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Solved: Repeat the seesaw problem in Example 9.1 with the

Chapter 9, Problem 5

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QUESTION:

Repeat the seesaw problem in Example \(9.1\) with the center of mass of the seesaw \(0.160 \mathrm{~m}\) to the left of the pivot (on the side of the lighter child) and assuming a mass of \(12.0 \mathrm{~kg}\) 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.

Equation Transcription:

Text Transcription:

9.1

0.160 m

12.0 kg

Questions & Answers

QUESTION:

Repeat the seesaw problem in Example \(9.1\) with the center of mass of the seesaw \(0.160 \mathrm{~m}\) to the left of the pivot (on the side of the lighter child) and assuming a mass of \(12.0 \mathrm{~kg}\) 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.

Equation Transcription:

Text Transcription:

9.1

0.160 m

12.0 kg

ANSWER:

Solution 5PE:

Step 1 of 5:-

Here our goal is to solve example 9.1, where,

        The mass of one child is, ,

        The mass of another child is, .

        The 1st child is at a certain distance from the centre of mass, which is, .

        By the definition of centre of mass for a two body system,

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