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Chapter 11: Problem 48

A force \(\vec{F}\) applied at the point \(x=2.0 \mathrm{m}, y=0 \mathrm{m}\) produces a torque \(4.6 \hat{k} \mathrm{N} \cdot\) mabout the origin. If the \(x\) -component of \(\vec{F}\) is \(3.1 \mathrm{N}\) what angle does it make with the \(x\) -axis?

Short Answer

Expert verified
The angle \( \theta \) is the inverse sine of \( \frac{4.6N\cdot m}{2.0m \times 3.1 / cos(\theta)} \).

Step by step solution

01

Analyze the given information

The position where the force is applied is (2.0m, 0). The force has a torque of \( 4.6 \hat{k} N\cdot m \) and its x-component is \( 3.1N \).
02

Calculate the magnitude of the vector r, and the force F

The magnitude of the vector r is the distance from the origin to the point where the force is applied, which is 2.0m. The x-component of the force F is given, but the full magnitude of the force is not known. However, we can express F as \( F = 3.1 / cos(\theta) \) where \( \theta \) is the angle between F and the x-axis.
03

Use the torque equation

The torque \( \tau \) is given by the cross product of the radius and the force, i.e. \( \tau = r \times F \). This can also be written as \( \tau = rF sin(\theta) \).
04

Solve the equation for theta

We already know that \( \tau = 4.6N\cdot m \), \( r = 2.0m \), and \( F = 3.1 / cos(\theta) \). We can plug in these values to the equation \( \tau = rF sin(\theta) \) and then solve for \( \theta \).
05

Translation into an arc sine

At this point, the equation will be in the form of \( sin(\theta) = some_number \). To isolate \( \theta \), take the arc sine (inverse sine) of both sides of the equation. The arc sine of a number n gives the angle whose sine is n. This will allow us to solve for \( \theta \).

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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 a mathematical operation used to find a vector that is perpendicular to two given vectors. When calculating torque, the cross product helps determine how a force acts at a point. Torque, denoted as \( \tau \), is the result of the cross product between the position vector \( \vec{r} \) and the force vector \( \vec{F} \).
For two-dimensional space, like in this problem, the cross product simplifies to:
  • \( \tau = \vec{r} \times \vec{F} = rF \sin(\theta) \)
Here, \( \theta \) is the angle between the force and the line from the origin to the point where the force is applied. This formula helps us understand the rotational effect that the force has about a chosen axis.
Vector Magnitude
The magnitude of a vector is a measure of its length. In the context of this problem, understanding vector magnitude helps identify how far the force is applied from the origin.
The magnitude of a position vector \( \vec{r} \) pointing from the origin to a point \((x, y)\) is calculated as:
  • \( |\vec{r}| = \sqrt{x^2 + y^2} \)
However, since the position is at \((2.0, 0)\), the magnitude is simply 2.0 meters. Calculating magnitudes is essential to simplify and solve problems involving vectors and helps in understanding the impact forces have based on their location.
Torque Equation
The torque equation is a fundamental tool for understanding the rotational effect of forces. Torque \( \tau \) is calculated by:
  • \( \tau = rF \sin(\theta) \)
In this formula:
  • \( r \) is the distance from the origin to the point of force application.
  • \( F \) is the force magnitude.
  • \( \theta \) is the angle between \( \vec{F} \) and the x-axis.
For this problem, the force鈥檚 x-component and torque are known, enabling us to form a relationship that helps solve for the angle. By rearranging and solving the torque equation, we can find the missing angle \( \theta \), revealing how the force opposes the rotational effect.
Inverse Trigonometric Functions
Inverse trigonometric functions are essential for solving equations involving angles. Once we express our equation from the torque problem in the form \( \sin(\theta) = n \), we use the inverse sine (arc sine) function to determine the angle \( \theta \).
This process involves:
  • Isolating \( \sin(\theta) \)
  • Taking the arc sine of both sides to find \( \theta = \sin^{-1}(n) \)
The result gives the needed angle, showing the direction the force vector makes with the x-axis. Inverse trigonometric functions simplify finding unknown angles, transforming complex relationships into solvable equations.

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Most popular questions from this chapter

You're choreographing your school's annual ice show. You call for eight \(60-\mathrm{kg}\) skaters to join hands and skate side by side in a line extending \(12 \mathrm{m}\). The skater at one end is to stop abruptly, so the line will rotate rigidly about that skater. For safety, you don't want the fastest skater to be moving at more than \(8.0 \mathrm{m} / \mathrm{s},\) and you don't want the force on that skater's hand to exceed \(300 \mathrm{N}\). What do you determine is the greatest speed the skaters can have before they execute their rotational maneuver?

A skater has rotational inertia \(4.2 \mathrm{kg} \cdot \mathrm{m}^{2}\) with his fists held to his chest and \(5.7 \mathrm{kg} \cdot \mathrm{m}^{2}\) with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5 -kg weight in each outstretched hand; the weights are \(76 \mathrm{cm}\) from his rotation axis. If he pulls his hands in to his chest, so they're essentially on his rotation axis, how fast will he be spinning?

A weightlifter's barbell consists of two 25 -kg masses on the ends of a \(15-\mathrm{kg}\) rod \(1.6 \mathrm{m}\) long. The weightlifter holds the rod at its center and spins it at 10 rpm about an axis perpendicular to the rod. What's the magnitude of the barbell's angular momentum?

A potter's wheel with rotational inertia \(6.40 \mathrm{kg} \cdot \mathrm{m}^{2}\) is spinning freely at 19.0 rpm. The potter drops a 2.70 -kg lump of clay onto the wheel, where it sticks \(46.0 \mathrm{cm}\) from the rotation axis. What's the wheel's subsequent angular speed?

As an automotive engineer, you're charged with redesigning a car's wheels with the goal of decreasing each wheel's angular momentum by \(30 \%\) for a given linear speed of the car. Other design considerations require that the wheel diameter go from \(38 \mathrm{cm}\) to \(35 \mathrm{cm} .\) If the old wheel had rotational inertia \(0.32 \mathrm{kg} \cdot \mathrm{m}^{2},\) what do you specify for the new rotational inertia?
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