Show on a diagram how your CM shifts when you move from a lying position to a sitting position.

Short Answer

Expert verified

The diagram showing the center of mass at lying and sitting positions is as given follows:

Step by step solution

01

Show on a diagram how your CM shifts when you move from a lying position to a sitting position.

The diagram showing the center of mass at lying and sitting positions is as given follows:

02

Let us understand the center of mass (CM)

The quantity center of mass is described as the point at which the entire mass of a certain object is concentrated. The center of mass is affected by the variation in the mass distribution of the component.

03

Show the center of mass at the lying position and sitting position

If a person is lying on a straight surface, his/her center of mass is located inside the body's volume, just below the navel. On the other hand, if the person is in a sitting position with extended legs, his/her center of mass will be slightly outside the body's volume in front of the midsection. Also, the center of mass while sitting will be slightly higher than lying because when you sit, most of your weight is concentrated on the upper half of your body.

The following is the diagram of the center of mass (CM) while lying down.

The following is the diagram of the center of mass in the sitting position.

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 Vaia!

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

Rain is falling at the rate of \({\bf{2}}{\bf{.5}}\;{{{\bf{cm}}} \mathord{\left/{\vphantom {{{\bf{cm}}} {\bf{h}}}} \right.\\} {\bf{h}}}\) and accumulates in a pan. If the raindrops hit at \({\bf{8}}{\bf{.0}}\;{{\bf{m}} \mathord{\left/{\vphantom {{\bf{m}} {\bf{s}}}} \right. \\} {\bf{s}}}\), estimate the force on the bottom of a \({\bf{1}}{\bf{.0}}\;{{\bf{m}}^{\bf{2}}}\) pan due to the impacting rain which we assume does not rebound. Water has a mass of \({\bf{1}}{\bf{.00 \times 1}}{{\bf{0}}^{\bf{3}}}\;{\bf{kg}}\;{\bf{per}}\;{{\bf{m}}^{\bf{3}}}\).

(I) The distance between a carbon atom \(\left( {{\bf{m = 12}}\;{\bf{u}}} \right)\) and an oxygen atom \(\left( {{\bf{m = 16}}\;{\bf{u}}} \right)\) in the CO molecule is \({\bf{1}}{\bf{.13 \times 1}}{{\bf{0}}^{{\bf{10}}}}\;{\bf{m}}\) How far from the carbon atom is the center of mass of the molecule?

(a) Calculate the impulse experienced when a 55-kg person lands on firm ground after jumping from a height of 2.8 m.

(b) Estimate the average force exerted on the person’s feet by the ground if the landing is stiff-legged, and again

(c) with bent legs. With stiff legs, assume the body moves 1.0 cm during impact, and when the legs are bent, about 50 cm. [Hint: The average net force on him, which is related to impulse, is the vector sum of gravity and the force exerted by the ground. See Fig. 7–34.] We will see in Chapter 9 that the force in (b) exceeds the ultimate strength of bone (Table 9–2).

FIGURE 7-34 Problem 24.

According to Eq. 7–4, the longer the impact time of an impulse, the smaller the force can be for the same momentum change, and hence the smaller the deformation of the object on which the force acts. On this basis, explain the value of air bags, which are intended to inflate during an automobile collision and reduce the possibility of fracture or death.

(I) A 110-kg tackler moving at 2.5 m/s meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

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.

Sign-up for free