Math Problem Statement

solve

Refer to Figure 3-4 The magnitudes of the forces as shown in the figure are: F1 = 80.0 N, F2 = 60.0 N, and F3 = 40.0 N. The resultant force acting on the particle O is given by.

Solution

To find the resultant force acting on the particle at point O, we need to determine the vector sum of the forces F1F_1, F2F_2, and F3F_3.

Step 1: Decompose the forces into their components.

  1. Force F1=80.0NF_1 = 80.0 \, \text{N}, directed at an angle of 3030^\circ below the positive x-axis.

    • F1x=F1cos(30)=80×32=40369.28NF_{1x} = F_1 \cos(30^\circ) = 80 \times \frac{\sqrt{3}}{2} = 40\sqrt{3} \approx 69.28 \, \text{N}
    • F1y=F1sin(30)=80×12=40NF_{1y} = F_1 \sin(30^\circ) = 80 \times \frac{1}{2} = 40 \, \text{N} (downward, so it's negative in the y-axis).
  2. Force F2=60.0NF_2 = 60.0 \, \text{N}, directed along the positive y-axis.

    • F2x=0F_{2x} = 0
    • F2y=60NF_{2y} = 60 \, \text{N}
  3. Force F3=40.0NF_3 = 40.0 \, \text{N}, directed along the negative x-axis.

    • F3x=40NF_{3x} = -40 \, \text{N}
    • F3y=0F_{3y} = 0

Step 2: Sum of components.

  1. Sum of x-components: Fresultant,x=F1x+F2x+F3x=403+04029.28NF_{\text{resultant}, x} = F_{1x} + F_{2x} + F_{3x} = 40\sqrt{3} + 0 - 40 \approx 29.28 \, \text{N}

  2. Sum of y-components: Fresultant,y=F1y+F2y+F3y=40+60+0=20NF_{\text{resultant}, y} = F_{1y} + F_{2y} + F_{3y} = -40 + 60 + 0 = 20 \, \text{N}

Step 3: Calculate the magnitude of the resultant force.

Fresultant=(Fresultant,x)2+(Fresultant,y)2F_{\text{resultant}} = \sqrt{(F_{\text{resultant}, x})^2 + (F_{\text{resultant}, y})^2} Fresultant=(29.28)2+(20)2856.0384+4001256.038435.45NF_{\text{resultant}} = \sqrt{(29.28)^2 + (20)^2} \approx \sqrt{856.0384 + 400} \approx \sqrt{1256.0384} \approx 35.45 \, \text{N}

Step 4: Calculate the direction of the resultant force.

θ=tan1(Fresultant,yFresultant,x)\theta = \tan^{-1} \left( \frac{F_{\text{resultant}, y}}{F_{\text{resultant}, x}} \right) θ=tan1(2029.28)34.56 above the positive x-axis\theta = \tan^{-1} \left( \frac{20}{29.28} \right) \approx 34.56^\circ \text{ above the positive x-axis}

Result

  • Magnitude of the resultant force: 35.45N35.45 \, \text{N}
  • Direction: 34.5634.56^\circ above the positive x-axis.

Would you like more details or have any questions?

Here are 5 related questions:

  1. How do we find the resultant force if there are more than three forces acting?
  2. What happens to the resultant force direction if one of the forces is reversed?
  3. How do we calculate the components if the forces are not perpendicular?
  4. Can we use graphical methods to find the resultant force? If yes, how?
  5. How would friction or other resistive forces affect the resultant force calculation?

Tip: Always resolve forces into components first, then combine them to simplify complex problems.

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Force Decomposition
Trigonometry

Formulas

F_{resultant} = \sqrt{(F_{resultant,x})^2 + (F_{resultant,y})^2}
\theta = \tan^{-1} \left( \frac{F_{resultant,y}}{F_{resultant,x}} \right)

Theorems

Vector addition theorem
Trigonometric identity for sine and cosine

Suitable Grade Level

Grades 11-12