Math Problem Statement

The force F acts on the bracket within the octant shown. If the magnitudes of the x and z components of F are Fx = 300 N and Fz = 600 N, respectively, and β = 60°, determine the magnitude of F and its y-component. Also, find the coordinate direction angles α and γ.

Solution

The problem involves finding the magnitude and the coordinate direction angles of a force vector F\mathbf{F} acting on a bracket. The given components of the force are Fx=300NF_x = 300 \, \text{N}, Fz=600NF_z = 600 \, \text{N}, and the angle β=60\beta = 60^\circ between FF and the xx-axis. We also need to find the direction angles α\alpha and γ\gamma.

Step 1: Find the Magnitude of the Force FF

Since β\beta is the angle with the xx-axis, the magnitude of FF is related to FxF_x by the cosine of β\beta: F=FxcosβF = \frac{F_x}{\cos \beta} F=300cos60=3000.5=600NF = \frac{300}{\cos 60^\circ} = \frac{300}{0.5} = 600 \, \text{N}

Step 2: Find FyF_y

Using the Pythagorean theorem on the components of F\mathbf{F}, we have: F=Fx2+Fy2+Fz2F = \sqrt{F_x^2 + F_y^2 + F_z^2} Plugging in the known values and solving for FyF_y: 600=3002+Fy2+6002600 = \sqrt{300^2 + F_y^2 + 600^2} 360000=90000+Fy2+360000360000 = 90000 + F_y^2 + 360000 Fy2=90000F_y^2 = 90000 Fy=300NF_y = 300 \, \text{N}

Step 3: Find the Coordinate Direction Angles α\alpha, β\beta, and γ\gamma

The coordinate direction angles can be found using: cosα=FxF\cos \alpha = \frac{F_x}{F} cosβ=FyF\cos \beta = \frac{F_y}{F} cosγ=FzF\cos \gamma = \frac{F_z}{F}

α=cos1(300600)=cos1(0.5)=60\alpha = \cos^{-1} \left( \frac{300}{600} \right) = \cos^{-1}(0.5) = 60^\circ β=cos1(300600)=cos1(0.5)=60\beta = \cos^{-1} \left( \frac{300}{600} \right) = \cos^{-1}(0.5) = 60^\circ γ=cos1(600600)=cos1(1)=0\gamma = \cos^{-1} \left( \frac{600}{600} \right) = \cos^{-1}(1) = 0^\circ

Hence, the magnitude of F\mathbf{F} is 600N600 \, \text{N}, and the coordinate direction angles are α=60\alpha = 60^\circ, β=60\beta = 60^\circ, and γ=0\gamma = 0^\circ.

Do you need further details or have any questions about these calculations?

Here are some related questions:

  1. What would the direction angles be if FxF_x or FzF_z were different?
  2. How would a negative component affect the direction angles?
  3. What is the significance of each direction angle in vector components?
  4. Can you explain how to resolve a vector into its components in three-dimensional space?
  5. How does changing the angle β\beta affect the magnitude and components of F\mathbf{F}?

Tip: To better understand vector operations, visualize them using vector component diagrams. This can help with comprehending the geometric relationships between the angles and the vector components.

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Math Problem Analysis

Mathematical Concepts

Vector Components
Trigonometry
Physics
Pythagorean Theorem

Formulas

F = sqrt(Fx^2 + Fy^2 + Fz^2)
cos(α) = Fx/F
cos(β) = Fy/F
cos(γ) = Fz/F

Theorems

Pythagorean Theorem
Vector Decomposition

Suitable Grade Level

University Level - Engineering or Physics