Math Problem Statement

A particle is in equilibrium under the action of three forces: 15 N to the right, 12 N downward, and an unknown force F. Find the magnitude of F and the angle it makes with the direction of the 15 N force.

Solution

To solve the problem, we'll use the conditions of equilibrium for a particle. A particle is in equilibrium when the vector sum of all forces acting on it is zero. This means that the sum of the forces in the horizontal direction (x-direction) and the vertical direction (y-direction) must each be zero.

Given forces:

  • 15 N force directed to the right (horizontal)
  • 12 N force directed downward (vertical)
  • FF with unknown magnitude and direction

Let's denote the angle that FF makes with the horizontal direction as θ\theta. We can resolve FF into its horizontal (FxF_x) and vertical (FyF_y) components: Fx=FcosθF_x = F \cos \theta Fy=FsinθF_y = F \sin \theta

The equilibrium conditions give us two equations:

  1. Sum of forces in the x-direction: Fx15=0F_x - 15 = 0 Fcosθ=15F \cos \theta = 15

  2. Sum of forces in the y-direction: Fy12=0F_y - 12 = 0 Fsinθ=12F \sin \theta = 12

We can solve these equations simultaneously to find the magnitude of FF and the angle θ\theta.

First, solve for FF: F=(Fcosθ)2+(Fsinθ)2F = \sqrt{(F \cos \theta)^2 + (F \sin \theta)^2} Substituting Fcosθ=15F \cos \theta = 15 and Fsinθ=12F \sin \theta = 12: F=152+122F = \sqrt{15^2 + 12^2} F=225+144F = \sqrt{225 + 144} F=369F = \sqrt{369} F=19.2NF = 19.2 \, \text{N}

Now, find the angle θ\theta: cosθ=15F=1519.2\cos \theta = \frac{15}{F} = \frac{15}{19.2} θ=cos1(1519.2)\theta = \cos^{-1}\left(\frac{15}{19.2}\right) θcos1(0.78125)\theta \approx \cos^{-1}(0.78125) θ38.7\theta \approx 38.7^\circ

So, the magnitude of FF is 19.2 N, and the angle it makes with the direction of the 15 N force is approximately 38.7 degrees.

Would you like more details or have any questions?

Here are five related questions for further exploration:

  1. How would the equilibrium conditions change if the forces were not coplanar?
  2. What would happen if an additional force was added to the system?
  3. Can you determine the components of a force if only its magnitude and angle are given?
  4. How would the system be analyzed if the angle of the force FF were given instead of the 12 N force?
  5. How can vector addition be used to find the resultant force in a non-equilibrium situation?

Tip: When dealing with equilibrium problems, always start by resolving all forces into their components and applying the equilibrium conditions to solve for unknowns.

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Math Problem Analysis

Mathematical Concepts

Equilibrium
Vector Resolution
Trigonometry

Formulas

F_x = F \cos \theta
F_y = F \sin \theta
F = \sqrt{(F_x)^2 + (F_y)^2}
tan \theta = \frac{F_y}{F_x}

Theorems

Conditions of Equilibrium

Suitable Grade Level

Grades 10-12