Math Problem Statement

Determine the resultant of the system of concurrent forces with the following force values and angles: F1 = 150 N, F2 = 200 N, F3 = 80 N, and F4 = 180 N at the respective angles.

Solution

The image presents a problem asking to determine the resultant of a system of concurrent forces in a two-dimensional plane, with the force magnitudes and angles given. Here's the breakdown:

Given:

  • F1=150NF_1 = 150 \, \text{N} at an angle of 30° above the positive xx-axis.
  • F2=200NF_2 = 200 \, \text{N} at an angle of 60° above the negative xx-axis.
  • F3=80NF_3 = 80 \, \text{N} at an angle of 45° below the negative xx-axis.
  • F4=180NF_4 = 180 \, \text{N} along the negative yy-axis.

Steps to solve:

  1. Resolve each force into its components:

    • F1F_1 (150 N, 30°):
      • F1x=150cos(30°)F_{1x} = 150 \cos(30°)
      • F1y=150sin(30°)F_{1y} = 150 \sin(30°)
    • F2F_2 (200 N, 60°):
      • F2x=200cos(60°)F_{2x} = -200 \cos(60°) (negative since it's along the negative xx-axis)
      • F2y=200sin(60°)F_{2y} = 200 \sin(60°)
    • F3F_3 (80 N, 45°):
      • F3x=80cos(45°)F_{3x} = -80 \cos(45°) (negative since it's along the negative xx-axis)
      • F3y=80sin(45°)F_{3y} = -80 \sin(45°) (negative since it's below the xx-axis)
    • F4F_4 (180 N, along negative yy-axis):
      • F4x=0F_{4x} = 0 (no xx-component)
      • F4y=180F_{4y} = -180 (entirely along negative yy-axis)
  2. Sum up all the components:

    • Fx=F1x+F2x+F3x+F4xF_x = F_{1x} + F_{2x} + F_{3x} + F_{4x}
    • Fy=F1y+F2y+F3y+F4yF_y = F_{1y} + F_{2y} + F_{3y} + F_{4y}
  3. Find the magnitude of the resultant force:

    • R=Fx2+Fy2R = \sqrt{F_x^2 + F_y^2}
  4. Find the angle of the resultant force:

    • θ=tan1(FyFx)\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)

I'll compute the values of the components and determine the resultant.The components and resultant for the system of forces are as follows:

  • Fx=26.67NF_x = -26.67 \, \text{N} (negative, meaning it's toward the negative xx-axis)
  • Fy=11.64NF_y = 11.64 \, \text{N} (positive, meaning it's toward the positive yy-axis)

The magnitude of the resultant force is:

  • R=29.09NR = 29.09 \, \text{N}

The angle of the resultant force is:

  • θR=156.42\theta_R = 156.42^\circ (measured counterclockwise from the positive xx-axis)

This angle indicates that the resultant force is directed primarily in the second quadrant (between the negative xx-axis and positive yy-axis).

Would you like further clarification on any steps, or more detailed explanation?

Here are some related questions for deeper exploration:

  1. How do the angles of forces affect their resultant direction?
  2. How would changing the magnitude of F1F_1 alter the resultant?
  3. What happens if we add another force along the positive yy-axis?
  4. How does the angle calculation change if we resolve forces using different reference axes?
  5. What are practical applications of determining the resultant force in engineering?

Tip: To simplify vector problems, always break forces into their components first before summing them up.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector addition
Trigonometry
Force components

Formulas

F_x = F \cos(\theta)
F_y = F \sin(\theta)
R = \sqrt{F_x^2 + F_y^2}
θ = \tan^{-1}(F_y / F_x)

Theorems

Pythagoras theorem
Trigonometric identities

Suitable Grade Level

High School to Early University (Grades 10-12, introductory physics)