/*! 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 85 The Iron Cross When a gymnast we... [FREE SOLUTION] | 91Ó°ÊÓ

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The Iron Cross When a gymnast weighing \(750 \mathrm{~N}\) executes the iron cross as in Figure \(\mathrm{P} 8.85 \mathrm{a}\), the primary muscles involved in supporting this position are the latissimus dorsi ("lats") and the pectoralis major ("pecs"). The rings exert an upward force on the arms and support the weight of the gymnast. The force exerted by the shoulder joint on the arm is labeled \(\overrightarrow{\mathbf{F}}_{x}\) while the two muscles exert a total force \(\overrightarrow{\mathbf{F}}_{m}\) on the arm. Estimate the magnitude of the force \(\overrightarrow{\mathbf{F}}_{m}\). Note that one ring supports half the weight of the gymnast, which is \(375 \mathrm{~N}\) as indicated in Figure \(\mathrm{P} 8.85 \mathrm{~b}\). Assume that the force \(\overrightarrow{\mathbf{F}}_{\mathrm{m}}\) acts at an angle of \(45^{\circ}\) below the horizontal at a distance of \(4.0 \mathrm{~cm}\) from the shoulder joint. In your estimate, take the distance from the shoulder joint to the hand to be \(70 \mathrm{~cm}\) and ignore the weight of the arm.

Short Answer

Expert verified
The estimated force exerted by the muscles (\(F_m\)) on the gymnast's arm to maintain the iron cross position is approximately \(9300 N\).

Step by step solution

01

Identify Relevant Forces

The problem describes multiple forces acting on the gymnast: the weight of the gymnast itself (\(F_g = 750 N\)), each ring supporting half the weight (\(F_r = 375 N\)), and the force exerted by muscles (\(F_m\)). It's also known that the force \(F_m\) acts at a 45 degree angle below the horizontal at a 4.0 cm distance from the shoulder joint.
02

Calculate Torque from Ring

The torque caused by the ring can be calculated using the formula `Torque = Force ⨉ distance ⨉ sin(angle)`. Because the force from ring is acting vertically and the distance is measured in the horizontal direction, these two vectors are at 90 degrees to each other. The sin(90) = 1, simplifying the equation to `\(τ_r = F_r ⨉ d\)`. Substituting \(F_r = 375 N\) and the distance from the shoulder joint to the hand \(d = 70 cm = 0.7 m\) (converting cm to m), we find that `\(τ_r = 375N ⨉ 0.7m = 262.5 N.m\)`
03

Calculate Torque from Muscles

The torque caused by muscles should balance out the torque from the ring to make the gymnast remain steady in the air. Therefore, `\(τ_m = τ_r\)`. Using the torque equation, `\(τ_m = F_m ⨉ d ⨉ sin(θ)\)`. Here \(F_m\) is the force by muscles, \(d = 4 cm = 0.04 m\) is the distance from shoulder joint to the point where \(F_m\) acts, and `\(θ = 45°\)` (sin(45)=0.7071). Setting values `\(τ_m = τ_r = 262.5 N.m\)' and `\(τ_m = F_m ⨉ 0.04 m ⨉ 0.7071\)`, we can solve for \(F_m\).
04

Calculate Force Exerted by Muscles

Solving `\(F_m = τ_m /(0.04 m ⨉ 0.7071)\)` and plugging `τ_m = 262.5 N.m`, we get `\(F_m = 262.5 N.m / (0.04 m ⨉ 0.7071) = 9300 N\)`

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Key Concepts

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

Torque Calculation
Understanding torque is crucial for analyzing movements in biomechanics. Torque, often referred to as the 'moment of force', is a measure of the turning force on an object such as a bone about an axis—which is often a joint in biomechanics. It is the product of the force applied and the perpendicular distance from the axis of rotation to the line of action of the force.

