Equilibrium with Multiple Objects: Videos & Practice Problems

In a seesaw problem, achieving rotational equilibrium involves balancing torques. The torque equation is expressed as mi>m1r=m2r. For a seesaw with a length of 4 meters, placing a 40 kg child closer to the fulcrum at 1.5 meters balances a 30 kg child sitting 2 meters away. This illustrates the principle of torque and the importance of mass distribution in rotational systems.

Position of second kid on seesaw

In this problem, we explore the concept of rotational equilibrium using a seesaw scenario involving two children of different masses. The seesaw is 4 meters long, with a uniform mass distribution and a mass of 50 kg, pivoted at its center. The child on the left has a mass of 30 kg, while the child on the right has a mass of 40 kg. To achieve balance, we need to determine how far the heavier child should sit from the fulcrum.

In rotational equilibrium, the sum of all torques acting on the system must equal zero. The torque (\( \tau \)) produced by a force is calculated using the formula:

\( \tau = r \cdot F \cdot \sin(\theta) \)

where \( r \) is the distance from the pivot to the point of force application, \( F \) is the force, and \( \theta \) is the angle between the force and the lever arm. In this case, the gravitational force acts vertically downward, making an angle of 90 degrees with the lever arm, which simplifies our calculations since \( \sin(90^\circ) = 1 \).

We denote the torque from the left child as \( \tau_1 \) and from the right child as \( \tau_2 \). The equation for equilibrium can be expressed as:

\( m_1 \cdot g \cdot r_1 = m_2 \cdot g \cdot r_2 \)

Here, \( m_1 \) is the mass of the left child (30 kg), \( m_2 \) is the mass of the right child (40 kg), \( r_1 \) is the distance from the fulcrum to the left child (2 meters), and \( r_2 \) is the distance we need to find for the right child.

Since gravity (\( g \)) appears on both sides of the equation, it cancels out, leading to:

\( m_1 \cdot r_1 = m_2 \cdot r_2 \

Rearranging this gives:

\( r_2 = \frac{m_1}{m_2} \cdot r_1 \

Substituting the known values:

\( r_2 = \frac{30 \, \text{kg}}{40 \, \text{kg}} \cdot 2 \, \text{m} = 1.5 \, \text{m} \

This result indicates that the heavier child must sit 1.5 meters from the fulcrum to balance the seesaw, confirming our initial understanding that the heavier child should be positioned closer to the pivot point. This fundamental relationship between mass and distance in seesaw problems is crucial for solving similar equilibrium scenarios.

A 20 kg, 5 m-long bar of uniform mass distribution is attached to the ceiling by a light string, as shown. Because the string is off-center (2 m from the right edge), the bar does not hang horizontally. To fix this, you place a small object on the right edge of the bar. What mass should this object have, to cause the bar to balance horizontally?

• A. 2 kg
• B. 4 kg
• C. 5 kg
• D. 10 kg

Answer: (C)

Two kids (m,_LEFT = 50 kg, m,_RIGHT = 40 kg) sit on the very ends of a 5 m-long, 30 kg seesaw. How far from the left end of the seesaw should the fulcrum be placed so the system is at equilibrium? (Remember the weight of the seesaw!)

• A. 2.29 m
• B. 2.50 m
• C. 2.71 m
• D. 4.58 m

Answer: (A)

Multiple objects hanging

In this problem, we are tasked with determining the missing masses and tensions in a system of objects connected by ropes and rods, all while ensuring the system is in both linear and rotational equilibrium. The key principles to remember are that the sum of all forces and the sum of all torques must equal zero. We will denote the masses as follows: mass A, mass B (given as 4 kg), and mass C. The gravitational acceleration is simplified to \( g = 10 \, \text{m/s}^2 \).

To begin, we identify the tensions in the system: \( T_A \) for mass A, \( T_B \) for mass B, \( T_C \) for mass C, and \( T_{BC} \) for the combined tension of masses B and C. Since mass B is known, we can calculate the force acting on it as \( m_B g = 4 \, \text{kg} \times 10 \, \text{m/s}^2 = 40 \, \text{N} \). This means that the tension \( T_B \) must also equal 40 N to maintain equilibrium.

