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A person is standing on a level floor. His head, upper torso, arms, and hands together weigh \(438 \mathrm{~N}\) and have a center of gravity that is \(1.28 \mathrm{~m}\) above the floor. His upper legs weigh \(144 \mathrm{~N}\) and have a center of gravity that is \(0.760 \mathrm{~m}\) above the floor. Finally, his lower legs and feet together weigh \(87 \mathrm{~N}\) and have a center of gravity that is \(0.250 \mathrm{~m}\) above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

Short Answer

Expert verified
The center of gravity is approximately 1.034 meters above the floor.

Step by step solution

01

Understanding the Problem

We need to find the overall center of gravity of the person. The individual's different parts (head & torso, upper legs, lower legs & feet) have specific weights and centers of gravity. We will calculate the center of gravity for the whole body using these values.
02

Calculate the Moment for Each Part

Calculate the moment for each body part using the formula: \( \text{Moment} = \text{Weight} \times \text{Height of Center of Gravity} \). For the head and torso, it is \( 438 \times 1.28 = 560.64 \). For the upper legs, it is \( 144 \times 0.76 = 109.44 \). For the lower legs and feet, it is \( 87 \times 0.25 = 21.75 \).
03

Sum of Moments

Add all the individual moments to get the total moment: \( 560.64 + 109.44 + 21.75 = 691.83 \).
04

Calculate Total Weight

Sum all the weights of the body parts to find the total weight. This is \( 438 + 144 + 87 = 669 \).
05

Compute the Center of Gravity

Calculate the center of gravity using the formula: \( \text{Center of Gravity} = \frac{\text{Total Moment}}{\text{Total Weight}} \). Substitute the values: \( \frac{691.83}{669} \approx 1.034 \).
06

Conclusion

Thus, the center of gravity of the entire body is approximately \( 1.034 \) meters above the floor.

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Key Concepts

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

Physics Problem-Solving
Physics problem-solving often involves breaking down a problem into manageable steps. To effectively solve these problems, especially those dealing with the center of gravity, it's important to understand the scenario and the physical significance behind the numbers. Let's dive into how to tackle such a problem.

First, identify all the different parts or segments involved. In our case, we looked at the head & torso, upper legs, and lower legs & feet separately. Recognize that each of these has a specific weight and distance from a reference point, often the ground or the base level.

Secondly, visualize what each parameter means. For instance, the weight acts downwards, reflecting the force due to gravity, and the height of the center of gravity tells you the average location of that weight. Once these are understood, the next step is applying these values into equations for moments to find the center of gravity for the entire setup.

Lastly, practice is essential. Using a structured approach will make tackling similar problems much easier. Having a methodical plan not only organizes your thoughts but also prevents careless mistakes.
Moment Calculation
Moment calculation is a crucial concept in physics, particularly in problems related to balance and center of gravity. In essence, a moment is the force applied multiplied by a distance from a reference point. It's measured in Newton-meters (Nm).

Consider each part of the body in the problem. For the head & torso weighing 438 N, if the center of gravity is 1.28 m above the ground, the moment is calculated as:
  • Moment = Weight * Height of Center of Gravity = 438 * 1.28 = 560.64 Nm
Repeat for the other parts:
  • Upper Legs: 144 * 0.76 = 109.44 Nm
  • Lower Legs & Feet: 87 * 0.25 = 21.75 Nm
Each part contributes a moment, reflecting how each segment's weight is distributed relative to the floor. Summing these moments gives us the total moment about the floor. This step is critical because it amalgamates all the individual positions into a single average position or center of gravity.
Mechanics
Mechanics is the branch of physics dealing with motion and forces. When solving problems in mechanics, like calculating centers of gravity, the forces we're dealing with are due to gravity. The center of gravity is an important concept because it's the balance point of an object where all of its weight can be considered to act.

In our exercise, we view the person's body as a combination of different segments each with its own center of gravity. By considering moments, we determine where the overall balance point lies. Remember, the center of gravity doesn't necessarily lie within the material of the object itself.

This might seem abstract, but think of a uniform seesaw with weights at various points. Its balance point depends on the arrangement of these weights. By calculating and combining moments about a pivot, you determine the entire seesaw's balance point.

This methodological approach sheds light on how forces interact in real-world structures, making mechanics foundational for solving large-scale engineering problems as well.

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

Interactive Solution 9.3 at presents a model for solving this problem. The wheel of a car has a radius of \(0.350 \mathrm{~m}\). The engine of the car applies a torque of \(295 \mathrm{~N} \cdot \mathrm{m}\) to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

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. One force has a magnitude of \(2.00 \mathrm{~N}\) and is applied perpendicular to the length of the stick at the free end. The other 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 -N force applied? Express this distance with respect to the end that is pinned.

Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius. One has the shape of a hoop and the other the shape of a solid disk. Each wheel starts from rest and has a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each makes the same number of revolutions in the same time. (a) Which wheel, if either, has the greater angular acceleration? (b) Which, if either, has the greater moment of inertia? (c) To which wheel, if either, is a greater net external torque applied? Explain your answers.

A wrecking ball (weight \(=4800 \mathrm{~N}\) ) is supported by a boom, which may be assumed to be uniform and has a weight of \(3600 \mathrm{~N}\). As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is \(32^{\circ}\), and the angle between the boom and the horizontal is \(48^{\circ} .\) Find (a) the tension in the support cable and (b) the magnitude of the force exerted on the lower end of the boom by the hinge at point \(P\).

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