Exercise 8.5 Hints: Getting Started | I'm Stuck Suppose a 427 kg alligator of length 3.4 m is stretched out on a board of the same length weighing 95 N. If the board is supported on the ends as in Figure 8.10, and the scale reads 1967 N, find the x-component of the alligator's center of gravity. X m Xcg, alligator = How many forces are exerted on the board? What is the torque exerted by each one? Since the board is at rest, the net torque must be zero.
Exercise 8.5 Hints: Getting Started | I'm Stuck Suppose a 427 kg alligator of length 3.4 m is stretched out on a board of the same length weighing 95 N. If the board is supported on the ends as in Figure 8.10, and the scale reads 1967 N, find the x-component of the alligator's center of gravity. X m Xcg, alligator = How many forces are exerted on the board? What is the torque exerted by each one? Since the board is at rest, the net torque must be zero.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Concept explainers
Angular speed, acceleration and displacement
Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.
Angular Position
Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.
Question
![**Exercise 8.5**
Suppose a 427 kg alligator of length 3.4 m is stretched out on a board of the same length weighing 95 N. If the board is supported on the ends as in Figure 8.10, and the scale reads 1967 N, find the x-component of the alligator's center of gravity.
\[ x_{cg, \text{alligator}} = \] ☐ m
How many forces are exerted on the board? What is the torque exerted by each one? Since the board is at rest, the net torque must be zero.
**Hints: Getting Started | I'm Stuck**
---
### Explanation for Educational Website:
This problem involves calculating the center of gravity of an alligator placed on a board. The alligator weighs 427 kg and has a length of 3.4 meters. The board it lies on is of the same length and weighs 95 N. The board is supported at both ends. To solve this problem, you will need to use the principles of equilibrium and torque, where the system should balance with a net torque of zero to remain at rest.
To find the x-component of the alligator's center of gravity (\(x_{cg, \text{alligator}}\)), consider all forces acting on the board and the torques they exert around a chosen pivot point. Use the reading from the scale (1967 N) as a force exerted on the board and establish equations that satisfy equilibrium conditions for both forces and torques.
By applying these principles, you'll better understand how forces and torque interact to maintain balance in static systems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a74cc31-62f4-4982-89c3-ce8d7d2cfbb2%2F2cffcea9-0855-4705-a691-a54b17eba207%2Fu9pl0wl_processed.png&w=3840&q=75)
Transcribed Image Text:**Exercise 8.5**
Suppose a 427 kg alligator of length 3.4 m is stretched out on a board of the same length weighing 95 N. If the board is supported on the ends as in Figure 8.10, and the scale reads 1967 N, find the x-component of the alligator's center of gravity.
\[ x_{cg, \text{alligator}} = \] ☐ m
How many forces are exerted on the board? What is the torque exerted by each one? Since the board is at rest, the net torque must be zero.
**Hints: Getting Started | I'm Stuck**
---
### Explanation for Educational Website:
This problem involves calculating the center of gravity of an alligator placed on a board. The alligator weighs 427 kg and has a length of 3.4 meters. The board it lies on is of the same length and weighs 95 N. The board is supported at both ends. To solve this problem, you will need to use the principles of equilibrium and torque, where the system should balance with a net torque of zero to remain at rest.
To find the x-component of the alligator's center of gravity (\(x_{cg, \text{alligator}}\)), consider all forces acting on the board and the torques they exert around a chosen pivot point. Use the reading from the scale (1967 N) as a force exerted on the board and establish equations that satisfy equilibrium conditions for both forces and torques.
By applying these principles, you'll better understand how forces and torque interact to maintain balance in static systems.
![## Example 8.5 Locating Your Lab Partner's Center of Gravity
### Goal
Use torque to find a center of gravity.
### Problem
In this example, we show how to find the location of a person's center of gravity. Suppose your lab partner has a height \( L \) of 176 cm (5 ft, 9 in) and a weight \( w \) of 696 N (156 lb). You can determine the position of his center of gravity by having him stretch out on a uniform board supported at one end by a scale, as shown in Figure 8.10.
*If the board's weight \( w_b \) is 44 N and the scale reading \( F \) is \( 3.06 \times 10^2 \) N, find the distance of your lab partner's center of gravity from the left end of the board.*
### Strategy
To find the position \( x_{cg} \) of the center of gravity, compute the torques using an axis through \( O \). Set the sum of the torques equal to zero and solve for \( x_{cg} \).
### Solution
Apply the second condition of equilibrium. There is no torque due to the normal force \( \vec{n} \) because its moment arm is zero.
Solve for \( x_{cg} \) and substitute known values.
\[
\Sigma \tau_0 = 0
\]
\[
-wx_{cg} - w_b(L/2) + FL = 0
\]
\[
x_{cg} = \frac{FL - w_b(L/2)}{w}
\]
\[
x_{cg} = \frac{(306.24 \, \text{N})(176 \, \text{cm}) - (44 \, \text{N})(88 \, \text{cm})}{696 \, \text{N}}
\]
\[
x_{cg} = 72 \, \text{cm}
\]
### Remarks
The given information is sufficient only to determine the \( x \) coordinate of the center of gravity. The other two coordinates can be estimated, based on the body's symmetry.
### Figure Description
The diagram shows a person lying on a uniform board. The board is supported at one end by a scale. The diagram indicates the placement of forces \( F \) on the scale, \( \vec{n}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a74cc31-62f4-4982-89c3-ce8d7d2cfbb2%2F2cffcea9-0855-4705-a691-a54b17eba207%2Fd08c15e_processed.png&w=3840&q=75)
Transcribed Image Text:## Example 8.5 Locating Your Lab Partner's Center of Gravity
### Goal
Use torque to find a center of gravity.
### Problem
In this example, we show how to find the location of a person's center of gravity. Suppose your lab partner has a height \( L \) of 176 cm (5 ft, 9 in) and a weight \( w \) of 696 N (156 lb). You can determine the position of his center of gravity by having him stretch out on a uniform board supported at one end by a scale, as shown in Figure 8.10.
*If the board's weight \( w_b \) is 44 N and the scale reading \( F \) is \( 3.06 \times 10^2 \) N, find the distance of your lab partner's center of gravity from the left end of the board.*
### Strategy
To find the position \( x_{cg} \) of the center of gravity, compute the torques using an axis through \( O \). Set the sum of the torques equal to zero and solve for \( x_{cg} \).
### Solution
Apply the second condition of equilibrium. There is no torque due to the normal force \( \vec{n} \) because its moment arm is zero.
Solve for \( x_{cg} \) and substitute known values.
\[
\Sigma \tau_0 = 0
\]
\[
-wx_{cg} - w_b(L/2) + FL = 0
\]
\[
x_{cg} = \frac{FL - w_b(L/2)}{w}
\]
\[
x_{cg} = \frac{(306.24 \, \text{N})(176 \, \text{cm}) - (44 \, \text{N})(88 \, \text{cm})}{696 \, \text{N}}
\]
\[
x_{cg} = 72 \, \text{cm}
\]
### Remarks
The given information is sufficient only to determine the \( x \) coordinate of the center of gravity. The other two coordinates can be estimated, based on the body's symmetry.
### Figure Description
The diagram shows a person lying on a uniform board. The board is supported at one end by a scale. The diagram indicates the placement of forces \( F \) on the scale, \( \vec{n}
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON