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Chapter 9: Problem 6

A square, \(0.40 \mathrm{m}\) on a side, is mounted so that it can rotate about an axis that passes through the center of the square. The axis is perpendicular to the plane of the square. A force of \(15 \mathrm{N}\) lies in this plane and is applied to the square. What is the magnitude of the maximum torque that such a force could produce?

Short Answer

Expert verified
The maximum torque is 4.245 Nm.

Step by step solution

01

Understand the problem

We need to find the maximum torque produced by an applied force on a rotating square. The torque is affected by the force, the distance from the rotation axis, and the angle between them.
02

Recall the formula for torque

Torque (\( \tau \)) is given by the formula: \(\tau = r \cdot F \cdot \sin(\theta)\). Here, \(r\) is the lever arm distance, \(F\) is the force applied, and \(\theta\) is the angle between force and lever arm.
03

Determine the lever arm distance

Since the force is applied in the plane of the square at a maximum distance from the axis (diagonal of the square), we use half the diagonal as the lever arm. The diagonal of the square is\(\sqrt{2} \cdot 0.40 \text{ m}\), so the lever arm is \(\frac{\sqrt{2} \cdot 0.40}{2} \text{ m}\).
04

Calculate the sine of the angle

For maximum torque, \(\theta = 90^{\circ}\), hence \(\sin(\theta) = 1\).
05

Calculate the diagonal of the square

The length of the diagonal is calculated as \(\sqrt{2} \times 0.40 \text{ m} = 0.40 \times 1.414 \approx 0.566 \text{ m}\). The lever arm thus becomes \(0.566 \text{ m} / 2 = 0.283 \text{ m}\).
06

Calculate the torque

Substitute the values into the torque formula: \(\tau = 0.283 \text{ m} \cdot 15 \text{ N} \cdot 1 = 4.245 \text{ Nm}\).
07

Conclusion

The maximum torque that the force can produce upon being applied to the square is \(4.245 \text{ Nm}\).

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Key Concepts

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

Lever Arm
The lever arm is a crucial component in calculating torque. It refers to the perpendicular distance from the rotation axis to the line of action of the force. In simpler terms, think of it as the 'reach' of the force as it tries to twist or spin an object around a pivot.
For a square, like in our exercise, the longest lever arm is along the diagonal. This is the furthest distance from the pivot point at the center to the corners. Using this maximum distance ensures the greatest torque possible. In this case, half of the diagonal is used, calculated as:
  • The diagonal of a square is \ \( \sqrt{2} \times \text{side length} \ \).
  • Thus, the lever arm is \ \( \frac{\sqrt{2} \times 0.40}{2} \ \text{m} \approx 0.283 \ \text{m} \ \).
Rotation Axis
The rotation axis is the imaginary line around which an object spins. It is fundamental to understanding rotational motion and torque.
In our problem, the axis goes through the center of the square and is perpendicular to its surface. This means the object spins evenly, much like a fan blade around its central hub.
A well-defined rotation axis helps in determining the lever arm and ensures calculations are based on consistent parameters.
Maximum Torque
Maximum torque occurs when a force is applied at the most effective angle and distance. Torque is a rotational force that depends heavily on these factors.
The formula for torque is: \ \( \tau = r \cdot F \cdot \sin(\theta) \ \)
To achieve maximum torque, the force must act perpendicularly (\ \( \theta = 90^{\circ} \ \)) to the lever arm. This maximizes the twisting effect, as \ \( \sin(90^{\circ}) = 1 \ \).
In the example, with the force of 15 N applied, and using the maximum lever arm of 0.283 m, the maximum torque is \ \( 4.245 \ \text{Nm} \ \). Exploiting the full potential of the lever arm and angle increases the efficiency of the applied force.
Force Equation
Understanding the force equation is key to calculating torque. The formula links the force exerted, the lever arm distance, and the angle to find the torque (\ \( \tau \ \)).
The equation is: \ \( \tau = r \cdot F \cdot \sin(\theta) \ \)
  • \ \( r \ \) represents the lever arm length.
  • \ \( F \ \) is the magnitude of force applied.
  • \ \( \sin(\theta) \ \) expresses the angle efficiency, with \ \( \theta = 90^{\circ} \ \) being ideal.
In practical terms, this equation allows you to determine how much rotational effect a given force will produce. Using all elements efficiently will optimize the torque output, as seen in our problem with a force of 15 N and a calculated torque of 4.245 Nm.

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Most popular questions from this chapter

A solid sphere is rolling on a surface. What fraction of its total kinetic energy is in the form of rotational kinetic energy about the center of mass?

Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius of \(4.0 \mathrm{kg}\) and \(0.35 \mathrm{m}\), respectively. One has the shape of a hoop and the other the shape of a solid disk. The wheels start from rest and have a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each turns through an angle of 13 rad in 8.0 s. Find the net external torque that acts on each wheel.

A man drags a 72 -kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted \(25^{\circ}\) above the horizontal, and the strap is inclined \(61^{\circ}\) above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap.

A thin rod has a length of \(0.25 \mathrm{m}\) and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of \(0.32 \mathrm{rad} / \mathrm{s}\) and a moment of inertia of \(1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2} .\) A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (mass \(=4.2 \times 10^{-3} \mathrm{kg}\) ) gets where it's going, what is the angular velocity of the rod?

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+\) \(M h^{2},\) where \(I_{\mathrm{cm}}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.
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