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Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle if the resultant force is directed vertically upward. 500 N 70000 y 30° 7000 -X

Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle if the resultant force is directed vertically upward. 500 N 70000 y 30° 7000 -X

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### Problem Statement on Equilibrium of Forces **Topic:** Vector Addition and Resultant Force Calculation #### Background: Two forces act on the screw eye. The objective is to determine the magnitude of the resultant force and the angle if the resultant force is directed vertically upward. #### Given Data: - Force \( F \) is 600 N. - There is an angle of 30° between force \( F \) and the vertical axis \( y \). - Another force of 500 N is applied at an angle \( \theta \) to the horizontal axis \( x \). #### Diagram Explained: The image depicts a screw eye anchored to a surface with two ropes extending from the screw eye, each exerting a force. - The first force is labeled as 500 N and is applied at an unknown angle \( \theta \) to the horizontal \( x \)-axis, directed upward and to the left. - The second force \( F \) is 600 N and is applied at a 30° angle from the vertical \( y \)-axis, directed upward and to the right. - The coordinate system is marked with \( x \)-axis (horizontal) and \( y \)-axis (vertical). #### Questions to Consider: 1. **What is the magnitude of the resultant force?** 2. **What is the angle \( \theta \) at which the 500 N force is applied to ensure that the resultant force is directed vertically upward?** #### Detailed Explanation: To solve for these unknowns, we can use the principles of vector addition and trigonometry. This involves breaking down each force into its horizontal and vertical components, then setting up equations to ensure that the resultant force is directed vertically upwards. #### Steps: 1. **Resolve each force into its components**: - For the 600 N force: - \( F_{y} = 600 \cos 30^\circ \) - \( F_{x} = 600 \sin 30^\circ \) - For the 500 N force: - \( F_{x'} = 500 \cos \theta \) - \( F_{y'} = 500 \sin \theta \) 2. **Equate the horizontal component to zero (since the resultant force is vertically upward)**: - \( F_{x} + F_{x'} = 0 \) This yields: \[ 600 \sin 30
Transcribed Image Text:### Problem Statement on Equilibrium of Forces **Topic:** Vector Addition and Resultant Force Calculation #### Background: Two forces act on the screw eye. The objective is to determine the magnitude of the resultant force and the angle if the resultant force is directed vertically upward. #### Given Data: - Force \( F \) is 600 N. - There is an angle of 30° between force \( F \) and the vertical axis \( y \). - Another force of 500 N is applied at an angle \( \theta \) to the horizontal axis \( x \). #### Diagram Explained: The image depicts a screw eye anchored to a surface with two ropes extending from the screw eye, each exerting a force. - The first force is labeled as 500 N and is applied at an unknown angle \( \theta \) to the horizontal \( x \)-axis, directed upward and to the left. - The second force \( F \) is 600 N and is applied at a 30° angle from the vertical \( y \)-axis, directed upward and to the right. - The coordinate system is marked with \( x \)-axis (horizontal) and \( y \)-axis (vertical). #### Questions to Consider: 1. **What is the magnitude of the resultant force?** 2. **What is the angle \( \theta \) at which the 500 N force is applied to ensure that the resultant force is directed vertically upward?** #### Detailed Explanation: To solve for these unknowns, we can use the principles of vector addition and trigonometry. This involves breaking down each force into its horizontal and vertical components, then setting up equations to ensure that the resultant force is directed vertically upwards. #### Steps: 1. **Resolve each force into its components**: - For the 600 N force: - \( F_{y} = 600 \cos 30^\circ \) - \( F_{x} = 600 \sin 30^\circ \) - For the 500 N force: - \( F_{x'} = 500 \cos \theta \) - \( F_{y'} = 500 \sin \theta \) 2. **Equate the horizontal component to zero (since the resultant force is vertically upward)**: - \( F_{x} + F_{x'} = 0 \) This yields: \[ 600 \sin 30
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