# What are the rectangular components of force ( F ) in the given diagram?

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**Created:** [Temmuz 24, 2024, 5:44ös UTC](https://sorumatik.co/t/what-are-the-rectangular-components-of-force-f-in-the-given-diagram/83842 "2024-07-24T17:44:09Z")\
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## **What are the rectangular components of force F in the given diagram?**

**Answer:**  
To determine the rectangular components of the force F in the given diagram, we need to resolve the force vector into its x, y, and z components. The force F is given as 75 N, and it’s directed at angles, with respect to the axes, provided in the diagram.

1. **Identify Given Angles:**

2. **Component in the z-direction F\_z:**

F\_z = F \cos(30^\circ)

Since \cos(30^\circ) = \frac{\sqrt{3}}{2} ,

F\_z = 75 \times \frac{\sqrt{3}}{2} = 75 \times 0.866 = 64.95 \, \text{N}

1. **Magnitude of the force in the x-y plane F\_{xy}:**  
Since the force makes a 30^\circ angle with the z-axis, the projection in the x-y plane is:

F\_{xy} = F \sin(30^\circ)

Since \sin(30^\circ) = \frac{1}{2} ,

F\_{xy} = 75 \times \frac{1}{2} = 37.5 \, \text{N}

1. **Component in the x-direction F\_x:**  
Given the angle in the x-y plane with respect to the x-axis is \theta\_{xy} = 45^\circ :

F\_x = F\_{xy} \cos(45^\circ)

Since \cos(45^\circ) = \frac{\sqrt{2}}{2},

F\_x = 37.5 \times \frac{\sqrt{2}}{2} = 37.5 \times 0.707 = 26.52 \, \text{N}

1. **Component in the y-direction F\_y:**

F\_y = F\_{xy} \sin(45^\circ)

Since \sin(45^\circ) = \frac{\sqrt{2}}{2} ,

F\_y = 37.5 \times \frac{\sqrt{2}}{2} = 37.5 \times 0.707 = 26.52 \, \text{N}

Therefore, the rectangular components of the force F are:

F\_x = 26.52 \, \text{N}

F\_y = 26.52 \, \text{N}

F\_z = 64.95 \, \text{N}