Question

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

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

Step-by-step solution

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.

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, \(τ = (mg) * (distance~to~pivot)\)

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. \((15~kg)(2.0~m) = (m_2)(1.0~m)\)

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 \(30~kg\).

Key concepts

Seesaw balancing

Seesaws are a fantastic way to learn about balance and physics in a fun, real-world setting. A seesaw is a long, plank-like object that rests on a central point called a pivot or fulcrum. When you place different weights at varying distances from the pivot, you can achieve balance, or seesaw equilibrium. Imagine two children of different weights sitting on opposite sides of a seesaw: to balance it, their distances from the pivot must be adjusted according to their weights. This principle is what makes seesaw physics intriguing and educational. By understanding and applying the concept of balance, you can predict how changes in weight and distance affect the overall equilibrium of the seesaw.

Rotational equilibrium

Rotational equilibrium is a key concept in physics that describes an object that is balanced so that it does not rotate around its axis. For a seesaw to be in rotational equilibrium, the torques on either side of the pivot must be equal. Torque is the rotational equivalent of force and is calculated by multiplying the force applied by the distance from the pivot: \[\tau = F \times \text{distance to pivot}\]In the seesaw problem, each child generates torque based on their weight and distance from the pivot. To achieve equilibrium, the torques they create must be equal. If one side has more torque, the seesaw will tip towards that side. Understanding rotational equilibrium allows us to solve problems relating to balance and stability effectively.

Physics problem-solving

Physics problems often require a step-by-step approach to break down concepts into manageable parts, especially with topics like balance and torque. In solving the seesaw balance problem, we start by identifying known values like the mass of the children and their distances from the pivot. We then apply important formulas, such as the torque equation, to understand how various factors affect balance. Using the equation \((m_1g)d_1 = (m_2g)d_2\),we can determine the necessary mass of the second child by ensuring both torques are equal, simplifying the problem to an algebraic expression. Lastly, always check the solution to ensure it makes sense within the physical context of the problem, like assessing if the calculated mass is reasonable.

Child mass calculation

Calculating the mass needed for balance on a seesaw is a valuable example of applying physics in real life. To balance a seesaw where one child weighs 15 kg and sits 2 meters from the pivot, the second child’s mass can be found using the torque equilibrium equation: \[m_2 = \frac{(m_1 \times d_1)}{d_2}\]Substitute the given values: \((15\, \text{kg}) \times (2\, \text{m}) = m_2 \times (1\, \text{m})\)So, \[m_2 = \frac{30}{1} = 30\, \text{kg}\]This calculation shows that the second child must weigh 30 kg to achieve balance, showcasing how theoretical physics concepts are used to solve practical problems. Knowing how to manipulate and solve using formulas helps demystify real-world applications like seesaw mechanics.