A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector 7 (2.00 m) î - (3.00 m)ĵ + (2.00 m) k, the force is F = Fxî + (7.00 N)ĵ - (6.00 N) Ê, and the corresponding torque about the origin is 7 = (4.00 N-m) î + (8.00 N-m) ĵ + (8.00 N-m) Ê. Determine F,x.
A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector 7 (2.00 m) î - (3.00 m)ĵ + (2.00 m) k, the force is F = Fxî + (7.00 N)ĵ - (6.00 N) Ê, and the corresponding torque about the origin is 7 = (4.00 N-m) î + (8.00 N-m) ĵ + (8.00 N-m) Ê. Determine F,x.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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: Torque and Force in an XYZ Coordinate System**
A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector:
\[
\vec{r} = (2.00 \, \text{m}) \hat{i} - (3.00 \, \text{m}) \hat{j} + (2.00 \, \text{m}) \hat{k}
\]
the force acting on the particle is:
\[
\vec{F} = F_x \hat{i} + (7.00 \, \text{N}) \hat{j} - (6.00 \, \text{N}) \hat{k}
\]
The corresponding torque about the origin is:
\[
\vec{\tau} = (4.00 \, \text{N} \cdot \text{m}) \hat{i} + (8.00 \, \text{N} \cdot \text{m}) \hat{j} + (8.00 \, \text{N} \cdot \text{m}) \hat{k}
\]
**Task:**
Determine the component \( F_x \) of the force vector.
**Input Fields:**
- **Number:** [Input box for the value of \( F_x \)]
- **Units:** [Dropdown or input box for units of \( F_x \)]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb493ecdd-cbdc-400d-a05a-de2010eb2d52%2F05cef266-e36b-4cc2-b5c6-15793ea413e9%2F77dzjo_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Torque and Force in an XYZ Coordinate System**
A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector:
\[
\vec{r} = (2.00 \, \text{m}) \hat{i} - (3.00 \, \text{m}) \hat{j} + (2.00 \, \text{m}) \hat{k}
\]
the force acting on the particle is:
\[
\vec{F} = F_x \hat{i} + (7.00 \, \text{N}) \hat{j} - (6.00 \, \text{N}) \hat{k}
\]
The corresponding torque about the origin is:
\[
\vec{\tau} = (4.00 \, \text{N} \cdot \text{m}) \hat{i} + (8.00 \, \text{N} \cdot \text{m}) \hat{j} + (8.00 \, \text{N} \cdot \text{m}) \hat{k}
\]
**Task:**
Determine the component \( F_x \) of the force vector.
**Input Fields:**
- **Number:** [Input box for the value of \( F_x \)]
- **Units:** [Dropdown or input box for units of \( F_x \)]
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