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One force acting on a machine part is \(\overrightarrow{F} = (-5.00 N)\hat{\imath} + (4.00 N)\hat{\jmath}\). The vector from the origin to the point where the force is applied is \(\overrightarrow{r} = (-0.450 m)\hat{\imath} + (0.150 m)\hat{\jmath}\). (a) In a sketch,show \(\overrightarrow{r}, \overrightarrow{F},\) and the origin. (b) Use the right- hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

Short Answer

Expert verified
The torque is \( -1.05\hat{k} \ Nm \), directed into the page.

Step by step solution

01

Understand the Problem

We are given a force vector \( \overrightarrow{F} = (-5.00 \ N)\hat{\imath} + (4.00 \ N)\hat{\jmath} \) and a position vector \( \overrightarrow{r} = (-0.450 \ m)\hat{\imath} + (0.150 \ m)\hat{\jmath} \). We need to find the torque produced by the force about the origin.
02

Sketch the Vectors

Draw a diagram with the origin marked as the reference point. Draw the vector \( \overrightarrow{r} \) originating from the origin and pointing to the point \((-0.450, 0.150)\) on the XY plane. Then, draw the vector \( \overrightarrow{F} \) starting from this point.
03

Determine Torque Direction Using Right-Hand Rule

To determine the torque direction using the right-hand rule, point your fingers in the direction of \( \overrightarrow{r} \) and curl them towards \( \overrightarrow{F} \). Your thumb will point in the direction of the torque, which is out of the page along the positive \( \hat{k} \) direction.
04

Calculate the Torque

The torque \( \overrightarrow{\tau} \) is given by the cross product \( \overrightarrow{r} \times \overrightarrow{F} \). Calculate the determinant:\[\overrightarrow{\tau} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \ -0.450 & 0.150 & 0 \ -5.00 & 4.00 & 0 \end{vmatrix} = (0.150 \times 0 - 0 \times 4.00)\hat{\imath} - (0 \times -5.00 - (-0.450) \times 0)\hat{\jmath} + ((-0.450) \times 4.00 - 0.150 \times -5.00)\hat{k}\]This results in:\[\overrightarrow{\tau} = (0)\hat{\imath} - (0)\hat{\jmath} + (-1.80 + 0.75)\hat{k} = -1.05\hat{k} \ Nm\]
05

Verify the Direction of Torque

The calculated torque \( -1.05\hat{k} \ Nm \) points in the negative \( \hat{k} \) direction, indicating it is into the page, opposite to our earlier right-hand rule prediction. This discrepancy shows a sign error in the right-hand rule prediction; whenever the vector calculations differ from initial predictions, recalibrate to ensure the computation is correct.

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Key Concepts

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

Vector Cross Product
The vector cross product is a crucial operation when calculating torque. It's mainly used to find a vector that is orthogonal to two given vectors. In mathematical terms, for two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), the cross product \( \overrightarrow{A} \times \overrightarrow{B} \) results in a vector that is perpendicular to the plane formed by \( \overrightarrow{A} \) and \( \overrightarrow{B} \).

In terms of physical interpretation, the cross product is vital for determining rotational forces, like torque, in physics. Torque, essentially the rotational equivalent of linear force, can be found by applying the cross product equation \( \overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F} \), where \( \overrightarrow{r} \) is the position vector and \( \overrightarrow{F} \) is the force vector.

The magnitude of the cross product depends on the sine of the angle between the two vectors and the magnitudes of the vectors themselves. This means the cross product is directly influenced by both the size and angle of interaction between the two vectors, making it foundational in torque calculations.
Right-Hand Rule
The right-hand rule is a simple way to determine the direction of the resultant vector in a cross product situation. It's particularly useful for finding the direction of torque, which is a pseudo-vector. For torque, specifically, you apply the rule to the position vector \( \overrightarrow{r} \) and the force vector \( \overrightarrow{F} \).

Here's how it works: point your fingers along the direction of \( \overrightarrow{r} \), and then curl them towards \( \overrightarrow{F} \). Your thumb will naturally point in the direction of the torque vector \( \overrightarrow{\tau} \).

This rule promotes a three-dimensional understanding, since torque often acts along the \( \hat{k} \) axis when \( \overrightarrow{r} \) and \( \overrightarrow{F} \) lie in the \( x-y \) plane. It's key to check the orientation of the angles and the axis for precision when predicting torque directions.
Force Vector
A force vector represents the magnitude and direction of force acting at a point. It's defined in a certain dimension, typically using the standard unit vectors \( \hat{\imath} \), \( \hat{\jmath} \), and \( \hat{k} \) in three-dimensional space. In this exercise, the force vector \( \overrightarrow{F} = (-5.00 \ N)\hat{\imath} + (4.00 \ N)\hat{\jmath} \) lies in the \( x-y \) plane.

This vector notation helps in breaking down the force into its components along each axis, which is essential for understanding its complete orientation and influence on an object. The negative \( \hat{\imath} \) component indicates force acting in the negative \( x \)-direction, while the positive \( \hat{\jmath} \) component means the force acts in the positive \( y \)-direction.

Knowing how to dissect these vectors is fundamental in physical calculations, as it helps in understanding how forces can generate movement or rotation.
Position Vector
The position vector, \( \overrightarrow{r} \), describes a point in space relative to the origin. It provides crucial information needed to calculate torque or understand an object's spatial orientation.

Here, the position vector is the origin point to the force's application point: \( \overrightarrow{r} = (-0.450 \ m)\hat{\imath} + (0.150 \ m)\hat{\jmath} \). This vector indicates that the point is moved to the left and slightly up from the origin in the 2D plane.

This vector is instrumental in detailing the 'lever arm' in torque problems, as torque depends on both length and the angle the force is applied. To compute torque, we use the position vector and the direction of the force vector in combination, through the cross product, to obtain an accurate description of the resulting rotational force.

Understanding position vectors allows for the precise calculations of spatial placement, paving the way for effective manipulation and understanding of forces in physics.

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Most popular questions from this chapter

A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.40). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks?

A hollow, thin-walled sphere of mass 12.0 kg and diameter 48.0 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t) = At^2 + Bt^4\), where \(A\) has numerical value 1.50 and \(B\) has numerical value 1.10. (a) What are the units of the constants \(A\) and \(B\)? (b) At the time 3.00 s, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

A doubling of the torque produces a greater angular acceleration. Which of the following would do this, assuming that the tension in the rope doesn't change? (a) Increasing the pulley diameter by a factor of \(\sqrt{ 2 }\); (b) increasing the pulley diameter by a factor of 2; (c) increasing the pulley diameter by a factor of 4; (d) decreasing the pulley diameter by a factor of \(\sqrt{ 2 }\).

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a \(neutron star\). The density of a neutron star is roughly 10\(^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0 \(\times 10^5\) km (comparable to our sun); its final radius is 16 km. If the original star rotated once in 30 days, find the angular speed of the neutron star.

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 kg, traveling perpendicular to the door at 12.0 m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

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