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Chapter 11: Problem 7

11.7 A \(15-\mathrm{kg}\) child sits on a playground seesaw, \(2.0 \mathrm{~m}\) from the pivot. A second child located \(1.0 \mathrm{~m}\) on the other side of the pivot would have to have a mass of _________ to lift the first child off the ground. a) greater than \(30 \mathrm{~kg} \quad\) c) equal to \(30 \mathrm{~kg}\) b) less than \(30 \mathrm{~kg}\)

Short Answer

Expert verified
Answer: The mass of the second child required to lift the first child off the ground is 30 kg.

Step by step solution

01

Identifying the given variables

The given variables in this problem are: 1. Mass of first child = \(15~kg\) 2. Distance of the first child from the pivot = \(2.0~m\) 3. Distance of the second child from the pivot = \(1.0~m\) Our goal is to find the mass of the second child so that the first child is lifted off the ground.
02

Balancing the Torque

In order to balance the torque, we follow the equation: \(τ_1 = τ_2\) where \(τ_1\) is the torque created by the first child and \(τ_2\) is the torque created by the second child. Torque can be calculated as: \(τ = (force) * (distance~to~pivot)\) Force is the product of mass and gravitational acceleration, \(F = mg\). Therefore, \(τ = (mg) * (distance~to~pivot)\)
03

Applying the Torque equation

Now, we need to balance the torque created by both children: \((m_1g)d_1 = (m_2g)d_2\) Where \(m_1 = 15~kg\), \(d_1 = 2.0~m\), \(d_2 = 1.0~m\) and we need to find \(m_2\). Gravitational acceleration \(g\) will be the same for both children and cancels out when equating torques. \((15~kg)(2.0~m) = (m_2)(1.0~m)\)
04

Solve for the mass of the second child

To find the mass of the second child, we can rearrange the equation: \(m_2 = \frac{(15~kg)(2.0~m)}{1.0~m}\) \(m_2 = 30~kg\) The mass of the second child required to lift the first child off the ground is equal to \(30~kg\). So, the correct answer is c) equal to \(30~kg\).

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Most popular questions from this chapter

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