Suppose a body has a force of 20 pounds acting on it to the right, 5 pounds acting on it – 140° from the horizontal, and 20 pounds acting on it directed 130° from the horizontal. What single force is the resultant force acting on the body? Round answers to 2 decimal places.
Suppose a body has a force of 20 pounds acting on it to the right, 5 pounds acting on it – 140° from the horizontal, and 20 pounds acting on it directed 130° from the horizontal. What single force is the resultant force acting on the body? Round answers to 2 decimal places.
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter2: Right Triangle Trigonometry
Section2.5: Vectors: A Geometric Approach
Problem 59PS
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![### Problem Statement
Suppose a body has a force of 20 pounds acting on it to the right, 5 pounds acting on it at -140° from the horizontal, and 20 pounds acting on it directed 130° from the horizontal. What single force is the resultant force acting on the body? Round answers to 2 decimal places.
**Answer:**
\[ \quad \] pounds \[ \quad \]° from the horizontal.
**Get Help:**
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### Explanation
In this problem, we are dealing with forces acting on a body in different directions. The goal is to find the resultant force, both in magnitude and direction.
#### Step-by-step Solution:
1. **Resolve each force into its horizontal (x) and vertical (y) components**:
- For the force of 20 pounds to the right: \( F_1 = 20 \) pounds at \( 0° \)
\[
F_{1x} = 20 \cos(0°) = 20 \quad , \quad F_{1y} = 20 \sin(0°) = 0
\]
- For the force of 5 pounds at -140°:
\[
F_{2x} = 5 \cos(-140°) \quad , \quad F_{2y} = 5 \sin(-140°)
\]
- For the force of 20 pounds at 130°:
\[
F_{3x} = 20 \cos(130°) \quad , \quad F_{3y} = 20 \sin(130°)
\]
2. **Calculate the x and y components for the remaining forces using trigonometric functions**:
- Using \( \cos(-140°) = \cos(140°) \) and \( \sin(-140°) = -\sin(140°) \):
\[
F_{2x} = 5 \cos(140°), \quad F_{2y} = -5 \sin(140°)
\]
\[
F_{2x} = -3.83, \quad F_{2y} = -3.21
\]
- For \( F_3 \):
\[
F_{3x} =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F416064f9-c57b-46ee-9b50-14b257515654%2Fa33e8cd9-248e-48e0-88d6-893bf4ac39e4%2F6g2t36n.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Suppose a body has a force of 20 pounds acting on it to the right, 5 pounds acting on it at -140° from the horizontal, and 20 pounds acting on it directed 130° from the horizontal. What single force is the resultant force acting on the body? Round answers to 2 decimal places.
**Answer:**
\[ \quad \] pounds \[ \quad \]° from the horizontal.
**Get Help:**
[VIDEO Button]
### Explanation
In this problem, we are dealing with forces acting on a body in different directions. The goal is to find the resultant force, both in magnitude and direction.
#### Step-by-step Solution:
1. **Resolve each force into its horizontal (x) and vertical (y) components**:
- For the force of 20 pounds to the right: \( F_1 = 20 \) pounds at \( 0° \)
\[
F_{1x} = 20 \cos(0°) = 20 \quad , \quad F_{1y} = 20 \sin(0°) = 0
\]
- For the force of 5 pounds at -140°:
\[
F_{2x} = 5 \cos(-140°) \quad , \quad F_{2y} = 5 \sin(-140°)
\]
- For the force of 20 pounds at 130°:
\[
F_{3x} = 20 \cos(130°) \quad , \quad F_{3y} = 20 \sin(130°)
\]
2. **Calculate the x and y components for the remaining forces using trigonometric functions**:
- Using \( \cos(-140°) = \cos(140°) \) and \( \sin(-140°) = -\sin(140°) \):
\[
F_{2x} = 5 \cos(140°), \quad F_{2y} = -5 \sin(140°)
\]
\[
F_{2x} = -3.83, \quad F_{2y} = -3.21
\]
- For \( F_3 \):
\[
F_{3x} =
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