Force F⃗=(-8.0 N) î+(6.0 N) j acts on a particle with position vector r⃗=(3.0 m) i+(4.0 m) ĵ

step 1 For this problem on the topic of angular momentum, we are told that a force of minus 8 Newton's

Force F⃗=(-8.0 N) î+(6.0 N) j acts on a particle with...

Force $\vec{F}=(-8.0 \mathrm{~N}) \hat{\mathrm{i}}+(6.0 \mathrm{~N}) \mathrm{j}$ acts on a particle with position vector $\vec{r}=(3.0 \mathrm{~m}) \mathrm{i}+(4.0 \mathrm{~m}) \hat{j}$ What are (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of $\vec{r}$ and $\vec{F} ?$

Force $\vec{F}=(-8.0 \mathrm{~N}) \hat{\mathrm{i}}+(6.0 \mathrm{~N}) \mathrm{j}$ acts on a particle with position vector $\vec{r}=(3.0 \mathrm{~m}) \mathrm{i}+(4.0 \mathrm{~m}) \hat{j}$ What are
(a) the torque on the particle about the origin, in unit-vector notation, and
(b) the angle between the directions of $\vec{r}$ and $\vec{F} ?$

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Key Concepts

Vector Cross Product

The vector cross product is an operation on two vectors in three-dimensional space that produces a third vector perpendicular to the plane formed by the original vectors. This operation is essential for calculating torque because it naturally includes the sine of the angle between the vectors, accounting for how effectively the force causes rotation.

Torque

Torque is a measure of the rotational effect a force exerts about a pivot point or axis. It is calculated as the cross product of the position vector (which defines the point of application relative to the axis) and the force vector, incorporating both the magnitude and the direction of the applied force.

Angle Between Vectors

Determining the angle between vectors involves using the dot product, which relates the product of the magnitudes of two vectors to the cosine of the angle between them. This relationship is useful for finding the orientation difference between two vectors, such as between the force and position vectors in rotational problems.

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