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Chapter 10: Problem 6

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m}\). (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) 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 \(\overrightarrow{\boldsymbol{F}} ?\)

Short Answer

Expert verified
The position vector \(\vec{r}\) is \(3.00\hat{i} + 4.00\hat{j}\) m. The torque is \(25.00 \hat{k} N.m\) and its magnitude is \(25.00Nm\). The direction of the torque is out of the plane, in the positive k-direction.

Step by step solution

01

Calculate position vector

The position vector \(\vec{r}\) in a 2-d plane, in terms of unit vectors \(i\) and \(j\), is given by \(\vec{r}\) = \(x\hat{i} + y\hat{j}\). Here, \(x = 3.00 m\) and \(y = 4.00 m\) are provided. Therefore, we have \(\vec{r} = 3.00\hat{i} + 4.00\hat{j}\) m.
02

Calculate Torque

Plugging our known values into the torque formula \(蟿 = rXF\), yields \(蟿 = (3.00\hat{i} + 4.00\hat{j}) X (7.00\hat{i} - 3.00\hat{j})\)
03

Cross Product Calculation

To compute torque, you have to use the cross product. Here, \(\hat{i}\) X \(\hat{i}\) = 0, \(\hat{j}\) X \(\hat{j}\) = 0, \(\hat{i}\) X \(\hat{j}\) = \(\hat{k}\), and \(\hat{j}\) X \(\hat{i}\) = -\(\hat{k}\). This will give us a \(蟿 = 3.00*7.00*0 + 4.00*(-3.00)*0 + 3.00*(-3.00)*(-\hat{k}) + 4.00*7.00* \hat{k} = 25.00 \hat{k} N.m\).
04

Compute Magnitude and Direction of Torque

The magnitude of the torque is given by the formula \(|蟿| = |r|*|F|*sin鈦�(胃)\). So, the magnitude of torque is \(|蟿| = 25.00Nm\). The direction of torque (assuming counterclockwise as positive direction) is in the positive k-direction or out of the 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 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.
Position Vector
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.
Magnitude and Direction of Torque
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.

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Most popular questions from this chapter

You live on a planet far from ours. Based on extensive communication with a physicist on earth, you have determined that all laws of physics on your planet are the same as ours and you have adopted the same units of seconds and meters as on earth. But you suspect that the value of \(g\), the acceleration of an object in free fall near the surface of your planet, is different from what it is on earth. To test this, you take a solid uniform cylinder and let it roll down an incline. The vertical height \(h\) of the top of the incline above the lower end of the incline can be varied. You measure the speed \(v_{\mathrm{cm}}\) of the center of mass of the cylinder when it reaches the bottom for various values of \(h .\) You plot \(v_{\mathrm{cm}}^{2}\) (in \(\mathrm{m}^{2} / \mathrm{s}^{2}\) ) versus \(h\) (in \(\mathrm{m}\) ) and find that your data lie close to a straight line with a slope of \(6.42 \mathrm{~m} / \mathrm{s}^{2}\). What is the value of \(g\) on your planet?

The mechanism shown in Fig. \(\mathbf{P} \mathbf{1 0 . 6 4}\) is used to raise a crate of supplies from a ship's hold. The crate has total mass \(50 \mathrm{~kg} .\) A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius \(0.25 \mathrm{~m}\) and moment of inertia \(I=2.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius \(0.12 \mathrm{~m},\) the cylinder turns, and the crate is raised. What magnitude of the force \(\vec{F}\) applied tangentially to the rotating crank is required to raise the crate with an acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2} ?\) (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

A large uniform horizontal turntable rotates freely about a vertical axle at its center. You measure the radius of the turntable to be \(3.00 \mathrm{~m} .\) To determine the moment of inertia \(I\) of the turntable about the axle, you start the turntable rotating with angular speed \(\omega\), which you measure. You then drop a small object of mass \(m\) onto the rim of the turntable. After the object has come to rest relative to the turntable, you measure the angular speed \(\omega_{\mathrm{f}}\) of the rotating turntable. You plot the quantity \(\left(\omega-\omega_{\mathrm{f}}\right) / \omega_{\mathrm{f}}\) (with both \(\omega\) and \(\omega_{\mathrm{f}}\) in rad \(\left./ \mathrm{s}\right)\) as a function of \(m\) (in kg). You find that your data lie close to a straight line that has slope \(0.250 \mathrm{~kg}^{-1}\). What is the moment of inertia \(I\) of the turntable?

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of \(18 \mathrm{~kg} \cdot \mathrm{m}^{2}\). She then tucks into a small ball, decreasing this moment of inertia to \(3.6 \mathrm{~kg} \cdot \mathrm{m}^{2}\). While tucked, she makes two complete revolutions in \(1.0 \mathrm{~s}\). If she hadn't tucked at all, how many revolutions would she have made in the \(1.5 \mathrm{~s}\) from board to water?

A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius \(0.600 \mathrm{~m}\) and mass \(1.60 \mathrm{~kg}\) and is initially at rest. A bullet with mass \(0.0200 \mathrm{~kg}\) is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of \(0.600 \mathrm{~m}\) from the axle. After being struck by the bullet, the disk rotates at \(4.00 \mathrm{rad} / \mathrm{s}\). What is the horizontal velocity of the bullet just before it strikes the disk?
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