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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 machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3.00 cm. It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by 22.5 rad/s ?

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

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00-m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 0.800-kg rod has a light basket at the other end to catch the 0.163-kg puck. The puck slides across the ice with velocity \(\overrightarrow{v}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 N is applied to the rim of the wheel. The wheel has radius 0.120 m. Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

The moment of inertia of the empty turntable is \(1.5 \mathrm{~kg} \mathrm{~m}^{2}\). With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m},\) the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s}\). What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (b) \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (c) \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (d) \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\).

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