Find the Resultant (Magnitude and Direction) of the forces below. F4 = 20 N 3 7 F3 = 60 N 14 3 20⁰ F₁ = 50 N 10⁰ F₂ = 70 N
Find the Resultant (Magnitude and Direction) of the forces below. F4 = 20 N 3 7 F3 = 60 N 14 3 20⁰ F₁ = 50 N 10⁰ F₂ = 70 N
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![### Resultant Forces in a Plane
**Objective:** Find the Resultant (Magnitude and Direction) of the forces shown in the diagram below.
#### Forces Description:
1. **F₁ = 50 N**
- Direction: 20° above the positive x-axis.
2. **F₂ = 70 N**
- Direction: 10° below the positive x-axis.
3. **F₃ = 60 N**
- Direction: In the third quadrant making a 7:3 ratio triangle indicating an angle to the negative x-axis (or equivalently the negative y-axis at its corresponding angle).
4. **F₄ = 20 N**
- Direction: In the second quadrant, making a 3:4 ratio triangle indicating an angle to the negative x-axis.
#### Diagram:
- The diagram shows a standard Cartesian plane with the four forces represented as vectors originating from the same origin.
- F₁ and F₂ are in the first quadrant but F₁ is directed upwards and F₂ downwards.
- F₃ is directed downwards and leftwards in the third quadrant.
- F₄ is directed leftwards and slightly upwards in the second quadrant.
#### Analysis:
To determine the resultant force (both in magnitude and direction), follow these steps:
1. **Resolve each force into its horizontal (x) and vertical (y) components.**
The horizontal (x) and vertical (y) component formulas for a force \( F \) at an angle \( \theta \) are:
\[ F_x = F \cos(\theta) \]
\[ F_y = F \sin(\theta) \]
2. **Calculate Components:**
- For \( F_1 = 50 \) N at \( 20^\circ \):
\[ F_{1x} = 50 \cos(20^\circ) \]
\[ F_{1y} = 50 \sin(20^\circ) \]
- For \( F_2 = 70 \) N at \( -10^\circ \):
\[ F_{2x} = 70 \cos(-10^\circ) \]
\[ F_{2y} = 70 \sin(-10^\circ) \]
- For \( F_3 = 60 \) N at an angle derived from the 7:3 ratio triangle:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24227ad7-0cee-4cb3-bf8f-36b2c099999b%2F7f3f6891-981e-4d49-945c-4fab9a7a0a27%2Foh9uira_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Resultant Forces in a Plane
**Objective:** Find the Resultant (Magnitude and Direction) of the forces shown in the diagram below.
#### Forces Description:
1. **F₁ = 50 N**
- Direction: 20° above the positive x-axis.
2. **F₂ = 70 N**
- Direction: 10° below the positive x-axis.
3. **F₃ = 60 N**
- Direction: In the third quadrant making a 7:3 ratio triangle indicating an angle to the negative x-axis (or equivalently the negative y-axis at its corresponding angle).
4. **F₄ = 20 N**
- Direction: In the second quadrant, making a 3:4 ratio triangle indicating an angle to the negative x-axis.
#### Diagram:
- The diagram shows a standard Cartesian plane with the four forces represented as vectors originating from the same origin.
- F₁ and F₂ are in the first quadrant but F₁ is directed upwards and F₂ downwards.
- F₃ is directed downwards and leftwards in the third quadrant.
- F₄ is directed leftwards and slightly upwards in the second quadrant.
#### Analysis:
To determine the resultant force (both in magnitude and direction), follow these steps:
1. **Resolve each force into its horizontal (x) and vertical (y) components.**
The horizontal (x) and vertical (y) component formulas for a force \( F \) at an angle \( \theta \) are:
\[ F_x = F \cos(\theta) \]
\[ F_y = F \sin(\theta) \]
2. **Calculate Components:**
- For \( F_1 = 50 \) N at \( 20^\circ \):
\[ F_{1x} = 50 \cos(20^\circ) \]
\[ F_{1y} = 50 \sin(20^\circ) \]
- For \( F_2 = 70 \) N at \( -10^\circ \):
\[ F_{2x} = 70 \cos(-10^\circ) \]
\[ F_{2y} = 70 \sin(-10^\circ) \]
- For \( F_3 = 60 \) N at an angle derived from the 7:3 ratio triangle:
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