/*! 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 55 A therapist tells a \(74 \mathrm... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 55

A therapist tells a \(74 \mathrm{~kg}\) patient with a broken leg that he must have his leg in a cast suspended horizontally. For minimum discomfort, the leg should be supported by a vertical strap attached at the center of mass of the leg-cast system (Fig. \(\mathbf{P} 11.55\) ). To comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper legs (thighs) together typically account for \(21.5 \%\) of body weight and the center of mass of each thigh is \(18.0 \mathrm{~cm}\) from the hip joint. The patient also reads that the two lower legs (including the feet) are \(14.0 \%\) of body weight, with a center of mass \(69.0 \mathrm{~cm}\) from the hip joint. The cast has a mass of \(5.50 \mathrm{~kg}\), and its center of mass is \(78.0 \mathrm{~cm}\) from the hip joint. How far from the hip joint should the supporting strap be attached to the cast?

Short Answer

Expert verified
The supporting strap should be attached \(53.66 cm\) from the hip joint.

Step by step solution

01

Calculate individual masses

Find out the mass of the upper leg and the lower leg using the given percentages of body weight. For the upper legs, it鈥檚 \(74 kg \times 21.5 / 100 = 15.91 kg\). For the lower legs, it鈥檚 \(74 kg \times 14 / 100 = 10.36 kg\). Also, notice that the given mass of the cast is \(5.50 kg\).
02

Calculate individual moments

Calculate the product of the mass and distance (from the hip joint) for each of the three components. For the upper legs, it鈥檚 \(15.91 kg \times 18 cm = 286.38 kg-cm\). For the lower legs, it鈥檚 \(10.36 kg \times 69 cm = 714.84 kg-cm\). For the cast, it鈥檚 \(5.50 kg \times 78 cm = 429.00 kg-cm\).
03

Calculate overall center of mass

Add the individual moments calculated in step 2, and divide by the sum of all the masses (i.e., the total mass of the leg-cast system) to get the center of mass. This would be \((286.38 kg-cm + 714.84 kg-cm + 429.00 kg-cm) / (15.91 kg + 10.36 kg + 5.50 kg) = 53.66 cm\). This is the distance from the hip joint where the supporting strap should be attached.

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Key Concepts

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

Physics Problem Solving
When tackling physics problems, understanding the underlying principles is crucial. One common type of problem involves finding the center of mass of an object or system. This is a fundamental concept in mechanics and requires a systematic approach to solve.

Firstly, it is important to gather all relevant data, such as masses and distances. In the textbook exercise presented, various pieces of information, such as percentages of body weight and distances from a fixed point, were given. With these, you can calculate the individual masses.

Then, you must apply the principles of mechanics to these values. For center of mass problems, this typically involves calculating the moment of each mass about a reference point鈥攚hich is the product of the mass and its distance from that point鈥攁nd then combining these moments to solve for the overall center of mass.

Solving such problems step-by-step helps break down complex questions into manageable parts, allowing for a more clear and logical path to the solution.
Mass Distribution
Mass distribution refers to how mass is spread out within an object or a system of objects. The distribution of mass directly influences the center of mass, which is essentially the average location of all the mass in the object.

In the case of the patient with a cast, understanding the mass distribution of different body parts 鈥� the upper leg, the lower leg, and the cast itself 鈥� was essential. Each part contributes differently to the overall mass and has a unique location relative to a reference point, the hip joint in this instance.

To determine the exact placement for minimal discomfort, the therapist's advice was to look at the center of mass of the leg-cast system. Significant in such problems is the proportionality: areas with more mass will have a greater influence on the position of the center of mass. By calculating individual masses and their respective distances from the reference point, we were able to solve for the optimized strap placement based on the mass distribution.
Moment of Force
The moment of force, also known as torque, is a measure of the tendency of a force to rotate an object about an axis or pivot. In simple terms, it is the rotational equivalent of force.

In the provided exercise, although no explicit force is mentioned, the concept of the moment is used in a comparative sense to determine the center of mass. We calculate the moment by multiplying the mass of an object by its distance from the pivot point, which in this case is the hip joint.

The summed moments of the upper leg, lower leg, and cast give us the necessary information to find the strap's optimal attachment point. This essentially balances the overall moment about the hip joint, ensuring the leg remains horizontal and stable with the least discomfort. This principle is widely applicable in designing systems that require balance and stability, such as cranes, bridges, or even to ensure proper posture when using prosthetics.

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Most popular questions from this chapter

(a) In Fig. P11.64 a \(6.00-\mathrm{m}\) -long, uniform beam is hanging from a point \(1.00 \mathrm{~m}\) to the right of its center. The beam weighs \(140 \mathrm{~N}\) and makes an angle of \(30.0^{\circ}\) with the vertical. At the right-hand end of the beam a \(100.0 \mathrm{~N}\) weight is hung; an unknown weight \(w\) hangs at the left end. If the system is in equilibrium, what is \(w ?\) You can ignore the thickness of the beam. (b) If the beam makes, instead, an angle of \(45.0^{\circ}\) with the vertical, what is \(w ?\)

A lead sphere has volume \(6.0 \mathrm{~cm}^{3}\) when it is resting on a lab table, where the pressure applied to the sphere is atmospheric pressure. The sphere is then placed in the fluid of a hydraulic press. What increase in the pressure above atmospheric pressure produces a \(0.50 \%\) decrease in the volume of the sphere?

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of \(90.8 \mathrm{~N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{~mm},\) what is the breaking stress of the alloy?

A uniform ladder \(5.0 \mathrm{~m}\) long rests against a frictionless, vertical wall with its lower end \(3.0 \mathrm{~m}\) from the wall. The ladder weighs \(160 \mathrm{~N}\). The coefficient of static friction between the foot of the ladder and the ground is \(0.40 .\) A man weighing \(740 \mathrm{~N}\) climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder. (a) What is the maximum friction force that the ground can exert on the ladder at its lower end? (b) What is the actual friction force when the man has climbed \(1.0 \mathrm{~m}\) along the ladder? (c) How far along the ladder can the man climb before the ladder starts to slip?

Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Fig. \(\mathbf{P 1 1 . 4 9}\) ). Suppose that an \(82.0 \mathrm{~kg}\) climber, who is \(1.90 \mathrm{~m}\) tall and has a center of gravity \(1.1 \mathrm{~m}\) from his feet, rappels down a vertical cliff with his body raised \(35.0^{\circ}\) above the horizontal. He holds the rope \(1.40 \mathrm{~m}\) from his feet, and it makes a \(25.0^{\circ}\) angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet. (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?
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