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Ch. 11 - Angular Momentum; General Rotation
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 11 - Angular Momentum; General RotationProblema 30
Capitolo 11, Problema 30

An engineer estimates that under the most adverse expected weather conditions, the total force on the highway sign in Fig. 11–33 will be F→\(\overrightarrow{F}\) = (± 2.4 î - 4.1 ĵ) kN, acting at the cm. What torque does this force exert about the base O?
Diagram of a highway sign labeled "BOSTON" with force vector acting on it, showing dimensions and coordinate axes.

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1
Identify the given quantities: The force vector is \( \mathbf{F} = (\pm 2.4 \mathbf{i} - 4.1 \mathbf{j}) \ \text{kN} \), and the position vector from the base \( O \) to the center of mass (cm) of the sign is \( \mathbf{r} \). The exact position vector \( \mathbf{r} \) should be determined from the diagram in Fig. 11–33.
Recall the formula for torque: \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \), where \( \mathbf{\tau} \) is the torque, \( \mathbf{r} \) is the position vector, and \( \mathbf{F} \) is the force vector. The cross product \( \mathbf{r} \times \mathbf{F} \) gives the torque vector.
Express the position vector \( \mathbf{r} \) in component form based on the diagram. For example, if the center of mass is located at \( (x, y) \) relative to the base \( O \), then \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} \).
Set up the cross product \( \mathbf{\tau} = \mathbf{r} \times \mathbf{F} \). Using the determinant method for cross products, \( \mathbf{\tau} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x & y & 0 \\ \pm 2.4 & -4.1 & 0 \end{vmatrix} \). Expand this determinant to find the components of the torque vector.
Simplify the determinant to calculate the torque vector \( \mathbf{\tau} \). The result will be in the form \( \mathbf{\tau} = \tau_z \mathbf{k} \), where \( \tau_z \) is the magnitude of the torque about the \( z \)-axis. Ensure the units are consistent (e.g., \( \text{kN} \cdot \text{m} \)).

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Torque

Torque is a measure of the rotational force applied to an object around a pivot point. It is calculated as the cross product of the position vector (from the pivot to the point of force application) and the force vector. The direction of torque is determined by the right-hand rule, and its unit is typically Newton-meters (Nm). Understanding torque is essential for analyzing how forces cause objects to rotate.
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Vector Components

Vector components break down a vector into its individual parts along specified axes, typically the x and y axes in two-dimensional problems. For example, a force vector can be expressed in terms of its horizontal (i) and vertical (j) components. This decomposition simplifies calculations, especially when determining resultant forces or torques, as it allows for the independent analysis of each component.
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Cross Product

The cross product is a mathematical operation that takes two vectors and produces a third vector that is perpendicular to the plane formed by the original vectors. In the context of torque, the cross product of the position vector and the force vector yields the torque vector. The magnitude of the torque is given by the product of the magnitudes of the two vectors and the sine of the angle between them, which is crucial for understanding rotational effects.
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Vector (Cross) Product and the Right-Hand-Rule
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