39. ssm The x vector component of a displacement vector F has a mag- nitude of 125 m and points along the negative x axis. The y vector compo- nent has a magnitude of 184 m and points along the negative y axis. Find the magnitude and direction of F. Specify the direction with respect to the negative x axis.
39. ssm The x vector component of a displacement vector F has a mag- nitude of 125 m and points along the negative x axis. The y vector compo- nent has a magnitude of 184 m and points along the negative y axis. Find the magnitude and direction of F. Specify the direction with respect to the negative x axis.
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem Statement
**Question 39:**
The x vector component of a displacement vector **r** has a magnitude of 125 m and points along the negative x-axis. The y vector component has a magnitude of 184 m and points along the negative y-axis. Find the magnitude and direction of **r**. Specify the direction with respect to the negative x-axis.
### Explanation
#### Given:
- x-component (r_x) = -125 m (negative x-axis)
- y-component (r_y) = -184 m (negative y-axis)
#### To Find:
1. The magnitude of vector **r**.
2. The direction of vector **r** with respect to the negative x-axis.
#### Solution:
a) **Magnitude of **r****:
Use the Pythagorean theorem to find the resultant vector **r**.
\[ | \mathbf{r} | = \sqrt{(r_x^2 + r_y^2)} \]
\[ | \mathbf{r} | = \sqrt{(-125\, \text{m})^2 + (-184\, \text{m})^2} \]
\[ | \mathbf{r} | = \sqrt{15625\, \text{m}^2 + 33856\, \text{m}^2} \]
\[ | \mathbf{r} | = \sqrt{49481\, \text{m}^2} \]
\[ | \mathbf{r} | \approx 222.46\, \text{m} \]
b) **Direction of **r****:
Calculate the direction (θ) with respect to the negative x-axis. The angle θ can be found using the tangent function:
\[ \tan \theta = \frac{r_y}{r_x} \]
\[ \tan \theta = \frac{-184\, \text{m}}{-125\, \text{m}} \]
\[ \theta = \tan^{-1}\left( \frac{184}{125} \right) \]
\[ \theta \approx 55.99^\circ \]
Since both components are along the negative axes, with respect to the negative x-axis, we add 180°:
\[ \text{Angle with respect to negative x-axis} = 180^\circ - 55.99^\circ \approx 124.01^\circ \]
### Final Answer:
- Magnitude of **r**:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2c8c9d8-cb7d-4bb3-b241-ba372e4ba26e%2F7012d48f-ad6b-454f-a4cd-ac020d0c6d9a%2Fmq1tkau_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Question 39:**
The x vector component of a displacement vector **r** has a magnitude of 125 m and points along the negative x-axis. The y vector component has a magnitude of 184 m and points along the negative y-axis. Find the magnitude and direction of **r**. Specify the direction with respect to the negative x-axis.
### Explanation
#### Given:
- x-component (r_x) = -125 m (negative x-axis)
- y-component (r_y) = -184 m (negative y-axis)
#### To Find:
1. The magnitude of vector **r**.
2. The direction of vector **r** with respect to the negative x-axis.
#### Solution:
a) **Magnitude of **r****:
Use the Pythagorean theorem to find the resultant vector **r**.
\[ | \mathbf{r} | = \sqrt{(r_x^2 + r_y^2)} \]
\[ | \mathbf{r} | = \sqrt{(-125\, \text{m})^2 + (-184\, \text{m})^2} \]
\[ | \mathbf{r} | = \sqrt{15625\, \text{m}^2 + 33856\, \text{m}^2} \]
\[ | \mathbf{r} | = \sqrt{49481\, \text{m}^2} \]
\[ | \mathbf{r} | \approx 222.46\, \text{m} \]
b) **Direction of **r****:
Calculate the direction (θ) with respect to the negative x-axis. The angle θ can be found using the tangent function:
\[ \tan \theta = \frac{r_y}{r_x} \]
\[ \tan \theta = \frac{-184\, \text{m}}{-125\, \text{m}} \]
\[ \theta = \tan^{-1}\left( \frac{184}{125} \right) \]
\[ \theta \approx 55.99^\circ \]
Since both components are along the negative axes, with respect to the negative x-axis, we add 180°:
\[ \text{Angle with respect to negative x-axis} = 180^\circ - 55.99^\circ \approx 124.01^\circ \]
### Final Answer:
- Magnitude of **r**:
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