Problem 6 - University Physics with Modern Physics

Question

A metal bar is in the \(x y-\) plane with one end of the bar at the origin. A force \(\vec{F}=(7.00 \mathrm{N}) \hat{\imath}+(-3.00 \mathrm{N}) \hat{J}\) is applied to the bar at the point \(x=3.00 \mathrm{m}, y=4.00 \mathrm{m} .\) (a) In terms of unit vectors \(\hat{\boldsymbol{i}}\) and \(\hat{\boldsymbol{J}},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\vec{\boldsymbol{F}}\) ?

Step-by-Step Solution

Answer

Position vector: \(3.00 \hat{\imath} + 4.00 \hat{\mathbf{j}}\), Torque magnitude: 37.00 N•m, direction: clockwise.

Step 1: Determine the Position Vector

The position vector \(\vec{r}\) for the point where the force is applied is determined by the coordinates \(x = 3.00 \text{ m}\) and \(y = 4.00 \text{ m}\). Thus, \(\vec{r} = 3.00 \hat{\imath} + 4.00 \hat{\mathbf{j}}\).

Step 2: Calculate the Cross Product

The torque \(\vec{\tau}\) is given by the cross product of \(\vec{r}\) and \(\vec{F}\). \[ \vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{\imath} & \hat{\mathbf{j}} & \hat{k} \ 3.00 & 4.00 & 0 \ 7.00 & -3.00 & 0 \end{vmatrix} \]Computing the determinant, \[ \vec{\tau} = (0 - 0)\hat{\imath} - (0 - 0)\hat{\mathbf{j}} + (3.00 \times -3.00 - 4.00 \times 7.00)\hat{k} \]\[ \vec{\tau} = 0\hat{\imath} - 0\hat{\mathbf{j}} - (9.00 + 28.00)\hat{k} \]\[ \vec{\tau} = -37.00 \hat{k} \]

Step 3: Determine Magnitude of the Torque

The magnitude of the torque vector \(\vec{\tau}\) can be determined as:\[ |\vec{\tau}| = |-37.00 \hat{k}| = 37.00 \text{ N}\cdot\text{m} \]

Step 4: Determine the Direction of the Torque

The torque direction corresponds to the \(\hat{k}\) component, indicating it is perpendicular to the plane of motion (\(x-y\) plane). The negative sign shows the rotation direction is clockwise with respect to the origin, as seen from above the \(x-y\) plane.

Key Concepts

Cross Product

The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors. In the context of torque, the cross product helps determine the torque vector produced by a force applied at a certain point. To calculate the cross product, we use a matrix determinant involving the unit vectors, along with the components of the position vector and the force vector. For example, with vectors \( \vec{r} = 3.00 \hat{\imath} + 4.00 \hat{\mathbf{j}} \) and \( \vec{F} = 7.00 \hat{\imath} + (-3.00) \hat{J} \), the cross product is determined using the determinant:\[\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{\imath} & \hat{\mathbf{j}} & \hat{k} 3.00 & 4.00 & 0 7.00 & -3.00 & 0 \end{vmatrix}\]Solving this yields a new vector along the \( \hat{k} \) direction, indicating the direction of torque. The calculation allows us to understand how the positioning and orientation of force contribute to rotational effects. This concept is vital in physics and engineering for analyzing systems involving rotation.

Position Vector

A position vector points from the origin of the coordinate system to the point where the force is applied. In this case, it is \( \vec{r} = 3.00 \hat{\imath} + 4.00 \hat{\mathbf{j}} \), where \( \hat{\imath} \) and \( \hat{\mathbf{j}} \) are the unit vectors in the x and y directions, respectively. The position vector is crucial as it helps define the line of action of the force concerning the rotational axis. This vector not only indicates the point of force application but also plays a role in calculating torques, depending on the force's distance from the pivot. It shows us how position affects torque, as the further the force is applied from the pivot, the larger the torque.

Direction of Torque

The direction of torque reveals how an applied force will rotate an object. In the torque vector \( \vec{\tau} = -37.00 \hat{k} \), the \( \hat{k} \) unit indicates the torque direction is perpendicular to the \( x-y \) plane. The negative sign implies that the direction of rotation is clockwise when viewed from a point above the \( x-y \) plane. This visualization aspect is essential in understanding how forces cause objects to rotate around a pivot point. Knowing the torque's direction helps in predicting and controlling rotational movements, which is vital in mechanical systems and structural engineering. It shows the intuitive nature of torque in real-world applications, providing insight into the motion's effects.