91影视

The cross product is fundamental when working with torque, because it allows us to determine the direction in which the force is acting around a pivot point. The cross product of two vectors, say \( \vec{A} \) and \( \vec{B} \), denoted as \( \vec{A} \times \vec{B} \), results in a new vector that is perpendicular to the plane formed by the original two vectors.

• The magnitude of this cross product vector is calculated as: \( |\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin\theta \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).

In our exercise, we cross the position vector \( \vec{r} \) and the force vector \( \vec{F} \) to find the torque, \( \tau \). Using the distributive property of the cross product and unit vector rules like \( \hat{i} \times \hat{j} = \hat{k} \) and \( \hat{j} \times \hat{i} = -\hat{k} \), allows us to compute the vector pointing along the \( z \)-axis, out of the plane. This result conveys the direction of the applied force's rotation.

A position vector is a powerful tool for locating a specific point in space relative to an origin, which is crucial when calculating torque. For a point in the 2-D plane, the position vector \( \vec{r} \) is expressed using components in the direction of the coordinate axes, represented as \( \vec{r} = x \hat{i} + y \hat{j} \).

• In this exercise, the position vector is aimed from the origin (where the pivot point is) to the point of force application at \( x = 3.00 \) m and \( y = 4.00 \) m. Therefore, \( \vec{r} \) is \( 3.00 \hat{i} + 4.00 \hat{j} \) m.

Understanding the position vector allows us to assess both the distance and angle of application of force, which are critical for evaluating torque. It's essentially the 'lever arm' component in the torque equation.

The magnitude and direction of torque tell us how strong and in which rotational direction the turning effect of a force is applied. Torque, denoted as \( \tau \), is determined by the product of three factors: the force's magnitude, the distance the force is applied from the pivot, and the sine of the angle between them. Mathematically, this is described by \( |\tau| = |\vec{r}||\vec{F}|\sin(\theta) \).

• In the problem, calculating the cross product yields \( \tau = 25.00 \hat{k} \) N.m, representing the torque's direction as out of the plane, in the positive \( \hat{k} \)-direction.

The magnitude of 25.00 N.m specifies the rotational strength. A positive \( \hat{k} \) direction implies that when looked upon from above the plane, the rotation is counterclockwise. This direction aspect of torque helps understand whether the motion will result in a push or pull effect relative to the pivot.