/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 If the competitor leans slightly... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the competitor leans slightly farther back (increasing the angle between his body and the vertical) but remains stationary in this new position, which statement is true? Assume that the rope remains horizontal. A. The difference between \(T_{1}\) and \(T_{2}\) increases, balancing the increased torque about his feet that his weight produces when he leans farther back. B. The difference between \(T_{1}\) and \(T_{2}\) decreases, balancing the increased torque about his feet that his weight produces when he leans farther back. C. Neither \(T_{1}\) nor \(T_{2}\) changes because no other forces are changing. D. Both \(T_{1}\) and \(T_{2}\) change, but the difference between them remains the same.

Short Answer

Expert verified
The difference between \(T_1\) and \(T_2\) increases, option A is correct.

Step by step solution

01

Identify the System

In this scenario, we have a competitor leaning back while two tension forces, \(T_1\) and \(T_2\), act on a horizontal rope held by the competitor. The system is assumed to remain stationary, meaning the net forces and net torques are zero.
02

Analyze the Forces

The competitor leaning back increases the horizontal distance from the center of mass to the pivot point (his feet). This increases the torque produced by his weight about his feet since torque \(\tau = r \cdot F \cdot \sin(\theta)\), where \(r\) is the distance and \(F\) is the force, his weight in this case.
03

Consider the Torque Balance

For the system to remain stationary, the torques must balance. The increased torque due to the lean needs to be balanced by a change in the torques due to \(T_1\) and \(T_2\). If the angle increases and the weight's torque increases, there must be compensatory adjustments in \(T_1\) and \(T_2\) to balance this.
04

Evaluate the Options

Given that the net torque must remain zero, the difference between \(T_1\) and \(T_2\) must adjust to balance the increased torque from the weight. So, we need a statement where the torque difference accommodates the change in body angle.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Torque Equilibrium
Torque equilibrium is a critical concept in physics when analyzing any system that is stationary and not rotating. Torque, often referred to as the moment of force, is calculated using the formula \( \tau = r \cdot F \cdot \sin(\theta) \), where \( \tau \) is the torque, \( r \) 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 line and the lever arm. This calculation helps determine how forces create rotational effects.
For a system in equilibrium, the sum of all torques acting on it must equal zero, meaning the system is not experiencing any net rotational acceleration. In the context of the exercise, the competitor's weight creates a torque about his feet, which needs to be balanced by adjustments in the forces from the tension in the rope.
  • Torque calculation helps understand rotational effects
  • Equilibrium means net torque is zero
  • Increased angle or distance from pivot affects torque magnitude
Horizontal Forces
Horizontal forces play an essential role in maintaining balance and stability in static systems like the one in the exercise. When the competitor leans back while holding the horizontal rope, two key horizontal forces are in action: the tension forces. These are represented as \( T_1 \) and \( T_2 \).

It is crucial to evaluate how these forces change when the competitor alters their position. A balance of horizontal forces means that their vector sum must equal zero for static equilibrium. This ensures the competitor remains stationary horizontally without sliding in either direction. As the angle of lean changes, \( T_1 \) and \( T_2 \) must adjust to maintain this balance.
  • Horizontal forces need to add up to zero for static equilibrium
  • Tension forces adjust as positions change to maintain balance
  • No net horizontal movement should occur to keep the system stationary
Static Equilibrium
Static equilibrium is a state where a system remains at rest, with no net force or torque causing motion. In physics, this concept applies when total forces and torques acting on a system are zero, preserving its state of rest or constant motion.

For the competitor, being in static equilibrium means ensuring that both the vertical and horizontal forces, as well as the torques, balance perfectly. Adjustments in tension forces (\( T_1 \) and \( T_2 \)) are responses to changes in position, ensuring that no motion occurs, either linear or rotational.
  • Equilibrium requires zero net force and torque
  • Balances both linear and rotational effects
  • Maintained through adjustments in forces and torques
Tension Forces in Ropes
Tension forces in ropes are simple yet powerful forces that play a significant role in maintaining equilibrium. Tension is the force imparted along a rope when it is pulled by forces acting from opposite ends. This force transmits between connected objects, impacting their equilibrium state.

In this scenario, the competitor's new body angle changes how the tensions \( T_1 \) and \( T_2 \) interact with the rope. The difference in these tensions interprets as an adjustment to counterbalance the additional torque created by the competitor leaning back. Therefore, the tension forces must change to ensure that the total torques and forces remain in equilibrium.
  • Tension transmits force through the rope
  • Adjusts to balance increased torque when leaning
  • Keeps the system in equilibrium by compensating imbalances

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(A 4 N\) and a 10 N force act on an object. The moment arm of the \(4 \mathrm{~N}\) force is \(0.2 \mathrm{~m}\). If the \(10 \mathrm{~N}\) force produces five times the torque of the \(4 \mathrm{~N}\) force, what is its moment arm?

Two people carry a heavy electric motor by placing it on a light board \(2.00 \mathrm{~m}\) long. One person lifts at one end with a force of \(400 \mathrm{~N}\), and the other lifts at the opposite end with a force of \(600 \mathrm{~N}\). (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{~N},\) with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

II Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a newtron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform. solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{~km}\) (comparable to our sun); its final radius is \(16 \mathrm{~km}\). If the original star rotated once in 30 days, find the angular speed of the neutron star.

A solid, uniform cylinder with mass \(8.00 \mathrm{~kg}\) and diameter \(0.20 \mathrm{~m}\) is spinning at \(20 \mathrm{rad} / \mathrm{s}\) on a thin, frictionless axle that passes along the cylinder axis. You bring the cylinder to rest by placing your hand on the rim. What is the friction force between your hand and the cylinder if the cylinder comes to rest in \(5 \mathrm{~s} ?\)

A small block on a frictionless horizontal surface has a mass of \(0.0250 \mathrm{~kg}\). It is attached to a mascless cord passing through a hole in the surface. (Sce Figure \(10.58 .\) ) The block is originally revolving at a distance of \(0.300 \mathrm{~m}\) from the hole with an angular speed of \(1.75 \mathrm{rad} / \mathrm{s}\). The cord is then pulled from below, shortening the radius of the circle in which the block revolves to \(0.150 \mathrm{~m}\). You may treat the block as a particle. (a) Is angular momentum conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinctic encrgy of the block. (d) How much work was done in pulling the cord?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.