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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 cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 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?

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When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_x\) and \(a_z\) are approximately zero and \(v_x\) and \(\omega_z\) are approximately constant. Rolling without slipping means \(v_x = r\omega_z\) and \(a_x = r\alpha_z\) . If an object is set in motion on a surface \(without\) these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_0\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_k\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_x\) of the center of mass and \(a_z\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_z = \omega_0\) but \(v_x =\) 0. Rolling without slipping sets in when \(v_x = r\omega_z\) . Calculate the \(distance\) the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

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