The formula for torque (\tau) is given by: \[\begin{equation}\tau = F \times d \times \text{sin}(\theta)\end{equation}\] where
  • F is the magnitude of the force applied,
  • d is the distance from the pivot point to the point where the force is applied, and
  • \theta is the angle between the force vector and the lever arm.
In the context of a gymnast performing an iron cross, torque calculation enables us to find the force that muscles must exert to maintain the pose. Here, both gravity and the muscle force create torques about the shoulder joint, and these torques must be equal for the gymnast to be in this static position. This concept is used to solve for the unknown muscle force in the textbook problem.
Biomechanical Forces
Biomechanical forces play a significant role in the study of human movement and are particularly important in sports and exercise science. In the case of the iron cross, the two main biomechanical forces to consider are the weight of the gymnast and the force exerted by the muscles.

It's important to note the direction and the point of application of these forces. For example, the gymnast's weight acts downwards through the body's center of gravity and is supported by the rings exerting an equal and opposite upward force. The muscular force, however, is not only about magnitude but also its direction, which is crucial when calculating torque since it's the component of the force perpendicular to the arm which contributes to the rotational effect.

In this scenario, understanding biomechanics helps us deduce that the shoulder muscles have to exert a large force to counteract the downward pull of gravity. This force has both a horizontal and vertical component, given its 45-degree angle application. The application of biomechanical principles in torque calculation highlights the complex interplay between muscle force and skeletal structure, essential for holding the iron cross pose.
Equilibrium in Physics
Equilibrium in physics is the state in which all the acting forces and torques on a system are in balance, leading to no change in motion. In static equilibrium, which applies to the gymnast's iron cross, objects are at rest, and there's no net force or torque acting on the body.

Applying the concept of equilibrium is central to solving problems involving biomechanics. For the gymnast to maintain the iron cross, the torques due to muscular force and the torque due to the gravitational force must be in equilibrium. That is, the clockwise torque due to the weight force must be balanced by the counter-clockwise torque from the muscular force. This balance can be expressed with the following equation: \[\begin{equation}\tau_{weight} = \tau_{muscle}\end{equation}\] The equilibrium condition provides us with a way to calculate the unknown quantities, as seen in the step-by-step solution given for the force exerted by the gymnast's muscles. By understanding that the gymnast's pose is a perfect example of static equilibrium, students can appreciate the practical application of physics in real-world biomechanics.

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Most popular questions from this chapter

Two astronauts (Fig. \(\mathrm{P} 8.72\) ), each having a mass of \(75.0 \mathrm{~kg}\), are connected by a \(10.0-\mathrm{m}\) rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum and (b) the rotational energy of the system. By pulling on the rope, the astronauts shorten the distance between them to \(5.00 \mathrm{~m}\). (c) What is the new angular momentum of the system? (d) What are their new speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronauts in shortening the rope?

A playground merry-go-round of radius \(2.00 \mathrm{~m}\) has a moment of inertia \(I=275 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is rotating about a frictionless vertical axle. As a child of mass \(25.0 \mathrm{~kg}\) stands at a distance of \(1.00 \mathrm{~m}\) from the axle, the system (merrygo-round and child) rotates at the rate of \(14.0 \mathrm{rev} / \mathrm{min}\). The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge?

A grinding wheel of radius \(0.350 \mathrm{~m}\) rotating on a frictionless axle is brought to rest by applying a constant friction force tangential to its rim. The constant torque produced by this force is \(76.0 \mathrm{~N} \cdot \mathrm{m}\). Find the magnitude of the friction force.

A giant swing at an amusement park consists of a 365 -kg uniform arm \(10.0 \mathrm{~m}\) long, with two seats of negligible mass connected at the lower end of the arm (Fig. P8.53). (a) How far from the upper end is the center of mass of the arm? (b) The gravitational potential energy of the arm is the same as if all its mass were concentrated at the center of mass. If the arm is raised through a \(45.0^{\circ}\) angle, find the gravitational potential energy, where the zero level is taken to be \(10.0 \mathrm{~m}\) below the axis. (c) The arm drops from rest from the position described in part (b). Find the gravitational potential energy of the system when it reaches the vertical orientation. (d) Find the speed of the seats at the bottom of the swing.

A person bending forward to lift a load "with his back" (Fig. P8.17a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure P8.17b of a person bending forward to lift a \(200-\mathrm{N}\) object. The spine and upper body are represented as a uniform horizontal rod of weight \(350 \mathrm{~N}\), pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is \(12.0^{\circ}\). Find the tension in the back muscle and the compressional force in the spine.

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