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Chapter 9: Problem 8

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of \(2.00 \mathrm{N}\) and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of \(6.00 \mathrm{N}\) and acts at a \(30.0^{\circ}\) angle with respect to the length of the stick. Where along the stick is the \(6.00-\mathrm{N}\) force applied? Express this distance with respect to the end of the stick that is pinned.

Short Answer

Expert verified
The 6.00-N force is applied 0.67 meters from the pinned end.

Step by step solution

01

Understand the Problem

We need to find the point along the meter stick where a second force of 6.00 N should be applied to ensure that the net torque is zero. The stick can rotate around one end (the pivot point), hence torques due to forces need to counterbalance each other.
02

Define Torque

Torque, denoted by \( \tau \), is given by the equation \( \tau = rF \sin(\theta) \), where \( r \) is the distance from the pivot point to the point where the force is applied, \( F \) is the magnitude of the force, and \( \theta \) is the angle between the force and the line extending from the pivot.
03

Apply Torque Concept for the First Force

For the first force of 2.00 N applied perpendicularly at the free end, the torque is \( \tau_1 = r_1 F_1 = 1.00 \times 2.00 = 2.00 \, \mathrm{N \cdot m} \). This torque acts in a direction that we'll consider positive.
04

Apply Torque Concept for the Second Force

For the second force of 6.00 N applied at an angle of \(30.0^{\circ}\), the torque is \( \tau_2 = r_2 F_2 \sin(30.0^{\circ}) \). Since \( \sin(30.0^{\circ}) = 0.5 \), the torque due to this force is \( \tau_2 = r_2 \times 6.00 \times 0.5 = 3.00 \, r_2 \, \mathrm{N \cdot m} \).
05

Set Net Torque to Zero

For the net torque to be zero, the torques must be equal and opposite, so \( 2.00 = 3.00 \, r_2 \). Solving for \( r_2 \), we get \( r_2 = \frac{2.00}{3.00} = 0.67 \).
06

Conclusion

The 6.00-N force must be applied 0.67 meters from the pivot (the pinned end) for the net torque on the stick to be zero.

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Key Concepts

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

Rotational Dynamics
Rotational dynamics involves understanding how rotational motion is influenced by forces and torques. In this context, torque is a measure of the turning force on an object around a pivot point. Just like linear dynamics describes how forces affect linear motion, rotational dynamics focuses on angular motion.

When dealing with problems in rotational dynamics, the most important element is identifying how and where forces are applied on an object like our meter stick. This helps determine the torque, which dictates how the object might rotate. To achieve balance or equilibrium in rotational dynamics, you need to make sure that the torques acting in different directions cancel each other out. This balancing act ensures that the object remains at rest or moves uniformly without rotating further.
Equilibrium
Equilibrium in rotational dynamics is a state where all the torques acting on an object sum to zero, ensuring no net angular acceleration. This is similar to how in linear dynamics, equilibrium means no net force causing linear acceleration.

For an object to be in rotational equilibrium:
  • The sum of all torques around any point must be zero.
  • Forces can be acting in different places and directions, but their rotation effects must cancel each other.
In our exercise, the stick remains in equilibrium when the torques produced by the forces balance out perfectly. This is achieved by applying the forces at specific points, taking into account both the distance to the pivot and the force's direction.
Force Analysis
Force analysis is crucial when solving problems in rotational dynamics and equilibrium. By breaking down the forces acting on an object, one can understand how they contribute to the overall motion or stability.

Particularly, when dealing with torque, consider:
  • The magnitude of the force.
  • The angle of the force with respect to the lever arm or line from the pivot.
  • The distance from the point of application to the pivot.
In our specific problem, understanding the angle at which the 6.00 N force is applied is key. It changes the effective component of the force that acts perpendicularly to the stick, which is what actually contributes to causing torque.
Physics Problem Solving
Physics problem-solving involves a methodical step-by-step approach to unravel complex scenarios. This typically means:
  • Understanding the problem by clearly identifying knowns and unknowns.
  • Applying relevant principles, such as torque and equilibrium conditions, to dissect the situation.
  • Performing calculations systematically to find necessary results.
  • Reasoning through the findings to ensure they make sense within the context.
In solving our exercise, we follow these steps to find the point where the 6.00 N force must be applied. We looked at torque equations, balanced them out for equilibrium, and calculated distances meticulously. Accurate problem-solving leads not only to the correct answer but to a deeper understanding of the concepts involved.

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Most popular questions from this chapter

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 \(\mu_{\mathrm{k}}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

A solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{kg} \cdot \mathrm{m}^{2}\). From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{kg} .\) After all the sand is in place, what is the angular velocity of the disk?

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 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?

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?

Two identical wheels are moving on horizontal surfaces. The center of mass of each has the same linear speed. However, one wheel is rolling, while the other is sliding on a frictionless surface without rolling. Each wheel then encounters an incline plane. One continues to roll up the incline, while the other continues to slide up. Eventually they come to a momentary halt, because the gravitational force slows them down. Each wheel is a disk of mass \(2.0 \mathrm{kg} .\) On the horizontal surfaces the center of mass of each wheel moves with a linear speed of \(6.0 \mathrm{m} / \mathrm{s}\). (a) What is the total kinetic energy of each wheel? (b) Determine the maximum height reached by each wheel as it moves up the incline.
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