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Chapter 9: Problem 8

Interactive LearningWare 9.1 at reviews the concepts that are important in this problem. One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop (see the drawing). Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of \(4.00 \mathrm{~N}\) and is applied perpendicular to the stick at the free end. The other force has a magnitude of \(6.00 \mathrm{~N}\) and acts at a \(60.0^{\circ}\) angle with respect to the stick. Where along the stick is the 6.00 -N force applied? Express this distance with respect to the axis of rotation.

Short Answer

Expert verified
The 6.00-N force is applied 0.385 meters from the axis of rotation.

Step by step solution

01

Understand the Problem

The problem involves a meter stick (a 1 meter long stick) with one end pinned, allowing it to rotate on a plane parallel to a tabletop. We have two forces acting such that their net torque about the pinned end is zero. One of these forces is known in magnitude and direction, so we need to find out where the other force is applied along the stick.
02

Calculate Torque from the First Force

Torque is given by \( \tau = F \cdot r \cdot \sin(\theta) \). The first force (\( F_1 = 4.00 \mathrm{~N} \)) acts perpendicularly at the free end (1 meter from the pivot), so \( \theta = 90^\circ \) and \( \sin(\theta) = 1 \). Thus, torque from the first force: \( \tau_1 = 4.00 \mathrm{~N} \cdot 1 \mathrm{~m} \cdot 1 = 4.00 \mathrm{~Nm} \).
03

Set Up Equation for Zero Net Torque

We have two opposing torques that must cancel each other out so that \( \tau_1 + \tau_2 = 0 \). This means \( \tau_2 = -\tau_1 \). Since \( \tau_1 = 4.00 \mathrm{~Nm} \), \( \tau_2 = -4.00 \mathrm{~Nm} \).
04

Express Torque from the Second Force

For the second force (\( F_2 = 6.00 \mathrm{~N} \)) at an angle \( \theta = 60^\circ \), the torque is \( \tau_2 = F_2 \cdot d \cdot \sin(60^\circ) \), where \( d \) is the unknown distance. We know \( \sin(60^\circ) = \sqrt{3}/2 \), so the torque becomes \( \tau_2 = 6.00 \mathrm{~N} \cdot d \cdot \frac{\sqrt{3}}{2} \).
05

Solve for the Unknown Distance

We equate \( \tau_2 = -4.00 \mathrm{~Nm} \). Substituting the expression for \( \tau_2 \), we have: \[ 6.00 \mathrm{~N} \cdot d \cdot \frac{\sqrt{3}}{2} = -4.00 \mathrm{~Nm} \]. Solving for \( d \), we get: \[ d = \frac{-4.00 \mathrm{~Nm}}{6.00 \mathrm{~N} \cdot \frac{\sqrt{3}}{2}} = \frac{-4.00}{3\sqrt{3}} \]. Simplifying, \( d \approx 0.385 \mathrm{~m} \). Hence the force is applied 0.385 meters from the axis.

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Key Concepts

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

Torque
Torque is a fundamental concept in rotational dynamics. Imagine trying to open a door; the further from the hinge you push, the easier it is. This is the idea behind torque: it measures how effectively a force can cause an object to rotate around an axis.
It is calculated using the formula:
  • \( \tau = F \cdot r \cdot \sin(\theta) \)
where \( \tau \) is the torque, \( F \) is the force applied, \( r \) is the distance from the axis of rotation (often called the "lever arm"), and \( \theta \) is the angle between the force direction and the lever arm.
The key takeaway is that torque tells us about the rotational "strength" or "twisting capability" of a force. It helps us understand how forces cause objects to spin.
Net Torque
Net torque is the sum of all the torques acting on an object. For an object to be in rotational equilibrium, the net torque must be zero. This means that any clockwise torques (positive) must be balanced by counterclockwise torques (negative). In our scenario, one torque comes from a 4.00 N force and another from a 6.00 N force. Because the net torque is zero, these forces perfectly counterbalance each other.
  • This concept ensures that the meter stick remains stationary and not spinning.
  • Net torque is a crucial concept for understanding balance and stability, particularly in engineering and physics.
Moment Arm
The moment arm, often referred to simply as "\( r \)," is the perpendicular distance from the axis of rotation to the line of action of the force. In other words, it is the lever arm that the force "acts" upon to cause rotation. In the problem:
  • For the 4.00 N force, the moment arm is 1 meter because it acts at the free end of the meter stick.
  • For the 6.00 N force, the moment arm is the distance \( d \) we're solving for, measured from the rotational axis to where the force is applied.
Understanding the moment arm helps us solve for unknowns in rotational dynamics scenarios because it directly influences the torque a force can exert.
Force Equilibrium
Force equilibrium occurs when all the forces acting on an object are balanced, resulting in no linear acceleration. However, in rotational dynamics, we're often more concerned with rotational equilibrium, where the net torque is zero. This concept is essential for maintaining stability in structures or mechanisms with rotating parts.
In the exercise, force equilibrium means that, despite having forces acting on the meter stick's different parts, their torques cancel out, causing it to remain in a state of "rotational rest." Understanding both force and rotational equilibrium is crucial for solving problems in mechanics, ensuring that objects or systems do not move or collapse under applied forces.

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Most popular questions from this chapter

See Multiple-Concept Example 12 to review some of the concepts that come into play here. The crane shown in the drawing is lifting a 180 -kg crate upward with an acceleration of \(1.2 \mathrm{~m} / \mathrm{s}^{2} .\) The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of \(130 \mathrm{~kg}\). The cable is then wound onto a hollow cylindrical drum that is mounted on the deck of the crane. The mass of the drum is \(150 \mathrm{~kg},\) and its radius is \(0.76 \mathrm{~m} .\) The engine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{~kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (see the drawing). The front brake is then applied for \(3.0 \mathrm{~s}\), and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{~m} .\) The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{k}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

A cylindrically shaped space station is rotating about the axis of the cylinder to create artificial gravity. The radius of the cylinder is \(82.5 \mathrm{~m}\). The moment of inertia of the station without people is \(3.00 \times 10^{9} \mathrm{~kg} \cdot \mathrm{m}^{2}\). Suppose 500 people, with an average mass of \(70.0 \mathrm{~kg}\) each, live on this station. As they move radially from the outer surface of the cylinder toward the axis, the angular speed of the station changes. What is the maximum possible percentage change in the station's angular speed due to the radial movement of the people?

The steering wheel of a car has a radius of \(0.19 \mathrm{~m}\), while the steering wheel of a truck has a radius of \(0.25 \mathrm{~m} .\) The same force is applied in the same direction to each. What is the ratio of the torque produced by this force in the truck to the torque produced in the car?

Two disks are rotating about the same axis. Disk A has a moment of inertia of \(3.4 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and an angular velocity of \(+7.2 \mathrm{rad} / \mathrm{s}\). Disk \(\mathrm{B}\) is rotating with an angular velocity of \(-9.8 \mathrm{rad} / \mathrm{s}\). The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of \(-2.4 \mathrm{rad} /\) s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?
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