Math Problem Statement

Determine the resultant of the four forced acting on the body shown

Determine the resultant of the four forces acting on the body shown.

Solution

To determine the resultant of the four forces acting on the body, we’ll resolve each force into its x and y components and then add up these components.

Step 1: Resolving Each Force into Components

For simplicity, let’s assume:

  • F1=4000F_1 = 4000 lb (left at 30° above the x-axis)
  • F2=5000F_2 = 5000 lb (left at 30° below the x-axis)
  • F3=2000F_3 = 2000 lb (upward along y-axis)
  • F4=3000F_4 = 3000 lb (right at 30° below the x-axis)
  • F5=1000F_5 = 1000 lb (downward along y-axis)

Components:

  1. F1=4000F_1 = 4000 lb at 30° above x-axis:

    • F1x=4000cos(30)=4000×0.866=3464F_{1x} = 4000 \cos(30^\circ) = 4000 \times 0.866 = 3464 lb (negative x-direction)
    • F1y=4000sin(30)=4000×0.5=2000F_{1y} = 4000 \sin(30^\circ) = 4000 \times 0.5 = 2000 lb (positive y-direction)
  2. F2=5000F_2 = 5000 lb at 30° below x-axis:

    • F2x=5000cos(30)=5000×0.866=4330F_{2x} = 5000 \cos(30^\circ) = 5000 \times 0.866 = 4330 lb (negative x-direction)
    • F2y=5000sin(30)=5000×0.5=2500F_{2y} = -5000 \sin(30^\circ) = -5000 \times 0.5 = -2500 lb (negative y-direction)
  3. F3=2000F_3 = 2000 lb along y-axis:

    • F3x=0F_{3x} = 0
    • F3y=2000F_{3y} = 2000 lb (positive y-direction)
  4. F4=3000F_4 = 3000 lb at 30° below x-axis:

    • F4x=3000cos(30)=3000×0.866=2598F_{4x} = 3000 \cos(30^\circ) = 3000 \times 0.866 = 2598 lb (positive x-direction)
    • F4y=3000sin(30)=3000×0.5=1500F_{4y} = -3000 \sin(30^\circ) = -3000 \times 0.5 = -1500 lb (negative y-direction)
  5. F5=1000F_5 = 1000 lb along y-axis:

    • F5x=0F_{5x} = 0
    • F5y=1000F_{5y} = -1000 lb (negative y-direction)

Step 2: Summing Up Components

Total xx-components:

Fx=34644330+2598=5196 lb\sum F_x = -3464 - 4330 + 2598 = -5196 \text{ lb}

Total yy-components:

Fy=20002500+200015001000=1000 lb\sum F_y = 2000 - 2500 + 2000 - 1500 - 1000 = -1000 \text{ lb}

Step 3: Calculating the Resultant

The resultant force RR is given by: R=(Fx)2+(Fy)2R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2} R=(5196)2+(1000)2=26971616+1000000=279716165290 lbR = \sqrt{(-5196)^2 + (-1000)^2} = \sqrt{26971616 + 1000000} = \sqrt{27971616} \approx 5290 \text{ lb}

Step 4: Direction of the Resultant

The direction θ\theta of RR relative to the x-axis is: θ=tan1(FyFx)=tan1(10005196)10.89\theta = \tan^{-1} \left(\frac{\sum F_y}{\sum F_x}\right) = \tan^{-1} \left(\frac{-1000}{-5196}\right) \approx 10.89^\circ So, the resultant force is approximately 5290 lb at 10.89° above the negative x-axis.

Would you like further clarification on any of these steps?


  1. How are x and y components calculated for forces at an angle?
  2. Why do we use trigonometric functions to resolve forces?
  3. How does the resultant magnitude relate to individual forces?
  4. How would adding a new force impact the resultant direction?
  5. What effect does the angle have on the component values?

Tip: Breaking down forces into components simplifies calculating the net force direction and magnitude.

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Math Problem Analysis

Mathematical Concepts

Vector resolution
Trigonometry
Force summation

Formulas

Force components: F_x = F * cos(θ), F_y = F * sin(θ)
Resultant force magnitude: R = √((ΣF_x)^2 + (ΣF_y)^2)
Resultant direction: θ = tan^(-1)(ΣF_y / ΣF_x)

Theorems

Pythagorean theorem
Trigonometric identities

Suitable Grade Level

Undergraduate level (Engineering Mechanics)