical components of F2, F1, F3, and F4) / (Sum of horizontal components of F1, F3, and F4)
Sum of vertical components of F2, F1, F3, and F4 = tan(39) * (Sum of horizontal components of F1, F3, and F4)
= 0.8098 * (2624.4 + 2673 - 4698.5)
= 0.8098 * 599.9
= 485.05 N
Since the overall force R, F1, and F3 are acting in NE, the vertical component of F4 will be negative. Thus, we can find the vertical component of F2 by subtracting the vertical components of F1, F3, and F4:
Vertical component of F2 = Vertical component of R - (Vertical components of F1, F3, and F4)
= 485.05 - 4000 * sin(49) - 3000 * sin(27) + 5000 * sin(20)
= 485.05 - 4000 * 0.7547 - 3000 * 0.4540 + 5000 * 0.3420
= -1971.10 N
The magnitude of F2 can then be found using the Pythagorean theorem:
F2 = sqrt[(Horizontal component of F2)^2 + (Vertical component of F2)^2]
= sqrt[(3623.17)^2 + (-1971.10)^2]
= sqrt[13127445.93 + 3882366.81]
= 12208.57 N
The direction of F2 can be found by taking the inverse tangent of its vertical and horizontal components:
Direction of F2 = atan(abs(Vertical component of F2) / abs(Horizontal component of F2))
= atan(1971.10/3623.17)
= 28.91 degrees
Since the horizontal component of F2 has a positive sign and the vertical component of F2 has a negative sign, that means the force F2 acts in the southeast direction. Therefore, F2 has a magnitude of 12208.57 N at an angle of 28.91° SE.
Final Answer:
F2 = 12208.57 N at 28.91° SE