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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 \(\vec{\boldsymbol{F}}\) ?

Short Answer

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

Step by step solution

01

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, \(\vec{r} = 3.00 \hat{\imath} + 4.00 \hat{\mathbf{j}}\).
02

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, \[ \vec{\tau} = (0 - 0)\hat{\imath} - (0 - 0)\hat{\mathbf{j}} + (3.00 \times -3.00 - 4.00 \times 7.00)\hat{k} \]\[ \vec{\tau} = 0\hat{\imath} - 0\hat{\mathbf{j}} - (9.00 + 28.00)\hat{k} \]\[ \vec{\tau} = -37.00 \hat{k} \]
03

Determine Magnitude of the Torque

The magnitude of the torque vector \(\vec{\tau}\) can be determined as:\[ |\vec{\tau}| = |-37.00 \hat{k}| = 37.00 \text{ N}\cdot\text{m} \]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg} .\) A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

A uniform disk with mass 40.0 \(\mathrm{kg}\) and radius 0.200 \(\mathrm{m}\) is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force \(F=30.0 \mathrm{N}\) is applied tangent to the rim of the disk. (a) What is the magnitude \(v\) of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution? (b) What is the magnitude \(a\) of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolution?

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of 7.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; (b) the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

Atwood's Machine. Figure Pl0.67 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C,\) and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be 4.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) , respectively, the moment of inertia of the wheel about its axis be \(0.300 \mathrm{kg} \cdot \mathrm{m}^{2},\) and the radius of the wheel be 0.120 \(\mathrm{m} .\)

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