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The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors. In physics, the cross product is instrumental when computing torque. Torque is the rotational equivalent of linear force. The cross product between two vectors, such as the position vector \( \vec{r} \) and the force vector \( \vec{F} \), results in another vector, that represents the torque \( \vec{\tau} \).
The formula for the cross product of \( \vec{r} = (-0.450 \ \mathrm{m}) \hat{\imath} + (0.150 \ \mathrm{m}) \hat{\boldsymbol{j}} \) and \( \vec{F} = (-5.00 \ \mathrm{N}) \hat{\imath} + (4.00 \ \mathrm{N}) \hat{\boldsymbol{j}} \), is calculated as:

• \( \vec{\tau} = \vec{r} \times \vec{F} = (0.150\cdot4.00) \hat{k} - (-0.450\cdot(-5.00)) \hat{k} \).
• The resulting torque is \( \vec{\tau} = -2.25 \hat{k} \), indicating its direction is along the z-axis.

Cross products are a fundamental tool particularly in situations involving rotational dynamics.

The right-hand rule is a useful mnemonic for determining the direction of vectors resulting from cross products. In the context of torque, it helps us find out where the torque is pointing relative to our coordinate system. Here's how it works:

• Start by pointing your fingers in the direction of the first vector, \( \vec{r} \) in this case.
• Then, curl your fingers in the direction of the second vector, \( \vec{F} \).
• Your thumb will then point in the direction of the resulting vector \( \vec{\tau} \).

In the given exercise, applying the right-hand rule should initially point your thumb out of the page, indicating a positive z-direction. However, the torque calculation pointed in the opposite direction, due to a sign error in interpretation.

Vector components break a vector into its fundamental parts, each aligned with the coordinate axes. This makes analyzing vectors simpler. Each vector in a two-dimensional space can be represented as a combination of the horizontal (x-axis) and vertical (y-axis) components.

• The force vector in the exercise is \( \vec{F} = (-5.00 \ \mathrm{N}) \hat{\imath} + (4.00 \ \mathrm{N}) \hat{\boldsymbol{j}} \).
• The position vector is \( \vec{r} = (-0.450 \ \mathrm{m}) \hat{\imath} + (0.150 \ \mathrm{m}) \hat{\boldsymbol{j}} \).

This breakdown into components is crucial for performing calculations such as the cross product, which are computed using these individual vector parts.

A coordinate system allows us to describe and analyze vectors geometrically. It provides a framework where the position, direction, and forces can be consistently interpreted. In most physics problems, including this one, we use an orthogonal coordinate system, typically the Cartesian coordinate system, which is defined by x, y, and z axes.

• The origin is the point where all the axes meet, where \( x = 0, y = 0, \) and \( z = 0 \).
• The x-axis runs horizontally, the y-axis vertically, and the z-axis extends perpendicular to both, usually illustrated as coming out of or going into the page.

In this exercise, the vectors \( \vec{r} \) and \( \vec{F} \) are plotted relative to this coordinate system, providing clarity on their direction and magnitude. Understanding this structure is essential for correctly applying concepts like the cross product and right-hand rule.