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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

You are designing a simple elevator system for an old warehouse that is being converted to loft apartments. A 22,500-N elevator is to be accelerated upward by connecting it to a counterweight by means of a light (but strong!) cable passing over a solid uniform disk- shaped pulley. The cable does not slip where it is in contact with the surface of the pulley. There is no appreciable friction at the axle of the pulley, but its mass is 875 kg and it is 1.50 m in diameter. (a) What mass should the counterweight have so that it will accelerate the elevator upward through 6.75 m in the first 3.00 s, starting from rest? (b) What is the tension in the cable on each side of the pulley?

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?

A thin uniform rod has a length of 0.500 m and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.400 rad/s and a moment of inertia about the axis of 3.00 \(\times 10^{-3}\) kg \(\cdot\) m\(^2\). A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.160 m/s. The bug can be treated as a point mass. What is the mass of (a) the rod; (b) the bug?

An engine delivers 175 hp to an aircraft propeller at 2400 rev/min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 38.0\(^\circ\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to 4.00 kg?

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