Repeat the seesaw problem in Example 9.1 with the center

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Problem 5PE Chapter 9

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 5PE

Repeat the seesaw problem in Example 9.1 with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium. Example 9.1 She Saw Torques On A Seesaw The two children shown in ?Figure 9.9 ?are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple—more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot.(a) If the second child has a mass of 32.0 kg, how far is she from the ?pivot? (b) What is ? p , the supporting force exerted by the pivot?

Step-by-Step Solution:

Solution 5PE Step-by-step solution Step 1 of 2 Net applied torque is zero here. So, Here, is the torque on right side from pivot, is the torque on the left side of pivot, w1 is the weight of fir st child and 1 is the distance of first child from p ivot point, s denotes the mass o f seesaw and r s is the distance of center of mass of seesaw from the pivot point, is the distance at which second child whose mass 32 kg sits and w2 is the weight of second child.

Step 2 of 2

Chapter 9, Problem 5PE is Solved
Textbook: Physics: Principles with Applications
Edition: 6th
Author: Douglas C. Giancoli
ISBN: 9780130606204

This full solution covers the following key subjects: mass, seesaw, pivot, Example, Child. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. The answer to “Repeat the seesaw problem in Example 9.1 with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium. Example 9.1 She Saw Torques On A Seesaw The two children shown in ?Figure 9.9 ?are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple—more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot.(a) If the second child has a mass of 32.0 kg, how far is she from the ?pivot? (b) What is ? p , the supporting force exerted by the pivot?” is broken down into a number of easy to follow steps, and 145 words. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. The full step-by-step solution to problem: 5PE from chapter: 9 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6th. Since the solution to 5PE from 9 chapter was answered, more than 556 students have viewed the full step-by-step answer.

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Repeat the seesaw problem in Example 9.1 with the center

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