Math Problem Statement

Compute

Solution

Let's compute the resultant force using the Trigonometric Method and Component/Analytical Method based on the information given. Since the image shows two forces F=450NF = 450 \, \text{N} at angles 3030^\circ and 4545^\circ relative to the horizontal axis, we'll assume there are two forces acting on a body and we need to find the resultant force.

1. Trigonometric Method:

For two forces F1F_1 and F2F_2 making angles θ1\theta_1 and θ2\theta_2 with the horizontal, the resultant force RR can be calculated using the Law of Cosines:

R=F12+F22+2F1F2cos(θ)R = \sqrt{F_1^2 + F_2^2 + 2 \cdot F_1 \cdot F_2 \cdot \cos(\theta)}

where θ\theta is the angle between the two forces.

Given:

  • F1=450NF_1 = 450 \, \text{N}
  • F2=450NF_2 = 450 \, \text{N}
  • θ=4530=15\theta = 45^\circ - 30^\circ = 15^\circ

Substitute the values into the formula:

R=4502+4502+2450450cos(15)R = \sqrt{450^2 + 450^2 + 2 \cdot 450 \cdot 450 \cdot \cos(15^\circ)}

R=202500+202500+24504500.9659R = \sqrt{202500 + 202500 + 2 \cdot 450 \cdot 450 \cdot 0.9659}

R=202500+202500+392122.5R = \sqrt{202500 + 202500 + 392122.5}

R=797122.5R = \sqrt{797122.5}

R892.83NR \approx 892.83 \, \text{N}

2. Component/Analytical Method:

Let's break each force into its xx and yy components.

For F1=450NF_1 = 450 \, \text{N} at 3030^\circ:

F1x=450cos(30)=4500.866=389.7NF_{1x} = 450 \cdot \cos(30^\circ) = 450 \cdot 0.866 = 389.7 \, \text{N} F1y=450sin(30)=4500.5=225NF_{1y} = 450 \cdot \sin(30^\circ) = 450 \cdot 0.5 = 225 \, \text{N}

For F2=450NF_2 = 450 \, \text{N} at 4545^\circ:

F2x=450cos(45)=4500.7071=318.2NF_{2x} = 450 \cdot \cos(45^\circ) = 450 \cdot 0.7071 = 318.2 \, \text{N} F2y=450sin(45)=4500.7071=318.2NF_{2y} = 450 \cdot \sin(45^\circ) = 450 \cdot 0.7071 = 318.2 \, \text{N}

Now sum the components:

Rx=F1x+F2x=389.7+318.2=707.9NR_x = F_{1x} + F_{2x} = 389.7 + 318.2 = 707.9 \, \text{N} Ry=F1y+F2y=225+318.2=543.2NR_y = F_{1y} + F_{2y} = 225 + 318.2 = 543.2 \, \text{N}

The magnitude of the resultant force RR is:

R=Rx2+Ry2=707.92+543.22=501137.7+295055.1=796192.8R = \sqrt{R_x^2 + R_y^2} = \sqrt{707.9^2 + 543.2^2} = \sqrt{501137.7 + 295055.1} = \sqrt{796192.8}

R892.29NR \approx 892.29 \, \text{N}

Finally, the direction θR\theta_R of the resultant is:

θR=tan1(RyRx)=tan1(543.2707.9)\theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{543.2}{707.9}\right)

θR37.99\theta_R \approx 37.99^\circ

Conclusion:

The magnitude of the resultant force is approximately 892.3N892.3 \, \text{N}, and it makes an angle of 3838^\circ with the horizontal axis.

Would you like to go over the graphical method or have any further questions?

Relative Questions:

  1. What happens if the angles between the forces increase or decrease?
  2. How does the resultant change if the forces are not equal in magnitude?
  3. Could you derive the trigonometric method formula using vector addition principles?
  4. How does the component method help in solving more complex systems?
  5. Can this method be applied to find the resultant of three or more forces?

Tip:

Always verify your angle measurements and trigonometric calculations, especially in vector addition problems, to ensure accuracy in determining the resultant force.

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Math Problem Analysis

Mathematical Concepts

Vectors
Trigonometry
Force Analysis

Formulas

Law of Cosines
Vector Components
Magnitude and Direction of Resultant Force

Theorems

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Suitable Grade Level

Advanced High School