Next, we analyze the torques acting on the system. For mass C, we can express the torque balance around a pivot point. The torque due to \( T_B \) must equal the torque due to \( T_C \). This can be expressed mathematically as:

\[ T_B \cdot r_B = T_C \cdot r_C \]

Given that \( T_B = 40 \, \text{N} \) and the distances \( r_B = 1 \, \text{m} \) and \( r_C = 2 \, \text{m} \), we can solve for \( T_C \):

\[ 40 \cdot 1 = T_C \cdot 2 \implies T_C = \frac{40}{2} = 20 \, \text{N} \]

Now that we have \( T_C \), we can find the mass C using the equation \( T_C = m_C g \):

\[ 20 = m_C \cdot 10 \implies m_C = 2 \, \text{kg} \]

With the mass of C known, we can now find the combined tension \( T_{BC} \) acting on both masses B and C:

\[ T_{BC} = T_B + T_C = 40 + 20 = 60 \, \text{N} \]

Next, we need to find the mass A. The tension \( T_A \) must balance the combined weight of masses B and C. Using the relationship of distances and masses, we can express the torque balance for mass A:

\[ T_A \cdot r_A = T_{BC} \cdot r_{BC} \]

Here, \( r_A = 4 \, \text{m} \) and \( r_{BC} = 1 \, \text{m} \), leading to:

\[ T_A \cdot 4 = 60 \cdot 1 \implies T_A = \frac{60}{4} = 15 \, \text{N} \]

Finally, we can find the mass A using the equation \( T_A = m_A g \):

\[ 15 = m_A \cdot 10 \implies m_A = 1.5 \, \text{kg} \]

In summary, the masses and tensions in the system are as follows: mass A is 1.5 kg, mass B is 4 kg, mass C is 2 kg, and the tensions are \( T_A = 15 \, \text{N} \), \( T_B = 40 \, \text{N} \), \( T_C = 20 \, \text{N} \), and \( T_{BC} = 60 \, \text{N} \). Understanding the relationships between forces and torques is crucial for solving such equilibrium problems efficiently.

Here’s what students ask on this topic:

How do you achieve rotational equilibrium in a seesaw problem?

To achieve rotational equilibrium in a seesaw problem, the torques on either side of the fulcrum must be balanced. Torque is calculated as the product of the force and the distance from the fulcrum. For two children with masses $m_1$ and $m_2$ sitting at distances $r_1$ and $r_2$ from the fulcrum, the equation $m_1{r}_1=m_2{r}_2$ is used. This means the heavier child must sit closer to the fulcrum to maintain balance. By ensuring the torques are equal and opposite, the seesaw will not rotate, achieving equilibrium.

What is the principle of torque in rotational dynamics?

The principle of torque in rotational dynamics is that torque is the rotational equivalent of force. It is calculated as the product of the force applied and the perpendicular distance from the axis of rotation. Mathematically, torque $\tau$ is expressed as $\tau =Fr\sin \theta$, where $F$ is the force, $r$ is the distance from the pivot, and $\theta$ is the angle between the force and the lever arm. In equilibrium, the sum of all torques acting on a system is zero, meaning there is no net rotation. This principle is crucial in understanding how objects balance and rotate.

Why does the heavier child need to sit closer to the fulcrum on a seesaw?

The heavier child needs to sit closer to the fulcrum on a seesaw to achieve rotational equilibrium. This is because torque, which causes rotation, is the product of force and distance from the fulcrum. For two children with different masses to balance, the torques they produce must be equal. The equation $m_1{r}_1=m_2{r}_2$ shows that if one child is heavier, they must sit closer to the fulcrum to reduce their torque, balancing the system.

How do you calculate the distance a child should sit from the fulcrum on a seesaw?

To calculate the distance a child should sit from the fulcrum on a seesaw, use the torque balance equation $m_1{r}_1=m_2{r}_2$. Here, $m_1$ and $m_2$ are the masses of the children, and $r_1$ and $r_2$ are their respective distances from the fulcrum. Rearrange the equation to solve for the unknown distance: $r_2=\; (m_1{r}_1)\; /m_2$. This calculation ensures the torques are balanced, achieving equilibrium.

What happens if the torques on a seesaw are not balanced?

If the torques on a seesaw are not balanced, the seesaw will rotate around the fulcrum, causing one side to go up and the other to go down. This occurs because the net torque is not zero, resulting in angular acceleration. The side with the greater torque will move downward, while the opposite side will rise. To prevent this, the torques must be equal and opposite, ensuring rotational equilibrium. This balance is achieved by adjusting the distances of the masses from the fulcrum according to the equation $m_1{r}_1=m_2{r}_2$.