A particle is located at the position vector r = (6.00î + 7.0oj) m and a force exerted on it is given by F = (4.00î + 3.00j) N. (a) What is the torque acting on the particle about the origin? N.m

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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:**

A particle is located at the position vector \( \vec{r} = (6.00 \hat{\imath} + 7.00 \hat{\jmath}) \) m and a force exerted on it is given by \( \vec{F} = (4.00 \hat{\imath} + 3.00 \hat{\jmath}) \) N.

(a) What is the torque acting on the particle about the origin?

\[ \tau = \text{______} \, \text{N} \cdot \text{m} \]

**Explanation:**

Here, we need to calculate the torque (\( \tau \)) acting on the particle about the origin. Torque is given by the cross-product of the position vector (\( \vec{r} \)) and the force vector (\( \vec{F} \)).

To find the torque \( \tau \):

\[ \tau = \vec{r} \times \vec{F} \] 

\[ \vec{r} = (6.00 \hat{\imath} + 7.00 \hat{\jmath}) \, \text{m} \]

\[ \vec{F} = (4.00 \hat{\imath} + 3.00 \hat{\jmath}) \, \text{N} \]

The torque \( \vec{\tau} \) is computed as:

\[ \vec{\tau} = (6.00 \hat{\imath} + 7.00 \hat{\jmath}) \times (4.00 \hat{\imath} + 3.00 \hat{\jmath}) \]

Solve the cross product to find the magnitude and direction.

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This transcribed text can be used in educational resources to provide students practice in vector cross product and understanding torque.
Transcribed Image Text:**Problem Statement:** A particle is located at the position vector \( \vec{r} = (6.00 \hat{\imath} + 7.00 \hat{\jmath}) \) m and a force exerted on it is given by \( \vec{F} = (4.00 \hat{\imath} + 3.00 \hat{\jmath}) \) N. (a) What is the torque acting on the particle about the origin? \[ \tau = \text{______} \, \text{N} \cdot \text{m} \] **Explanation:** Here, we need to calculate the torque (\( \tau \)) acting on the particle about the origin. Torque is given by the cross-product of the position vector (\( \vec{r} \)) and the force vector (\( \vec{F} \)). To find the torque \( \tau \): \[ \tau = \vec{r} \times \vec{F} \] \[ \vec{r} = (6.00 \hat{\imath} + 7.00 \hat{\jmath}) \, \text{m} \] \[ \vec{F} = (4.00 \hat{\imath} + 3.00 \hat{\jmath}) \, \text{N} \] The torque \( \vec{\tau} \) is computed as: \[ \vec{\tau} = (6.00 \hat{\imath} + 7.00 \hat{\jmath}) \times (4.00 \hat{\imath} + 3.00 \hat{\jmath}) \] Solve the cross product to find the magnitude and direction. --- This transcribed text can be used in educational resources to provide students practice in vector cross product and understanding torque.
**Instruction (f): Determine the Position Vector**

For this task, you are required to determine the position vector of a point located on the y-axis. 

**Vector Representation:**

- The vector is represented as **\(\vec{r} = \_\_\_\_\_\_ \, \text{m}\)**. 
- You need to specify the vector's value within the provided box to denote the position on the y-axis.

The position vector typically has the format \(\vec{r} = (x, y, z)\). Since the point is on the y-axis, x and z components would be zero, resulting in a vector of the form \((0, y, 0)\). Specify a value for \(y\) based on the context to complete the answer.
Transcribed Image Text:**Instruction (f): Determine the Position Vector** For this task, you are required to determine the position vector of a point located on the y-axis. **Vector Representation:** - The vector is represented as **\(\vec{r} = \_\_\_\_\_\_ \, \text{m}\)**. - You need to specify the vector's value within the provided box to denote the position on the y-axis. The position vector typically has the format \(\vec{r} = (x, y, z)\). Since the point is on the y-axis, x and z components would be zero, resulting in a vector of the form \((0, y, 0)\). Specify a value for \(y\) based on the context to complete the answer.
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